IB Analysis and Approaches HL · Continuous random variables and the normal distribution · End of topic

IB exam questions: topic 16

109 exercises with solutions · By Álvaro López

Past IB exam questions grouped by type and, within each type, from easiest to hardest. Try each one before opening the solution.

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Density function: constants and probabilities

Exercise 1 BasicIB · P1 November 1998

The probability density function of a random variable is given by

$f (x) = ⎩ ⎨ ⎧ \frac{x}{3} x^{2} - 1 0 1 \leq x \leq k otherwise$

Calculate the value of $k$ .

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Answer: $k = 2$

Show working and answer

The total integral must equal 1. With the substitution $u = x^{2} - 1$ , $d u = 2 x d x$ : $\int_{1}^{k} \frac{x}{3} x^{2} - 1 d x = \frac{1}{3} [\frac{( x ^{2} - 1 ) ^{3/2}}{3}]_{1}^{k} = \frac{( k ^{2} - 1 ) ^{3/2}}{3 3} = 1.$ Hence $(k^{2} - 1)^{3/2} = 33 = 3^{3/2}$ , that is, $k^{2} - 1 = 3$ and, since $k > 1$ , $k = 2$ .

Answer: $k = 2$

Exercise 2 MediumIB · P1 TZ2 May 2005

The probability density function $f (x)$ of a continuous random variable $X$ is defined on the interval $[0, a]$ as:

$f (x) = ⎩ ⎨ ⎧ \frac{x}{8} \frac{27}{8 x ^{2}} for 0 \leq x \leq 3 for 3 < x \leq a$

Find the value of $a$ .

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Answer: $a = \frac{54}{11}$

Show working and answer

The total area under the density is $1$ . The first piece contributes $\int_{0}^{3} \frac{x}{8} d x = [\frac{x ^{2}}{16}]_{0}^{3} = \frac{9}{16} .$ Therefore the second piece must contribute $1 - \frac{9}{16} = \frac{7}{16}$ : $\int_{3}^{a} \frac{27}{8 x ^{2}} d x = \frac{27}{8} [- \frac{1}{x}]_{3}^{a} = \frac{27}{8} (\frac{1}{3} - \frac{1}{a}) = \frac{7}{16} .$ So $\frac{1}{3} - \frac{1}{a} = \frac{7}{54}$ , hence $\frac{1}{a} = \frac{11}{54}$ and $a = \frac{54}{11} \approx 4.91$ .

Answer: $a = \frac{54}{11}$

Exercise 3 MediumIB · P1 TZ2 May 2009

A random variable has a probability density function given by

$f (x) = {k x (2 - x) 0 for 0 \leq x \leq 2 for all other values$

a) Find $k$ .

b) Find $E (X)$ .

Show answer

Answer: a) $k = \frac{3}{4}$ ; b) $E (X) = 1$

Show working and answer

a) The total area must be $1$ : $\int_{0}^{2} k x (2 - x) d x = k [x^{2} - \frac{x ^{3}}{3}]_{0}^{2} = k (4 - \frac{8}{3}) = \frac{4 k}{3} = 1 \Rightarrow k = \frac{3}{4} .$

b) $E (X) = \int_{0}^{2} x \cdot \frac{3}{4} x (2 - x) d x = \frac{3}{4} \int_{0}^{2} (2 x^{2} - x^{3}) d x = \frac{3}{4} [\frac{2 x ^{3}}{3} - \frac{x ^{4}}{4}]_{0}^{2} = \frac{3}{4} (\frac{16}{3} - 4) = 1.$ (This is to be expected: the density is symmetric about $x = 1$ .)

Answer: a) $k = \frac{3}{4}$ ; b) $E (X) = 1$

Exercise 4 MediumIB · P2 TZ2 May 2008

A continuous random variable $X$ has the following probability density function

$f (x) = {12 x^{2} (1 - x) 0 for 0 \leq x \leq 1 otherwise$

Find the probability that $X$ lies between the mean and the mode.

Show answer

Answer: $P (\frac{3}{5} < X < \frac{2}{3}) = \frac{1981}{16875} \approx 0.117$

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Mean. $μ = \int_{0}^{1} x \cdot 12 x^{2} (1 - x) d x = 12 \int_{0}^{1} (x^{3} - x^{4}) d x = 12 (\frac{1}{4} - \frac{1}{5}) = \frac{3}{5} .$

Mode. Differentiate $f (x) = 12 x^{2} - 12 x^{3}$ : $f^{'} (x) = 24 x - 36 x^{2} = 12 x (2 - 3 x) = 0$ , so $x = \frac{2}{3}$ (a maximum, since $f$ increases before it and decreases after it). Mode $= \frac{2}{3}$ .

Probability. $P (\frac{3}{5} < X < \frac{2}{3}) = \int_{3/5}^{2/3} (12 x^{2} - 12 x^{3}) d x = [4 x^{3} - 3 x^{4}]_{3/5}^{2/3} = \frac{16}{27} - \frac{297}{625} = \frac{1981}{16875} \approx 0.117.$

Answer: $P (\frac{3}{5} < X < \frac{2}{3}) = \frac{1981}{16875} \approx 0.117$

Exercise 5 MediumIB · P1 TZ2 May 2010

A continuous random variable $X$ has the following probability density function:

$f (x) = {c (x - x^{2}) 0 0 \leq x \leq 1 all other values$

a) Determine $c$ .

b) Find $E (X)$ .

Show answer

Answer: a) $c = 6$ ; b) $E (X) = \frac{1}{2}$

Show working and answer

a) Total area equal to $1$ : $c \int_{0}^{1} (x - x^{2}) d x = c (\frac{1}{2} - \frac{1}{3}) = \frac{c}{6} = 1 \Rightarrow c = 6.$

b) $E (X) = 6 \int_{0}^{1} (x^{2} - x^{3}) d x = 6 (\frac{1}{3} - \frac{1}{4}) = \frac{1}{2} .$ (The density is symmetric about $x = \frac{1}{2}$ .)

Answer: a) $c = 6$ ; b) $E (X) = \frac{1}{2}$

Exercise 6 MediumIB · P1 TZ2 May 2011

The random variable $X$ has the following probability density function $f$ :

$f (x) = {k x (x + 1) (2 - x) 0 0 \leq x \leq 2 all other values$

a) Sketch an approximate graph of the function. It is not necessary to find the coordinates of the maximum.

b) Find the value of $k$ .

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Answer: b) $k = \frac{3}{8}$

Show working and answer

a) Graph of the density f(x) = 3/8 · x(x+1)(2−x) for 0 ≤ x ≤ 2 and f(x) = 0 outside that interval: it starts at the origin, rises to a single maximum at x ≈ 1.22 with f ≈ 0.79, and falls until it crosses the horizontal axis at (2, 0). Axes with marks at 0, 1 and 2 on the x-axis, and at 0 and 0.5 on f(x)

The curve is an arc that meets the axis at $x = 0$ and $x = 2$ (the roots of the polynomial) and is positive between them, with a single maximum at around $x \approx 1.2$ .

b) The total area must be $1$ : $k \int_{0}^{2} x (x + 1) (2 - x) d x = k \int_{0}^{2} (- x^{3} + x^{2} + 2 x) d x = k [- \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + x^{2}]_{0}^{2} = k (- 4 + \frac{8}{3} + 4) = \frac{8 k}{3} = 1.$ So $k = \frac{3}{8}$ .

Answer: b) $k = \frac{3}{8}$

Exercise 7 MediumIB · P2 TZ2 May 2013

The length, $X$ metres, of the fish of a given species has the following probability density function:

$f (x) = ⎩ ⎨ ⎧ a x^{2} 0.5 a (1 - x) 0 for 0 \leq x \leq 0.5 for 0.5 \leq x \leq 1 otherwise$

a) Verify that $a = 9.6$ .

b) Sketch an approximate graph of the distribution.

c) Find $P (X < 0.6)$ .

Show answer

Answer: a) $a = 9.6$ (verified); c) $P (X < 0.6) = 0.616$

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a) The total area must be $1$ . First section: $\int_{0}^{0.5} a x^{2} d x = a \frac{0. 5 ^{3}}{3} = \frac{a}{24}$ . Second section (a triangle with base $0.5$ and height $0.5 a \cdot 0.5 = \frac{a}{4}$ ): $\int_{0.5}^{1} 0.5 a (1 - x) d x = 0.5 a \cdot \frac{0. 5 ^{2}}{2} = \frac{a}{16}$ . So $\frac{a}{24} + \frac{a}{16} = \frac{5 a}{48} = 1 \Rightarrow a = \frac{48}{5} = 9.6.$

b) With $a = 9.6$ : $f (x) = 9.6 x^{2}$ on $[0; 0.5]$ and $f (x) = 4.8 (1 - x)$ on $[0.5; 1]$ . At $x = 0.5$ both equal $2.4$ . Graph of the density f(x) for 0 ≤ x ≤ 1 and f(x) = 0 outside that interval: it starts at the origin with a parabolic shape, increasing as f = 9.6x² up to the maximum at x = 0.5, where f = 2.4; from there it decreases in a straight line f = 4.8(1 − x) down to (1, 0). The x-axis is in metres, with marks at 0, 0.5 and 1; the f(x)-axis has marks at 0, 1 and 2

c) $P (X < 0.6) = 1 - P (X > 0.6) = 1 - \int_{0.6}^{1} 4.8 (1 - x) d x = 1 - 4.8 \cdot \frac{0. 4 ^{2}}{2} = 1 - 0.384 = 0.616.$

Answer: a) $a = 9.6$ (verified); c) $P (X < 0.6) = 0.616$

Exercise 8 MediumIB · P1 TZ0 November 2014

The probability density function of a continuous random variable $T$ is given by

$f (t) = {∣2 - t ∣ 0 1 \leq t \leq 3 otherwise$

a) Sketch an approximate graph of $y = f (t)$ .

b) Find the interquartile range of T.

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Answer: b) IQR $= 2 \approx 1.41$

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a) The function is $f (t) = 2 - t$ for $1 \leq t \leq 2$ and $f (t) = t - 2$ for $2 \leq t \leq 3$ : two straight segments forming a V with vertex at $(2, 0)$ . Graph of y = f(t): an inverted V made of two straight segments, starting at the point (1, 1), falling in a straight line to the vertex (2, 0) and rising again in a straight line to (3, 1); outside the interval (1, 3) the graph is zero, on the horizontal axis. Axes marked at 1, 2 and 3 on t, and at 0 and 1 on y

b) The total area is $\frac{1}{2} + \frac{1}{2} = 1$ (two triangles). The density is symmetric about $t = 2$ , so the median is $2$ and the quartiles are symmetric. The first quartile $q_{1}$ satisfies $\int_{1}^{q_{1}} (2 - t) d t = \frac{1}{4}$ : $2 q_{1} - \frac{q _{1}^{2}}{2} - \frac{3}{2} = \frac{1}{4} \Rightarrow q_{1}^{2} - 4 q_{1} + \frac{7}{2} = 0 \Rightarrow q_{1} = 2 - \frac{2}{2} \approx 1.293$ (the root greater than $2$ is rejected). By symmetry, $q_{3} = 2 + \frac{2}{2} \approx 2.707$ . Then $IQR = q_{3} - q_{1} = 2 \approx 1.41.$

Answer: b) IQR $= 2 \approx 1.41$

Exercise 9 MediumIB · P1 May 1996

Let $X$ be a continuous random variable with probability density function

$f (x) = {\frac{x}{8} + c 0 0 \leq x \leq 4 otherwise$

Calculate the value of $c$ and the expected value of $X$ .

Show answer

Answer: $c = 0$ ; $E (X) = \frac{8}{3}$

Show working and answer

The total area is $1$ : $\int_{0}^{4} (\frac{x}{8} + c) d x = [\frac{x ^{2}}{16} + c x]_{0}^{4} = 1 + 4 c = 1 \Rightarrow c = 0.$ So $f (x) = \frac{x}{8}$ on $[0, 4]$ and $E (X) = \int_{0}^{4} x \cdot \frac{x}{8} d x = [\frac{x ^{3}}{24}]_{0}^{4} = \frac{64}{24} = \frac{8}{3} .$

Answer: $c = 0$ ; $E (X) = \frac{8}{3}$

Exercise 10 MediumIB · P1 November 1996

A probability density function is described below:

$f (x) = ⎩ ⎨ ⎧ 0 4 - 8 x 0 x < 0 0 \leq x \leq k x > k$

Calculate

a) the value of $k$ ;

b) the mean of the distribution.

Show answer

Answer: a) $k = \frac{1}{2}$ ; b) $E (X) = \frac{1}{6}$

Show working and answer

a) We need to check that $f \geq 0$ and that the area is $1$ : $\int_{0}^{k} (4 - 8 x) d x = [4 x - 4 x^{2}]_{0}^{k} = 4 k - 4 k^{2} = 1 \Rightarrow 4 k^{2} - 4 k + 1 = 0 \Rightarrow (2 k - 1)^{2} = 0 \Rightarrow k = \frac{1}{2} .$

b) $E (X) = \int_{0}^{1/2} x (4 - 8 x) d x = [2 x^{2} - \frac{8 x ^{3}}{3}]_{0}^{1/2} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} .$

Answer: a) $k = \frac{1}{2}$ ; b) $E (X) = \frac{1}{6}$

Exercise 11 MediumIB · P1 May 1998

The probability density function of a random variable $X$ is given by

$f (x) = {6 x (1 + x) 0 0 < x < k elsewhere$

a) Calculate $k$ .

b) Calculate the exact value of the mean of $X$ .

Show answer

Answer: a) $k = \frac{1}{2}$ ; b) $E (X) = \frac{11}{32}$

Show working and answer

a) The total integral is $1$ : $\int_{0}^{k} 6 x (1 + x) d x = [3 x^{2} + 2 x^{3}]_{0}^{k} = 3 k^{2} + 2 k^{3} = 1.$ Trying $k = \frac{1}{2}$ : $\frac{3}{4} + \frac{1}{4} = 1$ . As $3 k^{2} + 2 k^{3}$ is increasing for $k > 0$ , this is the only solution: $k = \frac{1}{2}$ .

b) $E (X) = \int_{0}^{1/2} x \cdot 6 x (1 + x) d x = \int_{0}^{1/2} (6 x^{2} + 6 x^{3}) d x = [2 x^{3} + \frac{3}{2} x^{4}]_{0}^{1/2} = \frac{1}{4} + \frac{3}{32} = \frac{11}{32} .$

Answer: a) $k = \frac{1}{2}$ ; b) $E (X) = \frac{11}{32}$

Exercise 12 MediumIB · P1 November 2001

A continuous random variable $X$ has the following probability density function

$f (x) = ⎩ ⎨ ⎧ \frac{4}{π ( 1 + x ^{2} )} 0 for 0 \leq x \leq 1 otherwise$

Calculate $E (X)$ .

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Answer: $E (X) = \frac{2 ln 2}{π} \approx 0.441$

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$E (X) = \int_{0}^{1} x \cdot \frac{4}{π ( 1 + x ^{2} )} d x = \frac{4}{π} \int_{0}^{1} \frac{x}{1 + x ^{2}} d x = \frac{4}{π} [\frac{1}{2} ln (1 + x^{2})]_{0}^{1} = \frac{2 ln 2}{π} .$

Answer: $E (X) = \frac{2 ln 2}{π} \approx 0.441$

Exercise 13 MediumIB · P1 TZ0 November 2004

A random variable $X$ has a probability density function given by

$f (x) = {k (2 x - x^{2}), 0, for 0 \leq x \leq 2 otherwise$

a) Find the value of $k$ .

b) Find $P (0.25 \leq X \leq 0.5)$ .

Show answer

Answer: a) $k = \frac{3}{4}$ ; b) $\frac{29}{256} \approx 0.113$

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a) $\int_{0}^{2} k (2 x - x^{2}) d x = k [x^{2} - \frac{x ^{3}}{3}]_{0}^{2} = k \cdot \frac{4}{3} = 1 \Rightarrow k = \frac{3}{4}$ .

b) $P (0.25 \leq X \leq 0.5) = \frac{3}{4} [x^{2} - \frac{x ^{3}}{3}]_{0.25}^{0.5} = \frac{3}{4} (\frac{5}{24} - \frac{11}{192}) = \frac{3}{4} \cdot \frac{29}{192} = \frac{29}{256} \approx 0.113.$

Answer: a) $k = \frac{3}{4}$ ; b) $\frac{29}{256} \approx 0.113$

Exercise 14 MediumIB · P2 TZ0 November 2006

The continuous random variable $X$ has the probability density function

$f (x) = {\frac{x}{1 + x ^{2}}, 0, for 0 \leq x \leq k otherwise$

a) Find the exact value of $k$ .

b) Find the mode of $X$ .

c) Calculate $P (1 \leq X \leq 2)$ .

Show answer

Answer: a) $k = e^{2} - 1$ ; b) mode $= 1$ ; c) $\frac{1}{2} ln \frac{5}{2} \approx 0.458$

Show working and answer

a) We must have $\int_{0}^{k} \frac{x}{1 + x ^{2}} d x = 1$ : $[\frac{1}{2} ln (1 + x^{2})]_{0}^{k} = \frac{1}{2} ln (1 + k^{2}) = 1 \Rightarrow 1 + k^{2} = e^{2} \Rightarrow k = e^{2} - 1 \approx 2.53.$

b) The mode is the maximum of $f$ . Differentiating: $f^{'} (x) = \frac{( 1 + x ^{2} ) - 2 x ^{2}}{( 1 + x ^{2} ) ^{2}} = \frac{1 - x ^{2}}{( 1 + x ^{2} ) ^{2}} = 0 \Rightarrow x = 1.$ The derivative is positive before $1$ and negative after it, so this is a maximum: $f (1) = 0.5$ (and at the endpoint $f (k) \approx 0.34 < 0.5$ ).

c) $P (1 \leq X \leq 2) = \int_{1}^{2} \frac{x}{1 + x ^{2}} d x = \frac{1}{2} [ln (1 + x^{2})]_{1}^{2} = \frac{1}{2} (ln 5 - ln 2) = \frac{1}{2} ln \frac{5}{2} \approx 0.458.$

Answer: a) $k = e^{2} - 1$ ; b) mode $= 1$ ; c) $\frac{1}{2} ln \frac{5}{2} \approx 0.458$

Exercise 15 MediumIB · P1 TZ1 May 2010

A continuous random variable $X$ has probability density function

$f (x) = {0, a e^{- a x}, x < 0 x \geq 0$

It is known that $P (X < 1) = 1 - \frac{1}{2}$ .

a) Show that $a = \frac{1}{2} ln 2$ .

b) Find the median of $X$ .

c) Calculate the probability that $X < 3$ given that $X > 1$ .

Show answer

Answer: a) shown: $a = \frac{1}{2} ln 2$ ; b) median $= 2$ ; c) $\frac{1}{2}$

Show working and answer

a) Since $P (X < 1) = \int_{0}^{1} a e^{- a x} d x = [- e^{- a x}]_{0}^{1} = 1 - e^{- a}$ , we equate: $1 - e^{- a} = 1 - \frac{1}{2} \Rightarrow e^{- a} = 2^{- 1/2} \Rightarrow - a = - \frac{1}{2} ln 2 \Rightarrow a = \frac{1}{2} ln 2.$

b) The median $m$ satisfies $P (X < m) = \frac{1}{2}$ , that is, $1 - e^{- am} = \frac{1}{2}$ : $e^{- am} = \frac{1}{2} \Rightarrow am = ln 2 \Rightarrow m = \frac{ln 2}{\frac{1}{2} ln 2} = 2.$

c) $P (X < 3 ∣ X > 1) = \frac{P ( 1 < X < 3 )}{P ( X > 1 )} = \frac{e ^{- a} - e ^{- 3 a}}{e ^{- a}} = 1 - e^{- 2 a} = 1 - e^{- l n 2} = 1 - \frac{1}{2} = \frac{1}{2} .$

Answer: a) shown: $a = \frac{1}{2} ln 2$ ; b) median $= 2$ ; c) $\frac{1}{2}$

Exercise 16 MediumIB · P1 TZ1 May 2022

The continuous random variable $X$ has probability density function

$f (x) = ⎩ ⎨ ⎧ \frac{k}{4 - 3 x ^{2}}, 0, 0 \leq x \leq 1 otherwise$

a) Calculate the value of $k$ .

b) Calculate $E (X)$ .

Show answer

Answer: a) $k = \frac{3 3}{π}$ ; b) $E (X) = \frac{3}{π} \approx 0.551$

Show working and answer

a) Since $\int \frac{d x}{4 - 3 x ^{2}} = \frac{1}{3} arcsin \frac{3 x}{2}$ : $\int_{0}^{1} \frac{k d x}{4 - 3 x ^{2}} = \frac{k}{3} arcsin \frac{3}{2} = \frac{k}{3} \cdot \frac{π}{3} = 1 \Rightarrow k = \frac{3 3}{π} .$

b) $E (X) = k \int_{0}^{1} \frac{x}{4 - 3 x ^{2}} d x = k [- \frac{1}{3} 4 - 3 x^{2}]_{0}^{1} = k (- \frac{1}{3} + \frac{2}{3}) = \frac{k}{3} = \frac{3}{π} \approx 0.551.$

Answer: a) $k = \frac{3 3}{π}$ ; b) $E (X) = \frac{3}{π} \approx 0.551$

Exercise 17 HardIB · P2 TZ0 November 2015

The continuous random variable $X$ has the probability density function

$f (x) = A sin (ln (x)), 1 \leq x \leq 5.$

a) Find the value of $A$ , correct to three decimal places.

b) Find the mode of $X$ .

c) Find the value of $P (X \leq 3 ∣ X \geq 2)$ .

Show answer

Answer: a) $A \approx 0.323$ ; b) mode $= e^{π /2} \approx 4.81$ ; c) $0.288$

Show working and answer

a) The integral of the density must equal $1$ : $A \int_{1}^{5} sin (ln x) d x = 1.$ Using the GDC, $\int_{1}^{5} sin (ln x) d x \approx 3.0947$ , so $A = \frac{1}{3.0947} \approx 0.323$ .

b) The mode is the value of $x$ where $f$ is maximum. As $A > 0$ , we need to maximise $sin (ln x)$ , which equals $1$ when $ln x = \frac{π}{2}$ : $x = e^{π /2} \approx 4.81 \in [1, 5] .$

c) $P (X \leq 3 ∣ X \geq 2) = \frac{P ( 2 \leq X \leq 3 )}{P ( X \geq 2 )} = \frac{A \int _{2}^{3} sin ( ln x ) d x}{A \int _{2}^{5} sin ( ln x ) d x} = \frac{0.2533}{0.8805} \approx 0.288.$

Answer: a) $A \approx 0.323$ ; b) mode $= e^{π /2} \approx 4.81$ ; c) $0.288$

Exercise 18 HardIB · P2 TZ2 May 2012

The probability density function of a continuous random variable $X$ is given by

$f (x) = \frac{1}{1 + x ^{4}}, 0 \leq x \leq a .$

a) Find the value of $a$ .

b) Find the mean of $X$ .

Show answer

Answer: a) $a \approx 1.40$ (1.3956); b) $E (X) \approx 0.548$

Show working and answer

a) We must have $\int_{0}^{a} \frac{1}{1 + x ^{4}} d x = 1$ . The antiderivative is not elementary; using the GDC (solving the equation with numerical integration) we obtain $a \approx 1.3956.$

b) $E (X) = \int_{0}^{a} \frac{x}{1 + x ^{4}} d x = [\frac{1}{2} arctan (x^{2})]_{0}^{a} = \frac{1}{2} arctan (a^{2}) = \frac{1}{2} arctan (1.9478) \approx 0.548.$ (The same value is obtained by integrating numerically with the GDC.)

Answer: a) $a \approx 1.40$ (1.3956); b) $E (X) \approx 0.548$

Exercise 19 HardIB · P1 TZ2 May 2019

The random variable $X$ has probability density function $f$ , given by

$f (x) = {k (π - arcsin x) 0 0 \leq x \leq 1 otherwise$

where $k$ is a positive constant.

a) State the mode of $X$ .

b) (i) Find $\int arcsin x d x$ .

(ii) Hence, show that $k = \frac{2}{2 + π}$ .

c) Given that $y = (\frac{x ^{2}}{2}) arcsin x - (\frac{1}{4}) arcsin x + (\frac{x}{4}) 1 - x^{2}$ , show that

(i) $\frac{d y}{d x} = x arcsin x$

(ii) $E (X) = \frac{3 π}{4 ( π + 2 )}$ .

Show answer

Answer: a) mode $= 0$ ; b) $k = \frac{2}{π + 2}$ ; c) $E (X) = \frac{3 π}{4 ( π + 2 )}$

Show working and answer

a) On $[0, 1]$ , $arcsin x$ increases, so $f (x) = k (π - arcsin x)$ decreases, and its maximum is at $x = 0$ . Mode $= 0$ .

b) i) Integrating by parts ( $u = arcsin x$ , $d v = d x$ ): $\int arcsin x d x = x arcsin x - \int \frac{x}{1 - x ^{2}} d x = x arcsin x + 1 - x^{2} + C .$

(ii) $\int_{0}^{1} arcsin x d x = [x arcsin x + 1 - x^{2}]_{0}^{1} = \frac{π}{2} - 1$ . As the total area is $1$ : $k (π - \frac{π}{2} + 1) = k \frac{π + 2}{2} = 1 \Rightarrow k = \frac{2}{π + 2} .$

c) i) Differentiating $y = \frac{x ^{2}}{2} arcsin x - \frac{1}{4} arcsin x + \frac{x}{4} 1 - x^{2}$ : $\frac{d y}{d x} = x arcsin x + \frac{x ^{2}}{2 1 - x ^{2}} - \frac{1}{4 1 - x ^{2}} + \frac{1}{4} 1 - x^{2} - \frac{x ^{2}}{4 1 - x ^{2}} .$ The last three terms, with common denominator $4 1 - x^{2}$ , give $\frac{2 x ^{2} - 1 + ( 1 - x ^{2} ) - x ^{2}}{4 1 - x ^{2}} = 0$ , so $\frac{d y}{d x} = x arcsin x$ .

(ii) By the previous part, $\int_{0}^{1} x arcsin x d x = y (1) - y (0) = \frac{π}{4} - \frac{π}{8} - 0 = \frac{π}{8}$ . Then $E (X) = k \int_{0}^{1} x (π - arcsin x) d x = k (\frac{π}{2} - \frac{π}{8}) = \frac{3 π k}{8} = \frac{3 π}{8} \cdot \frac{2}{π + 2} = \frac{3 π}{4 ( π + 2 )} .$

Answer: a) mode $= 0$ ; b) $k = \frac{2}{π + 2}$ ; c) $E (X) = \frac{3 π}{4 ( π + 2 )}$

Exercise 20 HardIB · P2 TZ1 May 2018

The continuous random variable $X$ has probability density function $f$ given by

$f (x) = ⎩ ⎨ ⎧ 3 a x, a (2 - x), 0, 0 \leq x < 0.5 0.5 \leq x < 2 otherwise$

a) Show that $a = \frac{2}{3}$ .

b) Calculate $P (X < 1)$ .

c) Given that $P (s < X < 0.8) = 2 \times P (2 s < X < 0.8)$ , and that $0.25 < s < 0.4$ , find the value of $s$ .

Show answer

Answer: a) $a = \frac{2}{3}$ ; b) $P (X < 1) = \frac{2}{3}$ ; c) $s \approx 0.274$

Show working and answer

a) The integral of $f$ over its whole domain is $1$ : $\int_{0}^{0.5} 3 a x d x + \int_{0.5}^{2} a (2 - x) d x = \frac{3 a}{8} + a [2 x - \frac{x ^{2}}{2}]_{0.5}^{2} = \frac{3 a}{8} + \frac{9 a}{8} = \frac{3 a}{2} = 1.$ Hence $a = \frac{2}{3}$ .

b) With $a = \frac{2}{3}$ the density is $2 x$ on $[0; 0.5)$ and $\frac{2}{3} (2 - x)$ on $[0.5; 2)$ : $P (X < 1) = \int_{0}^{0.5} 2 x d x + \int_{0.5}^{1} \frac{2}{3} (2 - x) d x = \frac{1}{4} + \frac{2}{3} (\frac{3}{2} - \frac{7}{8}) = \frac{1}{4} + \frac{5}{12} = \frac{2}{3} .$

c) As $s < 0.4$ and $0.25 < s$ , we have $0.25 < s < 0.4$ , so $2 s \in (0.5; 0.8)$ . Let $F (u) = P (X < u)$ . For $u < 0.5$ , $F (u) = u^{2}$ , and for $0.5 \leq u < 2$ , $F (u) = \frac{1}{4} + \frac{2}{3} (2 u - \frac{u ^{2}}{2} - \frac{7}{8})$ , so $F (0.8) = 0.52$ .

The condition $F (0.8) - F (s) = 2 [F (0.8) - F (2 s)]$ gives $0.52 - s^{2} = 2 [0.52 - \frac{1}{4} - \frac{2}{3} (4 s - 2 s^{2} - \frac{7}{8})] \Rightarrow 11 s^{2} - 16 s + 3.56 = 0.$ The solutions are $s = \frac{16 \pm 99.36}{22}$ , that is, $s \approx 0.274$ or $s \approx 1.18$ . Only $0.274$ lies in $(0.25; 0.4)$ .

Answer: a) $a = \frac{2}{3}$ ; b) $P (X < 1) = \frac{2}{3}$ ; c) $s \approx 0.274$

Exercise 21 HardIB · P1 TZ1 May 2017

The continuous random variable $X$ has probability density function

$f (x) = {k sin (\frac{π x}{6}) 0 0 \leq x \leq 6 otherwise$

a) Calculate the value of $k$ .

b) Considering the graph of $f$ , write down:

(i) the mean of $X$ ;

(ii) the median of $X$ ;

(iii) the mode of $X$ .

c) (i) Show that $P (0 \leq X \leq 2) = \frac{1}{4}$ .

(ii) Using the above, state the interquartile range of $X$ .

d) Calculate $P (X \leq 4 ∣ X \geq 3)$ .

Show answer

Answer: a) $k = \frac{π}{12}$ ; b) mean $= 3$ , median $= 3$ , mode $= 3$ ; c) ii) interquartile range $= 2$ ; d) $\frac{1}{2}$

Show working and answer

a) The total integral is $1$ : $k \int_{0}^{6} sin \frac{π x}{6} d x = k [- \frac{6}{π} cos \frac{π x}{6}]_{0}^{6} = k \cdot \frac{12}{π} = 1 \Rightarrow k = \frac{π}{12} .$

Graph of f(x) = (π/12) · sin(πx/6) for 0 ≤ x ≤ 6: a single sine arch above the horizontal axis, symmetric about the line x = 3, zero at x = 0 and x = 6; its maximum height π/12 ≈ 0.26 is reached at x = 3. Horizontal axis marked at 0, 1, 2, 3, 4, 5 and 6

b) The graph is symmetric about $x = 3$ and reaches its maximum there, so:

(i) mean $= 3$ ; ii) median $= 3$ ; iii) mode $= 3$ .

c) i) $P (0 \leq X \leq 2) = \frac{π}{12} [- \frac{6}{π} cos \frac{π x}{6}]_{0}^{2} = \frac{1}{2} (1 - cos \frac{π}{3}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} .$

(ii) As $P (X \leq 2) = \frac{1}{4}$ , the first quartile is $Q_{1} = 2$ . By symmetry, $Q_{3} = 6 - 2 = 4$ . So the interquartile range is $Q_{3} - Q_{1} = 4 - 2 = 2$ .

d) By symmetry, $P (3 \leq X \leq 4) = P (2 \leq X \leq 3) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ , and $P (X \geq 3) = \frac{1}{2}$ . So $P (X \leq 4 ∣ X \geq 3) = \frac{P ( 3 \leq X \leq 4 )}{P ( X \geq 3 )} = \frac{1/4}{1/2} = \frac{1}{2} .$

Answer: a) $k = \frac{π}{12}$ ; b) mean $= 3$ , median $= 3$ , mode $= 3$ ; c) ii) interquartile range $= 2$ ; d) $\frac{1}{2}$

Exercise 22 HardIB · P2 November 1999

The probability density function of a continuous random variable $X$ is given by

$f (x) = {e - k e^{k x} 0 0 \leq x \leq 1 otherwise$

a) Show that $k = 1$ .

b) Find the probability that the random variable $X$ takes a value between $\frac{1}{4}$ and $\frac{1}{2}$ . Give your answer in terms of $e$ .

c) Calculate the mean and the variance of the distribution. Give your answer exactly in terms of $e$ .

Show answer

Answer: a) $k = 1$ ; b) $\frac{e}{4} - e^{1/2} + e^{1/4} \approx 0.315$ ; c) $E (X) = \frac{e}{2} - 1 \approx 0.359$ , $Var (X) = 1 + \frac{e}{3} - \frac{e ^{2}}{4} \approx 0.0588$

Show working and answer

a) The total integral must equal 1: $\int_{0}^{1} (e - k e^{k x}) d x = [e x - e^{k x}]_{0}^{1} = e - e^{k} + 1 = 1 \Rightarrow e^{k} = e \Rightarrow k = 1.$

b) With $k = 1$ , $f (x) = e - e^{x}$ and an antiderivative is $e x - e^{x}$ : $P (\frac{1}{4} \leq X \leq \frac{1}{2}) = (\frac{e}{2} - e^{1/2}) - (\frac{e}{4} - e^{1/4}) = \frac{e}{4} - e^{1/2} + e^{1/4} \approx 0.315.$

c) Integrating by parts, $E (X) = \int_{0}^{1} x (e - e^{x}) d x = \frac{e}{2} - [x e^{x} - e^{x}]_{0}^{1} = \frac{e}{2} - 1 \approx 0.359.$ $E (X^{2}) = \int_{0}^{1} x^{2} (e - e^{x}) d x = \frac{e}{3} - [x^{2} e^{x} - 2 x e^{x} + 2 e^{x}]_{0}^{1} = \frac{e}{3} - (e - 2) = 2 - \frac{2 e}{3} .$ $Var (X) = 2 - \frac{2 e}{3} - (\frac{e}{2} - 1)^{2} = 1 + \frac{e}{3} - \frac{e ^{2}}{4} \approx 0.0588.$

Answer: a) $k = 1$ ; b) $\frac{e}{4} - e^{1/2} + e^{1/4} \approx 0.315$ ; c) $E (X) = \frac{e}{2} - 1 \approx 0.359$ , $Var (X) = 1 + \frac{e}{3} - \frac{e ^{2}}{4} \approx 0.0588$

Exercise 23 HardIB · P2 TZ1 May 2015

The probability density function of a continuous random variable $X$ is given by

$f (x) = ⎩ ⎨ ⎧ 0, \frac{sin x}{4}, a (x - π), 0, x < 0 0 \leq x \leq π π < x \leq 2 π 2 π < x$

a) Sketch the graph of $y = f (x)$ .

b) Find $P (X \leq π)$ .

c) Show that $a = \frac{1}{π ^{2}}$ .

d) Write down the median of $X$ .

e) Calculate the mean of $X$ .

f) Calculate the variance of $X$ .

g) Calculate $P (\frac{π}{2} \leq X \leq \frac{3 π}{2})$ .

h) Given that $\frac{π}{2} \leq X \leq \frac{3 π}{2}$ , calculate the probability that $π \leq X \leq 2 π$ .

Show answer

Answer: b) $\frac{1}{2}$ ; c) shown: $a = \frac{1}{π ^{2}}$ ; d) median $= π$ ; e) $\frac{13 π}{12} \approx 3.40$ ; f) $\frac{71 π ^{2}}{144} - 1 \approx 3.87$ ; g) $0.375$ ; h) $\frac{1}{3}$

Show working and answer

a) Piecewise: $0$ if $x < 0$ ; an arc of $\frac{sin x}{4}$ on $[0, π]$ (maximum $0.25$ at $x = π /2$ ); a straight line with gradient $a$ on $[π, 2 π]$ , from $(π, 0)$ to $(2 π, aπ)$ ; and $0$ if $x > 2 π$ (with $a = 1/ π^{2}$ it reaches $\frac{1}{π} \approx 0.32$ ).

Graph of the density f(x) piecewise: 0 for x < 0; an arc of sin x / 4 on (0, π) rising from the origin to its maximum 0.25 at x = π/2 and back to 0 at x = π; a rising straight segment from (π, 0) to (2π, 1/π ≈ 0.32) on (π, 2π); and 0 for x > 2π. x-axis with marks at π and 2π; f(x) axis with marks at 0.25 and 1/π ≈ 0.32

b) $P (X \leq π) = \int_{0}^{π} \frac{sin x}{4} d x = \frac{1}{4} [- cos x]_{0}^{π} = \frac{1}{4} \cdot 2 = \frac{1}{2}$ .

c) The area of the second part must be $1 - \frac{1}{2} = \frac{1}{2}$ : $\int_{π}^{2 π} a (x - π) d x = a [\frac{( x - π ) ^{2}}{2}]_{π}^{2 π} = \frac{a π ^{2}}{2} = \frac{1}{2} \Rightarrow a = \frac{1}{π ^{2}} .$

d) Up to $π$ , exactly $\frac{1}{2}$ has accumulated, so the median is $x = π$ .

e) $E (X) = \int_{0}^{π} \frac{x sin x}{4} d x + \int_{π}^{2 π} \frac{x ( x - π )}{π ^{2}} d x = \frac{π}{4} + \frac{5 π}{6} = \frac{13 π}{12} \approx 3.40.$

f) $E (X^{2}) = \int_{0}^{π} \frac{x ^{2} sin x}{4} d x + \int_{π}^{2 π} \frac{x ^{2} ( x - π )}{π ^{2}} d x = (\frac{π ^{2}}{4} - 1) + \frac{17 π ^{2}}{12} = \frac{5 π ^{2}}{3} - 1.$ $Var (X) = \frac{5 π ^{2}}{3} - 1 - (\frac{13 π}{12})^{2} = \frac{71 π ^{2}}{144} - 1 \approx 3.87.$

g) $P (\frac{π}{2} \leq X \leq \frac{3 π}{2}) = \int_{π /2}^{π} \frac{sin x}{4} d x + \int_{π}^{3 π /2} \frac{x - π}{π ^{2}} d x = \frac{1}{4} + \frac{1}{8} = \frac{3}{8} = 0.375.$

h) Conditional probability: the intersection is $π \leq X \leq \frac{3 π}{2}$ , which has probability $\frac{1}{8}$ : $\frac{P ( π \leq X \leq \frac{3 π}{2} )}{P ( \frac{π}{2} \leq X \leq \frac{3 π}{2} )} = \frac{1/8}{3/8} = \frac{1}{3} .$

Answer: b) $\frac{1}{2}$ ; c) shown: $a = \frac{1}{π ^{2}}$ ; d) median $= π$ ; e) $\frac{13 π}{12} \approx 3.40$ ; f) $\frac{71 π ^{2}}{144} - 1 \approx 3.87$ ; g) $0.375$ ; h) $\frac{1}{3}$

Exercise 24 HardIB · P2 TZ1 May 2021

The continuous random variable $X$ has a probability density function given by

$f (x) = ⎩ ⎨ ⎧ \frac{x}{( x ^{2} + k ) ^{3}}, 0, 0 \leq x \leq 4 otherwise where k \in R^{+} .$

a) Show that $16 + k - k = k 16 + k$ .

b) Calculate the value of $k$ .

Show answer

Answer: a) shown; b) $k = 0.645$

Show working and answer

a) An antiderivative is $\int \frac{x}{( x ^{2} + k ) ^{3/2}} d x = - \frac{1}{x ^{2} + k}$ (this can be checked by differentiating). The condition for a density function is: $\int_{0}^{4} f (x) d x = [- \frac{1}{x ^{2} + k}]_{0}^{4} = \frac{1}{k} - \frac{1}{16 + k} = 1.$ Multiplying by $k 16 + k$ : $16 + k - k = k 16 + k .$

b) Using the GDC, solve the equation $\frac{1}{k} - \frac{1}{16 + k} = 1$ for $k > 0$ : $k = 0.645$ .

Answer: a) shown; b) $k = 0.645$

Exercise 25 HardIB · P2 TZ0 November 2022

The continuous random variable $X$ has the following probability density function:

$f (x) = {a x e^{x}, 0, 0 \leq x \leq b otherwise$

a) Find an expression for $a$ in terms of $b$ .

b) In the case where $a = b = 1$ , find the median of $X$ .

Show answer

Answer: a) $a = \frac{1}{b e ^{b} - e ^{b} + 1}$ ; b) median $= 0.768$

Show working and answer

a) By parts, $\int x e^{x} d x = x e^{x} - e^{x}$ , so $\int_{0}^{b} a x e^{x} d x = a [x e^{x} - e^{x}]_{0}^{b} = a (b e^{b} - e^{b} + 1) = 1 \Rightarrow a = \frac{1}{b e ^{b} - e ^{b} + 1} .$

b) With $a = b = 1$ : $a = \frac{1}{e - e + 1} = 1$ , and $f (x) = x e^{x}$ on $[0, 1]$ . The median $m$ satisfies: $\int_{0}^{m} x e^{x} d x = m e^{m} - e^{m} + 1 = \frac{1}{2} .$ Using the GDC (solve the equation): $m = 0.768$ .

Answer: a) $a = \frac{1}{b e ^{b} - e ^{b} + 1}$ ; b) median $= 0.768$

Expected value, variance, median and mode of a density

Exercise 26 BasicIB · P1 TZ1 May 2008

The random variable $T$ has probability density function $f (t) = \frac{π}{4} cos (\frac{π t}{2})$ , $- 1 \leq t \leq 1$ .

Calculate

a) $P (T = 0)$ ;

b) the interquartile range.

Show answer

Answer: a) $P (T = 0) = 0$ ; b) interquartile range $= \frac{2}{3}$

Show working and answer

a) For a continuous random variable the probability of any particular value is zero (the area under a point is $0$ ): $P (T = 0) = 0$ .

b) The distribution function is, for $- 1 \leq t \leq 1$ : $F (t) = \int_{- 1}^{t} \frac{π}{4} cos \frac{π s}{2} d s = \frac{1}{2} [sin \frac{π s}{2}]_{- 1}^{t} = \frac{1}{2} sin \frac{π t}{2} + \frac{1}{2} .$

First quartile: $F (Q_{1}) = \frac{1}{4} \Rightarrow sin \frac{π Q _{1}}{2} = - \frac{1}{2} \Rightarrow \frac{π Q _{1}}{2} = - \frac{π}{6} \Rightarrow Q_{1} = - \frac{1}{3}$ .
Third quartile: $F (Q_{3}) = \frac{3}{4} \Rightarrow sin \frac{π Q _{3}}{2} = \frac{1}{2} \Rightarrow Q_{3} = \frac{1}{3}$ .

The interquartile range is $Q_{3} - Q_{1} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ .

Answer: a) $P (T = 0) = 0$ ; b) interquartile range $= \frac{2}{3}$

Exercise 27 MediumIB · P2 TZ0 November 2008

A continuous random variable $X$ has a probability density function given by

$f (x) = ⎩ ⎨ ⎧ \frac{( x + 1 ) ^{3}}{60} 0 for 1 \leq x \leq 3 otherwise$

Find:

a) $P (1.5 \leq X \leq 2.5)$

b) $E (X)$

c) The median of $X$ .

Show answer

Answer: a) $\frac{37}{80} \approx 0.463$ ; b) $E (X) = \frac{173}{75} \approx 2.31$ ; c) median $\approx 2.41$

Show working and answer

a) $P (1.5 \leq X \leq 2.5) = \int_{1.5}^{2.5} \frac{( x + 1 ) ^{3}}{60} d x = [\frac{( x + 1 ) ^{4}}{240}]_{1.5}^{2.5} = \frac{3. 5 ^{4} - 2. 5 ^{4}}{240} = \frac{111}{240} = \frac{37}{80} \approx 0.463.$

b) With $u = x + 1$ (and $x = u - 1$ ): $E (X) = \int_{1}^{3} \frac{x ( x + 1 ) ^{3}}{60} d x = \frac{1}{60} \int_{2}^{4} (u^{4} - u^{3}) d u = \frac{1}{60} [\frac{u ^{5}}{5} - \frac{u ^{4}}{4}]_{2}^{4} = \frac{173}{75} \approx 2.31.$

c) The distribution function is $F (m) = \frac{( m + 1 ) ^{4} - 16}{240}$ . With $F (m) = \frac{1}{2}$ : $(m + 1)^{4} = 136 \Rightarrow m = 4136 - 1 \approx 2.41.$

Answer: a) $\frac{37}{80} \approx 0.463$ ; b) $E (X) = \frac{173}{75} \approx 2.31$ ; c) median $\approx 2.41$

Exercise 28 MediumIB · P2 TZ0 November 2009

The annual losses, related to climate factors, of an insurance company are modelled by a random variable $X$ , whose probability density function is:

$f (x) = ⎩ ⎨ ⎧ \frac{2.5 ( 200 ) ^{2.5}}{x ^{3.5}} 0 x \geq 200 all other values$

Find the median.

Show answer

Answer: median $\approx 263.9$

Show working and answer

The distribution function, for $m \geq 200$ , is $F (m) = \int_{200}^{m} 2.5 \cdot 20 0^{2.5} x^{- 3.5} d x = [- 20 0^{2.5} x^{- 2.5}]_{200}^{m} = 1 - (\frac{200}{m})^{2.5} .$ The median satisfies $F (m) = \frac{1}{2}$ : $(\frac{200}{m})^{2.5} = \frac{1}{2} \Rightarrow m = 200 \cdot 2^{1/2.5} = 200 \cdot 2^{0.4} \approx 263.9.$

Answer: median $\approx 263.9$

Exercise 29 MediumIB · P1 November 1997

The random variable $X$ has probability density function $f (x)$ , given by

$f (x) = ⎩ ⎨ ⎧ \frac{1}{4} x \frac{1}{4} \frac{1}{12} (6 - x) 0 0 \leq x < 1 1 \leq x < 3 3 \leq x \leq 6 otherwise$

Calculate the median of $X$ .

Show answer

Answer: median $= \frac{5}{2}$

Show working and answer

Cumulative areas:

On $[0; 1)$ : $\int_{0}^{1} \frac{x}{4} d x = \frac{1}{8}$ .
On $[1; 3)$ : the density is $\frac{1}{4}$ , and the area up to $3$ is $\frac{1}{8} + \frac{2}{4} = \frac{5}{8} > \frac{1}{2}$ .

The median $m$ therefore lies in $[1; 3)$ : $\frac{1}{8} + \frac{1}{4} (m - 1) = \frac{1}{2} \Rightarrow m - 1 = \frac{3}{2} \Rightarrow m = \frac{5}{2} .$

Answer: median $= \frac{5}{2}$

Exercise 30 MediumIB · P1 November 2002

The probability density function $f (x)$ of a continuous random variable $X$ is defined by

$f (x) = ⎩ ⎨ ⎧ \frac{1}{4} x (4 - x^{2}), 0, 0 \leq x \leq 2 otherwise$

Calculate the value of the median.

Show answer

Answer: median $= 4 - 22 \approx 1.08$

Show working and answer

The median $m$ satisfies $\int_{0}^{m} f (x) d x = \frac{1}{2}$ : $\int_{0}^{m} \frac{1}{4} (4 x - x^{3}) d x = \frac{1}{4} [2 x^{2} - \frac{x ^{4}}{4}]_{0}^{m} = \frac{m ^{2}}{2} - \frac{m ^{4}}{16} = \frac{1}{2} .$ Multiplying by 16: $m^{4} - 8 m^{2} + 8 = 0$ , so $m^{2} = 4 \pm 22$ . Since $0 \leq m \leq 2$ , we take $m^{2} = 4 - 22$ : $m = 4 - 22 \approx 1.08.$

Answer: median $= 4 - 22 \approx 1.08$

Exercise 31 MediumIB · P1 November 2003

A continuous random variable, $X$ , has probability density function

$f (x) = sin x, 0 \leq x \leq \frac{π}{2}$

Find the median of $X$ .

Show answer

Answer: median $= \frac{π}{3}$

Show working and answer

The median $m$ satisfies $\int_{0}^{m} sin x d x = \frac{1}{2}$ : $[- cos x]_{0}^{m} = 1 - cos m = \frac{1}{2} \Rightarrow cos m = \frac{1}{2} .$ In $[0; \frac{π}{2}]$ this gives $m = \frac{π}{3}$ .

Answer: median $= \frac{π}{3}$

Exercise 32 MediumIB · P2 TZ0 May 2006

The time, $T$ minutes, required by students to answer a question in a mathematics exam has probability density function

$f (t) = ⎩ ⎨ ⎧ \frac{1}{72} (12 t - t^{2} - 20), 0, for 4 \leq t \leq 10 otherwise$

a) Find

(i) $μ$ , the expected value of $T$ .

(ii) $σ^{2}$ , the variance of $T$ .

b) A student is chosen at random. Find the probability that the time this student takes to answer the question lies in the interval $[μ - σ, μ]$ .

Show answer

Answer: a) $μ = 6.5$ ; $σ^{2} = 2.15$ ; b) $0.321$

Show working and answer

a) i) $μ = \frac{1}{72} \int_{4}^{10} t (12 t - t^{2} - 20) d t = \frac{1}{72} [4 t^{3} - \frac{t ^{4}}{4} - 10 t^{2}]_{4}^{10} = \frac{13}{2} = 6.5.$

(ii) $E (T^{2}) = \frac{1}{72} \int_{4}^{10} t^{2} (12 t - t^{2} - 20) d t = \frac{1}{72} [3 t^{4} - \frac{t ^{5}}{5} - \frac{20 t ^{3}}{3}]_{4}^{10} = \frac{222}{5} = 44.4.$ $σ^{2} = E (T^{2}) - μ^{2} = \frac{43}{20} = 2.15.$

b) $σ = 2.15 \approx 1.466$ , so the interval is $[5.03; 6.5]$ and $P (μ - σ \leq T \leq μ) = \int_{5.03}^{6.5} f (t) d t \approx 0.321.$

Answer: a) $μ = 6.5$ ; $σ^{2} = 2.15$ ; b) $0.321$

Exercise 33 MediumIB · P1 TZ1 May 2007

A continuous random variable $X$ has probability density function $f$ defined by

$f (x) = {e^{x}, 0, 0 \leq x \leq ln 2 otherwise$

Calculate the exact value of $E (X)$ .

Show answer

Answer: $E (X) = 2 ln 2 - 1$

Show working and answer

$E (X) = \int_{0}^{l n 2} x e^{x} d x .$ By parts ( $u = x$ , $d v = e^{x} d x$ ): $\int x e^{x} d x = x e^{x} - e^{x} .$ Evaluating between $0$ and $ln 2$ : $E (X) = [x e^{x} - e^{x}]_{0}^{l n 2} = (2 ln 2 - 2) - (0 - 1) = 2 ln 2 - 1 \approx 0.386.$

Answer: $E (X) = 2 ln 2 - 1$

Exercise 34 MediumIB · P1 TZ2 May 2007

The probability density function $f$ of a continuous random variable $X$ is given by

$f (x) = ⎩ ⎨ ⎧ \frac{8}{π ( x ^{2} + 4 )}, 0, 0 \leq x \leq 2 otherwise$

a) State the mode of $X$ .

b) Find the exact value of $E (X)$ .

Show answer

Answer: a) mode $x = 0$ ; b) $E (X) = \frac{4}{π} ln 2$

Show working and answer

a) The function $\frac{8}{π ( x ^{2} + 4 )}$ is decreasing on $[0, 2]$ , so its maximum is at the left-hand endpoint: the mode is $x = 0$ .

b) $E (X) = \int_{0}^{2} x \cdot \frac{8}{π ( x ^{2} + 4 )} d x = \frac{8}{π} \cdot \frac{1}{2} [ln (x^{2} + 4)]_{0}^{2} = \frac{4}{π} (ln 8 - ln 4) = \frac{4}{π} ln 2 \approx 0.883.$

Answer: a) mode $x = 0$ ; b) $E (X) = \frac{4}{π} ln 2$

Exercise 35 MediumIB · P1 TZ0 November 2011

A continuous random variable $X$ has the following probability density function:

$f (x) = {k sin x, 0, 0 \leq x \leq \frac{π}{2} otherwise$

a) Find the value of $k$ .

b) Find $E (X)$ .

c) Find the median of $X$ .

Show answer

Answer: a) $k = 1$ ; b) $E (X) = 1$ ; c) median $= \frac{π}{3}$

Show working and answer

a) $\int_{0}^{π /2} k sin x d x = k [- cos x]_{0}^{π /2} = k = 1$ , so $k = 1$ .

b) By parts ( $u = x$ , $d v = sin x d x$ ): $E (X) = \int_{0}^{π /2} x sin x d x = [- x cos x + sin x]_{0}^{π /2} = 1.$

c) The median $m$ satisfies $\int_{0}^{m} sin x d x = 1 - cos m = \frac{1}{2}$ , that is, $cos m = \frac{1}{2}$ . In $[0, π /2]$ this gives $m = \frac{π}{3}$ .

Answer: a) $k = 1$ ; b) $E (X) = 1$ ; c) median $= \frac{π}{3}$

Exercise 36 MediumIB · P1 TZ0 November 2012

The continuous random variable $X$ has probability density function given by

$f (x) = {a e^{- x}, 0, 0 \leq x \leq 1 otherwise$

a) State the mode of $X$ .

b) Determine the value of $a$ .

c) Calculate $E (X)$ .

Show answer

Answer: a) mode $x = 0$ ; b) $a = \frac{e}{e - 1}$ ; c) $E (X) = \frac{e - 2}{e - 1} \approx 0.418$

Show working and answer

a) $f (x) = a e^{- x}$ is decreasing, so the maximum is at $x = 0$ : the mode is $x = 0$ .

b) $\int_{0}^{1} a e^{- x} d x = a (1 - e^{- 1}) = 1 \Rightarrow a = \frac{1}{1 - e ^{- 1}} = \frac{e}{e - 1}$ .

c) By parts: $E (X) = a \int_{0}^{1} x e^{- x} d x = a [- x e^{- x} - e^{- x}]_{0}^{1} = a (1 - 2 e^{- 1}) = \frac{e}{e - 1} \cdot \frac{e - 2}{e} = \frac{e - 2}{e - 1} \approx 0.418.$

Answer: a) mode $x = 0$ ; b) $a = \frac{e}{e - 1}$ ; c) $E (X) = \frac{e - 2}{e - 1} \approx 0.418$

Exercise 37 MediumIB · P1 TZ1 May 2013

The probability density function of the random variable $X$ is defined as

$f (x) = {sin x, 0, 0 \leq x \leq \frac{π}{2} otherwise$

Find $E (X)$ .

Show answer

Answer: $E (X) = 1$

Show working and answer

By parts ( $u = x$ , $d v = sin x d x$ ): $E (X) = \int_{0}^{π /2} x sin x d x = [- x cos x + sin x]_{0}^{π /2} = (0 + 1) - (0 + 0) = 1.$

Answer: $E (X) = 1$

Exercise 38 HardIB · P2 TZ1 May 2004

(ii) Let $f (x)$ be the probability density function of a random variable $X$ , where

$f (x) = {k x^{2} 0 for 0 \leq x \leq 2 otherwise$

a) Show that $k = \frac{3}{8}$ .

b) Calculate:

(i) $E (X)$ .

(ii) The median of $X$ .

Show answer

Answer: a) $k = \frac{3}{8}$ ; b) i) $E (X) = \frac{3}{2}$ ; ii) median $= 34 \approx 1.59$

Show working and answer

a) The total area must be $1$ : $\int_{0}^{2} k x^{2} d x = k \frac{2 ^{3}}{3} = \frac{8 k}{3} = 1 \Rightarrow k = \frac{3}{8} .$

b) i) $E (X) = \int_{0}^{2} x \cdot \frac{3}{8} x^{2} d x = \frac{3}{8} [\frac{x ^{4}}{4}]_{0}^{2} = \frac{3}{8} \cdot 4 = \frac{3}{2} .$

(ii) The median $m$ satisfies $P (X \leq m) = \frac{1}{2}$ : $\int_{0}^{m} \frac{3}{8} x^{2} d x = \frac{m ^{3}}{8} = \frac{1}{2} \Rightarrow m^{3} = 4 \Rightarrow m = 34 \approx 1.587.$

Answer: a) $k = \frac{3}{8}$ ; b) i) $E (X) = \frac{3}{2}$ ; ii) median $= 34 \approx 1.59$

Exercise 39 HardIB · P2 TZ1 May 2014

A random variable $X$ has probability density function

$f (x) = {a x + b 0 2 \leq x \leq 3 otherwise$

with $a, b \in R$ .

a) Show that $5 a + 2 b = 2$ .

Let $E (X) = μ$ .

b) (i) Show that $a = 12 μ - 30$ .

(ii) Find a similar expression for $b$ in terms of $μ$ .

Let $2.3$ be the median of the distribution.

c) (i) Find the value of $μ$ .

(ii) Find the value of the standard deviation of $X$ .

Show answer

Answer: b) ii) $b = 76 - 30 μ$ ; c) i) $μ \approx 2.34$ ; ii) $σ \approx 0.241$

Show working and answer

a) Total area equal to $1$ : $\int_{2}^{3} (a x + b) d x = [\frac{a x ^{2}}{2} + b x]_{2}^{3} = \frac{5 a}{2} + b = 1 \Rightarrow 5 a + 2 b = 2.$

b) i) $μ = \int_{2}^{3} x (a x + b) d x = [\frac{a x ^{3}}{3} + \frac{b x ^{2}}{2}]_{2}^{3} = \frac{19 a}{3} + \frac{5 b}{2} .$ From a), $b = 1 - \frac{5 a}{2}$ . Substituting: $μ = \frac{19 a}{3} + \frac{5}{2} - \frac{25 a}{4} = \frac{a}{12} + \frac{5}{2}$ , hence $a = 12 μ - 30$ .

(ii) $b = 1 - \frac{5 a}{2} = 1 - \frac{5}{2} (12 μ - 30) = 76 - 30 μ$ .

c) i) With median $2.3$ : $\int_{2}^{2.3} (a x + b) d x = \frac{a ( 2. 3 ^{2} - 2 ^{2} )}{2} + 0.3 b = 0.645 a + 0.3 b = \frac{1}{2}$ . Substituting $a = 12 μ - 30$ and $b = 76 - 30 μ$ : $0.645 (12 μ - 30) + 0.3 (76 - 30 μ) = 0.5 \Rightarrow - 1.26 μ + 3.45 = 0.5 \Rightarrow μ = \frac{295}{126} \approx 2.34.$

(ii) Then $a = 12 μ - 30 = - \frac{40}{21}$ and $b = 76 - 30 μ = \frac{121}{21}$ . Using the GDC (numerical integration): $E (X^{2}) - μ^{2} = \int_{2}^{3} x^{2} (a x + b) d x - μ^{2} \approx 0.0581, σ = 0.0581 \approx 0.241.$

Answer: b) ii) $b = 76 - 30 μ$ ; c) i) $μ \approx 2.34$ ; ii) $σ \approx 0.241$

Exercise 40 HardIB · P2 TZ2 May 2014

The probability density function of a random variable $X$ is given by:

$f (x) = {a x cos x 0 0 \leq x \leq \frac{π}{2} otherwise where a \in R$

a) Show that $a = \frac{2}{π - 2}$ .

b) Find $P (X < \frac{π}{4})$ .

c) Find:

(i) the mode of $X$ .

(ii) the median of $X$ .

d) Find $P (X < \frac{π}{8} ∣ X < \frac{π}{4})$ .

Show answer

Answer: a) verified; b) $0.460$ ; c) i) mode $\approx 0.860$ ; ii) median $\approx 0.826$ ; d) $0.283$

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a) We must have $a \int_{0}^{π /2} x cos x d x = 1$ . Integrating by parts, $\int x cos x d x = x sin x + cos x$ : $\int_{0}^{π /2} x cos x d x = [x sin x + cos x]_{0}^{π /2} = \frac{π}{2} - 1 = \frac{π - 2}{2} .$ So $a \cdot \frac{π - 2}{2} = 1$ and $a = \frac{2}{π - 2}$ .

b) The distribution function is $F (t) = a (t sin t + cos t - 1)$ . Then $P (X < \frac{π}{4}) = \frac{2}{π - 2} (\frac{π}{4} \cdot \frac{2}{2} + \frac{2}{2} - 1) \approx 0.460.$

c) i) The mode is the maximum of $f$ : $f^{'} (x) = a (cos x - x sin x) = 0 \Rightarrow x tan x = 1$ . Using the GDC: $x \approx 0.860$ .

(ii) The median $m$ satisfies $F (m) = \frac{1}{2}$ : $a (m sin m + cos m - 1) = \frac{1}{2}$ . Using the GDC: $m \approx 0.826$ .

d) $P (X < \frac{π}{8} ∣ X < \frac{π}{4}) = \frac{F ( π /8 )}{F ( π /4 )} = \frac{0.1299}{0.4598} \approx 0.283.$

Answer: a) verified; b) $0.460$ ; c) i) mode $\approx 0.860$ ; ii) median $\approx 0.826$ ; d) $0.283$

Exercise 41 HardIB · P2 TZ2 May 2017

A continuous random variable $X$ has probability density function $f$

$f (x) = ⎩ ⎨ ⎧ \frac{x ^{2}}{a} + b 0 0 \leq x \leq 4 otherwise$

where $a$ and $b$ are positive constants. It is known that $P (X \geq 2) = 0.75$ .

a) Show that $a = 32$ and $b = \frac{1}{12}$ .

b) Find $E (X)$ .

c) Find $Var (X)$ .

d) Find the median of $X$ .

Eight independent observations of $X$ are made, and the random variable $Y$ is the number of observations such that $X \geq 2$ .

e) Find $E (Y)$ .

f) Find $P (Y \geq 3)$ .

Show answer

Answer: b) $\frac{8}{3}$ ; c) $\frac{16}{15}$ ; d) $2.91$ ; e) $6$ ; f) $0.996$

Show working and answer

a) The total integral is $1$ and $P (X \geq 2) = 0.75$ : $\int_{0}^{4} (\frac{x ^{2}}{a} + b) d x = \frac{64}{3 a} + 4 b = 1, \int_{2}^{4} (\frac{x ^{2}}{a} + b) d x = \frac{56}{3 a} + 2 b = 0.75.$ Multiplying the second by 2: $\frac{112}{3 a} + 4 b = 1.5$ . Subtracting the first: $\frac{48}{3 a} = \frac{16}{a} = 0.5$ , so $a = 32$ . Then $\frac{64}{96} + 4 b = 1$ , hence $b = \frac{1}{12}$ .

b) $E (X) = \int_{0}^{4} x (\frac{x ^{2}}{32} + \frac{1}{12}) d x = [\frac{x ^{4}}{128} + \frac{x ^{2}}{24}]_{0}^{4} = 2 + \frac{2}{3} = \frac{8}{3} .$

c) $E (X^{2}) = \int_{0}^{4} x^{2} (\frac{x ^{2}}{32} + \frac{1}{12}) d x = [\frac{x ^{5}}{160} + \frac{x ^{3}}{36}]_{0}^{4} = \frac{32}{5} + \frac{16}{9} = \frac{368}{45}$ . $Var (X) = \frac{368}{45} - (\frac{8}{3})^{2} = \frac{368}{45} - \frac{320}{45} = \frac{48}{45} = \frac{16}{15} .$

d) The median $m$ satisfies $\int_{0}^{m} f = \frac{m ^{3}}{96} + \frac{m}{12} = \frac{1}{2}$ , that is, $m^{3} + 8 m - 48 = 0$ . Using the GDC: $m \approx 2.91$ .

e) $Y \sim B (8; 0.75)$ , since $P (X \geq 2) = 0.75$ . Then $E (Y) = 8 \cdot 0.75 = 6$ .

f) Using the GDC (binomial cumulative): $P (Y \geq 3) = 1 - P (Y \leq 2) \approx 0.996$ .

Answer: b) $\frac{8}{3}$ ; c) $\frac{16}{15}$ ; d) $2.91$ ; e) $6$ ; f) $0.996$

Exercise 42 HardIB · P2 TZ2 May 2016

A continuous random variable $T$ has the following probability density function $f$ :

$f (t) = ⎩ ⎨ ⎧ \frac{t ∣ sin 2 t ∣}{π} 0 if 0 \leq t \leq π otherwise$

a) Sketch an approximate graph of $y = f (t)$ .

b) Use this approximate graph to find the mode of $T$ .

c) Find the mean of $T$ .

d) Find the variance of $T$ .

e) Find the probability that the value of $T$ lies between the mean and the mode.

f) (i) Find $\int_{0}^{T} f (t) d t$ , where $0 \leq T \leq \frac{π}{2}$ .

(ii) Hence, verify that the first quartile of $T$ is $\frac{π}{2}$ .

Show answer

Answer: b) mode $\approx 2.46$ ; c) $E (T) \approx 2.04$ ; d) $Var (T) \approx 0.516$ ; e) $\approx 0.285$ ; f) i) $- \frac{T cos 2 T}{2 π} + \frac{sin 2 T}{4 π}$ ; ii) $Q_{1} = \frac{π}{2}$

Show working and answer

a) Graph of y = f(t) = t·|sin 2t|/π for 0 ≤ t ≤ π: the curve starts at the origin and forms two humps separated by a zero at t = π/2 (a small one, with height about 0.29 near t ≈ 1.01, and a larger one, with height 0.77 at t ≈ 2.46), returning to zero at t = π; the point (2.46, 0.77) is marked. The t-axis has marks at 0.5, 1, π/2, 2, 2.5 and π, and the y-axis at 0.5 and 1

The function is $f (t) = \frac{t ∣ sin 2 t ∣}{π}$ : it is zero at $t = 0$ , $\frac{π}{2}$ and $π$ , and the negative part of the sine is reflected upwards by the absolute value.

b) The mode is the abscissa of the highest peak of the graph, which lies in the second hump: $t \approx 2.46$ .

c) Using the GDC (numerical integration): $E (T) = \int_{0}^{π} t \cdot \frac{t ∣ sin 2 t ∣}{π} d t \approx 2.04.$

d) $E (T^{2}) = \int_{0}^{π} t^{2} f (t) d t \approx 4.67$ , so $Var (T) = E (T^{2}) - [E (T)]^{2} \approx 4.67 - 2.0 4^{2} \approx 0.516$ .

e) $P (2.04 < T < 2.46) = \int_{2.04}^{2.46} f (t) d t \approx 0.285$ (calculated with the unrounded values).

f) i) On $[0; \frac{π}{2}]$ , $∣ sin 2 t ∣ = sin 2 t$ . Integrating by parts: $\int_{0}^{T} \frac{t sin 2 t}{π} d t = \frac{1}{π} [- \frac{t cos 2 t}{2} + \frac{sin 2 t}{4}]_{0}^{T} = - \frac{T cos 2 T}{2 π} + \frac{sin 2 T}{4 π} .$

(ii) For $T = \frac{π}{2}$ : $cos π = - 1$ and $sin π = 0$ , so the integral equals $\frac{π /2}{2 π} = \frac{1}{4}$ . As $P (T \leq \frac{π}{2}) = \frac{1}{4}$ , the first quartile is $\frac{π}{2}$ .

Answer: b) mode $\approx 2.46$ ; c) $E (T) \approx 2.04$ ; d) $Var (T) \approx 0.516$ ; e) $\approx 0.285$ ; f) i) $- \frac{T cos 2 T}{2 π} + \frac{sin 2 T}{4 π}$ ; ii) $Q_{1} = \frac{π}{2}$

Exercise 43 HardIB · P2 May 2003

A businessman spends $X$ hours per day talking on the telephone. The probability density function of $X$ is given by

$f (x) = ⎩ ⎨ ⎧ \frac{1}{12} (8 x - x^{3}), 0, for 0 \leq x \leq 2 otherwise$

a) i) Write down an integral whose value is $E (X)$ .

(ii) Hence calculate the value of $E (X)$ .

b) i) Show that the median, $m$ , of $X$ satisfies the equation $m^{4} - 16 m^{2} + 24 = 0$ .

(ii) Hence calculate the value of $m$ .

c) Calculate the mode of $X$ .

Show answer

Answer: a) $E (X) = \frac{56}{45} \approx 1.24$ ; b) $m \approx 1.29$ ; c) mode $\approx 1.63$

Show working and answer

a) i) $E (X) = \int_{0}^{2} x \cdot \frac{1}{12} (8 x - x^{3}) d x$ .

(ii) $E (X) = \frac{1}{12} \int_{0}^{2} (8 x^{2} - x^{4}) d x = \frac{1}{12} [\frac{8 x ^{3}}{3} - \frac{x ^{5}}{5}]_{0}^{2} = \frac{1}{12} (\frac{64}{3} - \frac{32}{5}) = \frac{56}{45} \approx 1.24.$

b) i) The median satisfies $\int_{0}^{m} \frac{1}{12} (8 x - x^{3}) d x = \frac{1}{2}$ : $\frac{1}{12} [4 x^{2} - \frac{x ^{4}}{4}]_{0}^{m} = \frac{4 m ^{2}}{12} - \frac{m ^{4}}{48} = \frac{1}{2} .$ Multiplying by 48: $16 m^{2} - m^{4} = 24$ , that is, $m^{4} - 16 m^{2} + 24 = 0$ .

(ii) $m^{2} = 8 \pm 40$ ; since $m \leq 2$ , $m^{2} = 8 - 210$ , and $m \approx 1.29$ .

c) The mode is the maximum of $f$ : $f^{'} (x) = \frac{1}{12} (8 - 3 x^{2}) = 0 \Rightarrow x = 8/3 = \frac{2 6}{3} \approx 1.63$ . It is a maximum because $f^{'}$ changes from positive to negative.

Answer: a) $E (X) = \frac{56}{45} \approx 1.24$ ; b) $m \approx 1.29$ ; c) mode $\approx 1.63$

Exercise 44 HardIB · P2 TZ0 November 2007

The probability density function of the continuous random variable $X$ is given by

$f (x) = ⎩ ⎨ ⎧ \frac{c}{4 + x ^{2}}, 0, for - \frac{2}{3} \leq x \leq 23 otherwise$

a) Find the exact value of the constant $c$ in terms of $π$ .

b) Sketch an approximate graph of $f (x)$ and, hence, state the mode of the distribution.

c) Find the exact value of $E (X)$ .

Show answer

Answer: a) $c = \frac{4}{π}$ ; b) mode $= 0$ ; c) $E (X) = \frac{2}{π} ln 3$

Show working and answer

a) The integral of $f$ must equal $1$ . Since $\int \frac{d x}{4 + x ^{2}} = \frac{1}{2} arctan \frac{x}{2}$ : $c \cdot \frac{1}{2} [arctan \frac{x}{2}]_{- 2/ 3}^{23} = \frac{c}{2} (arctan 3 - arctan (- \frac{1}{3})) = \frac{c}{2} (\frac{π}{3} + \frac{π}{6}) = \frac{c π}{4} = 1.$ Hence $c = \frac{4}{π}$ .

b) $f (x) = \frac{4}{π ( 4 + x ^{2} )}$ is an even function with a maximum at $x = 0$ , where $f (0) = \frac{1}{π} \approx 0.32$ ; it decreases on both sides. At the endpoints: $f (- \frac{2}{3}) = \frac{3}{4 π} \approx 0.24$ and $f (23) = \frac{1}{4 π} \approx 0.08$ . The mode is $x = 0$ .

Graph of f(x) = 4/(π(4 + x²)) on the interval −2/√3 ≤ x ≤ 2√3 (from approximately −1.15 to 3.46), outside which the graph is zero: smooth low bell-shaped curve with maximum at x = 0, where f(0) = 1/π ≈ 0.32; it decreases on both sides, slightly more slowly on the left (f ≈ 0.24 at the left endpoint and f ≈ 0.08 at the right endpoint)

c) $E (X) = \frac{4}{π} \int_{- 2/ 3}^{23} \frac{x}{4 + x ^{2}} d x = \frac{4}{π} \cdot \frac{1}{2} [ln (4 + x^{2})]_{- 2/ 3}^{23} = \frac{2}{π} (ln 16 - ln \frac{16}{3}) = \frac{2}{π} ln 3 \approx 0.699.$

Answer: a) $c = \frac{4}{π}$ ; b) mode $= 0$ ; c) $E (X) = \frac{2}{π} ln 3$

Exercise 45 HardIB · P2 TZ1 May 2011

A continuous random variable $X$ has a probability density function given by the function $f (x)$ , where

$f (x) = ⎩ ⎨ ⎧ k (x + 2)^{2}, k, 0, - 2 \leq x < 0 0 \leq x \leq \frac{4}{3} otherwise$

a) Find the value of $k$ .

b) Hence, find

(i) the mean of $X$ ;

(ii) the median of $X$ .

Show answer

Answer: a) $k = \frac{1}{4}$ ; b) i) $E (X) = - \frac{1}{9}$ ; ii) median $= 36 - 2 \approx - 0.183$

Show working and answer

a) The total integral must equal $1$ : $\int_{- 2}^{0} k (x + 2)^{2} d x + \int_{0}^{4/3} k d x = k \cdot \frac{8}{3} + k \cdot \frac{4}{3} = 4 k = 1 \Rightarrow k = \frac{1}{4} .$

b) i) $E (X) = \int_{- 2}^{0} \frac{x ( x + 2 ) ^{2}}{4} d x + \int_{0}^{4/3} \frac{x}{4} d x = - \frac{1}{3} + \frac{2}{9} = - \frac{1}{9} .$

(ii) The area of the first part is $\int_{- 2}^{0} \frac{( x + 2 ) ^{2}}{4} d x = \frac{8}{12} = \frac{2}{3} > \frac{1}{2}$ , so the median $m$ lies in $[- 2, 0]$ : $\int_{- 2}^{m} \frac{( x + 2 ) ^{2}}{4} d x = \frac{( m + 2 ) ^{3}}{12} = \frac{1}{2} \Rightarrow (m + 2)^{3} = 6 \Rightarrow m = 36 - 2 \approx - 0.183.$

Answer: a) $k = \frac{1}{4}$ ; b) i) $E (X) = - \frac{1}{9}$ ; ii) median $= 36 - 2 \approx - 0.183$

Exercise 46 HardIB · P2 TZ1 May 2012

The probability density function of the continuous random variable $X$ is given by

$f (x) = ⎩ ⎨ ⎧ \frac{1}{k 2 ^{x}}, 0, 1 \leq x \leq 2 otherwise$

where $k$ is a constant. Find the expected value of $X$ .

Show answer

Answer: $E (X) = \frac{1}{ln 2} \approx 1.44$

Show working and answer

Constant $k$ : since $\int_{1}^{2} 2^{- x} d x = [- \frac{2 ^{- x}}{ln 2}]_{1}^{2} = \frac{\frac{1}{2} - \frac{1}{4}}{ln 2} = \frac{1}{4 ln 2}$ , the condition $\frac{1}{k} \cdot \frac{1}{4 ln 2} = 1$ gives $k = \frac{1}{4 ln 2}$ . Then $f (x) = 4 ln 2 \cdot 2^{- x}$ .
Expected value, by parts: $E (X) = 4 ln 2 \int_{1}^{2} x 2^{- x} d x = 4 ln 2 [- \frac{x 2 ^{- x}}{ln 2} - \frac{2 ^{- x}}{( ln 2 ) ^{2}}]_{1}^{2} .$ Evaluating: $[\cdot]_{1}^{2} = (- \frac{1}{2 ln 2} - \frac{1}{4 ( ln 2 ) ^{2}}) - (- \frac{1}{2 ln 2} - \frac{1}{2 ( ln 2 ) ^{2}}) = \frac{1}{4 ( ln 2 ) ^{2}}$ .
Therefore $E (X) = 4 ln 2 \cdot \frac{1}{4 ( ln 2 ) ^{2}} = \frac{1}{ln 2} \approx 1.44$ .

Answer: $E (X) = \frac{1}{ln 2} \approx 1.44$

Exercise 47 HardIB · P2 May TZ2 2021Proof

Let $X$ be a continuous random variable and let $f_{n}$ be its probability density function, which is given by

$f_{n} (x) = {(n + 1) x^{n}, 0, 0 \leq x \leq 1 otherwise where n \in R, n \geq 0.$

a) Show that $E (X) = \frac{n + 1}{n + 2}$ .

b) Show that $Var (X) = \frac{n + 1}{( n + 2 ) ^{2} ( n + 3 )}$ .

Show answer

Answer: it is shown that $E (X) = \frac{n + 1}{n + 2}$ and $Var (X) = \frac{n + 1}{( n + 2 ) ^{2} ( n + 3 )}$ .

Show working and answer

a) We calculate the expected value: $E (X) = \int_{0}^{1} x (n + 1) x^{n} d x = (n + 1) \int_{0}^{1} x^{n + 1} d x = (n + 1) [\frac{x ^{n + 2}}{n + 2}]_{0}^{1} = \frac{n + 1}{n + 2} .$

b) First, $E (X^{2})$ : $E (X^{2}) = \int_{0}^{1} x^{2} (n + 1) x^{n} d x = (n + 1) [\frac{x ^{n + 3}}{n + 3}]_{0}^{1} = \frac{n + 1}{n + 3} .$ Then $Var (X) = E (X^{2}) - [E (X)]^{2} = \frac{n + 1}{n + 3} - \frac{( n + 1 ) ^{2}}{( n + 2 ) ^{2}} = \frac{( n + 1 ) [ ( n + 2 ) ^{2} - ( n + 1 ) ( n + 3 ) ]}{( n + 2 ) ^{2} ( n + 3 )} .$ Since $(n + 2)^{2} - (n + 1) (n + 3) = n^{2} + 4 n + 4 - (n^{2} + 4 n + 3) = 1$ , this gives $Var (X) = \frac{n + 1}{( n + 2 ) ^{2} ( n + 3 )} .$

Answer: it is shown that $E (X) = \frac{n + 1}{n + 2}$ and $Var (X) = \frac{n + 1}{( n + 2 ) ^{2} ( n + 3 )}$ .

Exercise 48 HardIB · P2 TZ0 November 2021

The continuous random variable $X$ has a probability density function given by

$f (x) = {arccos x, 0, 0 \leq x \leq 1 otherwise$

The median of this distribution is $m$ .

a) Determine the value of $m$ .

b) Given that $P (∣ X - m ∣ \leq a) = 0.3$ , calculate the value of $a$ .

Show answer

Answer: a) $m = 0.360$ ; b) $a = 0.125$

Show working and answer

a) The distribution function, found by parts, is $F (x) = \int_{0}^{x} arccos t d t = x arccos x - 1 - x^{2} + 1$ . The median $m$ satisfies $F (m) = \frac{1}{2}$ : $m arccos m - 1 - m^{2} + 1 = \frac{1}{2} .$ Using the GDC (solve the equation): $m = 0.360$ .

b) The condition $P (∣ X - m ∣ \leq a) = 0.3$ is equivalent to $F (m + a) - F (m - a) = 0.3$ . Using the GDC (solve the equation, with $m = 0.3600$ ): $a = 0.125.$

Answer: a) $m = 0.360$ ; b) $a = 0.125$

Exercise 49 HardIB · P1 TZ2 May 2022

A continuous random variable $X$ has the following probability density function:

$f (x) = ⎩ ⎨ ⎧ \frac{2}{( b - a ) ( c - a )} (x - a), \frac{2}{( b - a ) ( b - c )} (b - x), 0, a \leq x \leq c c < x \leq b otherwise$

The following figure shows the graph of $y = f (x)$ for $a \leq x \leq b$ .

Graph of y = f(x) for a ≤ x ≤ b: a triangle resting on the x-axis between a and b, with its vertex above a point c to the right of the midpoint (roughly four-fifths of the way from a to b); the line rises from (a, 0) to the vertex and falls to (b, 0). The vertex carries no label

Given that $c \geq \frac{a + b}{2}$ , find an expression for the median of $X$ in terms of $a$ , $b$ and $c$ .

Show answer

Answer: median $= a + \frac{( b - a ) ( c - a )}{2}$

Show working and answer

The figure shows a triangle with its vertex at $x = c$ , to the right of the midpoint of $[a, b]$ .

Area of the left-hand part (up to $c$ ): $F (c) = \int_{a}^{c} \frac{2 ( x - a )}{( b - a ) ( c - a )} d x = \frac{( c - a ) ^{2}}{( b - a ) ( c - a )} = \frac{c - a}{b - a} .$
Since $c \geq \frac{a + b}{2}$ , we have $c - a \geq \frac{b - a}{2}$ , that is $F (c) \geq \frac{1}{2}$ : the median $m$ lies in the first part, $a \leq m \leq c$ .
We impose $F (m) = \frac{1}{2}$ : $\frac{( m - a ) ^{2}}{( b - a ) ( c - a )} = \frac{1}{2} \Rightarrow (m - a)^{2} = \frac{( b - a ) ( c - a )}{2} \Rightarrow m = a + \frac{( b - a ) ( c - a )}{2} .$

Answer: median $= a + \frac{( b - a ) ( c - a )}{2}$

Normal distribution: probabilities

Exercise 50 BasicIB · P1 May 1998

The average test score in a mathematics class was 60, with a standard deviation of 10. Assuming that the test scores follow a normal distribution, find the proportion of students who score more than 80 on the given test.

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Answer: $P (X > 80) \approx 0.0228$ (a $2.28 %$ of the students)

Show working and answer

Let $X \sim N (60; 1 0^{2})$ . Standardising, $z = \frac{80 - 60}{10} = 2$ , so $P (X > 80) = P (Z > 2) = 1 - 0.97725 \approx 0.0228.$ Using the GDC (normal cdf, limits 80 and a large value) the same value is obtained.

Answer: $P (X > 80) \approx 0.0228$ (a $2.28 %$ of the students)

Exercise 51 BasicIB · P1 November 1999

The random variable $X$ has a normal distribution with mean 30 and standard deviation 2. Calculate $P (27 \leq X \leq 34)$ .

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Answer: $P (27 \leq X \leq 34) \approx 0.910$

Show working and answer

Using the GDC (normal cdf, $μ = 30$ , $σ = 2$ ): $P (27 \leq X \leq 34) = P (- 1.5 \leq Z \leq 2) \approx 0.910$ .

Answer: $P (27 \leq X \leq 34) \approx 0.910$

Exercise 52 BasicIB · P1 May 2002

The weights of a certain species of bird are normally distributed, with a mean of 0.8 kg and a standard deviation of 0.12 kg. Find the probability that the weight of a bird of the species, chosen at random, is between 0.74 kg and 0.95 kg.

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Answer: $P (0.74 \leq X \leq 0.95) \approx 0.586$

Show working and answer

Using the GDC (normal cdf, $μ = 0.8$ , $σ = 0.12$ ): $P (0.74 \leq X \leq 0.95) = P (- 0.5 \leq Z \leq 1.25) \approx 0.586$ .

Answer: $P (0.74 \leq X \leq 0.95) \approx 0.586$

Exercise 53 BasicIB · P1 TZ2 May 2004

The following figure shows the probability density function for the random variable $X$ , which is normally distributed with mean 250 and standard deviation 50.

Bell curve of the normal distribution with mean 250 and standard deviation 50, with horizontal axis x and vertical axis f(x) without numerical scale; only 180, 250 and 280 are marked on the horizontal axis, and the area under the curve between x = 180 and x = 280 is shaded with diagonal lines

Find the probability represented by the shaded region.

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Answer: $P (180 \leq X \leq 280) \approx 0.645$

Show working and answer

The shaded region is $P (180 \leq X \leq 280)$ with $X \sim N (250; 5 0^{2})$ . Standardising: $z = \frac{180 - 250}{50} = - 1.4$ and $z = \frac{280 - 250}{50} = 0.6$ . Using the GDC (normal cdf): $P (180 \leq X \leq 280) = P (- 1.4 \leq Z \leq 0.6) \approx 0.645.$

Answer: $P (180 \leq X \leq 280) \approx 0.645$

Exercise 54 MediumIB · P1 November 1996

Smith & Co. produce jam jars in such a way that the net weight of a jam jar is normally distributed with a mean of 450 g and a standard deviation of 5 g.

What percentage of jars have a net weight of less than 422 g?

Show answer

Answer: $P (X < 422) \approx 1.07 \cdot 1 0^{- 8}$ , or $1.07 \cdot 1 0^{- 6} %$ of the jars.

Show working and answer

Let $X \sim N (450, 5^{2})$ . Standardise: $z = \frac{422 - 450}{5} = - 5.6$ .

Using the GDC (cumulative normal): $P (X < 422) = P (Z < - 5.6) \approx 1.07 \cdot 1 0^{- 8} .$ As a percentage: $1.07 \cdot 1 0^{- 8} \cdot 100 = 1.07 \cdot 1 0^{- 6} %$ , that is, practically no jars.

Answer: $P (X < 422) \approx 1.07 \cdot 1 0^{- 8}$ , or $1.07 \cdot 1 0^{- 6} %$ of the jars.

Exercise 55 MediumIB · P2 TZ2 May 2013

It is believed that the lifetimes of Manx cats follow a normal distribution, with a mean of 13.5 years and a variance of $9.5 years^{2}$ .

a) For these Manx cats, calculate the range of lifetimes that lie less than one standard deviation from the mean.

b) In a population of 10,000 Manx cats, estimate how many will have a lifetime of less than 10 years. Give your answer to the nearest whole number.

Show answer

Answer: a) between $10.42$ and $16.58$ years; b) $1281$ cats

Show working and answer

The standard deviation is $σ = 9.5 = 3.082$ years.

a) Less than one standard deviation from the mean: $13.5 \pm 3.082$ , that is, from $10.42$ to $16.58$ years.

b) Using the GDC (cumulative normal): $P (X < 10) = P (Z < \frac{10 - 13.5}{3.082}) = P (Z < - 1.136) = 0.1281$ . The expected number is $10000 \cdot 0.1281 = 1281$ .

Answer: a) between $10.42$ and $16.58$ years; b) $1281$ cats

Exercise 56 MediumIB · P2 TZ0 November 2009

Tim goes to a well-known restaurant where tables cannot be reserved. The waiting times until a table is obtained have been found to follow a normal distribution, with a mean of 18 minutes and a standard deviation of 4 minutes.

a) Tim says he will leave if, 25 minutes after arriving at the restaurant, he still has not got a table. Find the probability that Tim leaves the restaurant without having got a table.

b) Tim has been waiting for 15 minutes. Find the probability that Tim gets a table within the next five minutes.

Show answer

Answer: a) $0.0401$ ; b) $0.601$

Show working and answer

Let $T \sim N (18, 4^{2})$ be the waiting time in minutes.

a) Using the GDC (cumulative normal): $P (T > 25) = P (Z > 1.75) = 0.0401$ .

b) Tim has already waited 15 minutes, so we want $P (T < 20 ∣ T > 15)$ (he gets a table between minutes 15 and 20): $P (15 < T < 20 ∣ T > 15) = \frac{P ( 15 < T < 20 )}{P ( T > 15 )} = \frac{0.6915 - 0.2266}{0.7734} = \frac{0.4649}{0.7734} \approx 0.601.$

Answer: a) $0.0401$ ; b) $0.601$

Exercise 57 MediumIB · P2 TZ0 November 2014

The wingspans of birds of a certain species can be modelled by a normal distribution with mean 60.2 cm and standard deviation 2.4 cm. According to this model, 99% of the birds have a wingspan greater than $x$ cm.

a) Find the value of $x$ .

In a field experiment, a research team studies a large sample of these birds. They measure the wingspan of each bird, rounded to the nearest 0.1 cm.

b) Find the probability that a bird chosen at random has a measured wingspan of 60.2 cm.

Show answer

Answer: a) $x \approx 54.6$ cm; b) $0.0166$

Show working and answer

Let $X \sim N (60.2; 2. 4^{2})$ .

a) $P (X > x) = 0.99$ is equivalent to $P (X < x) = 0.01$ . Using the GDC (inverse normal): $x = 54.6$ cm.

b) A measured wingspan of $60.2$ cm corresponds to an actual wingspan between $60.15$ and $60.25$ cm. Using the GDC (normal cdf): $P (60.15 < X < 60.25) = 0.0166.$

Answer: a) $x \approx 54.6$ cm; b) $0.0166$

Exercise 58 MediumIB · P2 TZ1 May 2018

The age, $L$ , in years, of a wolf can be modelled by the normal distribution $L \sim N (8, 5)$ , that is, with mean 8 and variance 5.

a) Find the probability that a randomly selected wolf is at least 5 years old.

Eight wolves are selected at random and their ages recorded.

b) Find the probability that more than six of these wolves are at least 5 years old.

Show answer

Answer: a) $0.910$ ; b) $0.843$

Show working and answer

Here $L \sim N (8, 5)$ : the variance is $5$ , so $σ = 5$ .

a) Using the GDC (normal cdf): $P (L \geq 5) = P (Z \geq \frac{5 - 8}{5}) = P (Z \geq - 1.342) = 0.910$ .

b) Let $Y$ be the number of wolves, out of 8, that are at least 5 years old: $Y \sim B (8; 0.9101)$ . "More than six" means $Y \geq 7$ . Using the GDC (binomial): $P (Y \geq 7) = P (Y = 7) + P (Y = 8) = 0.843.$

Answer: a) $0.910$ ; b) $0.843$

Exercise 59 MediumIB · P1 TZ1 May 2019

Let $X$ be a random variable that follows a normal distribution with mean $μ$ . Given that $P (X < μ - 5) = 0.2$ , calculate:

a) $P (X > μ + 5)$ ;

b) $P (X < μ + 5 ∣ X > μ - 5)$ .

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Answer: a) $0.2$ ; b) $\frac{3}{4}$

Show working and answer

a) The normal distribution is symmetric about $μ$ , so $P (X > μ + 5) = P (X < μ - 5) = 0.2$ .

b) $P (X < μ + 5 ∣ X > μ - 5) = \frac{P ( μ - 5 < X < μ + 5 )}{P ( X > μ - 5 )} .$ Numerator: $1 - 0.2 - 0.2 = 0.6$ . Denominator: $1 - 0.2 = 0.8$ . Therefore $\frac{0.6}{0.8} = \frac{3}{4} .$

Answer: a) $0.2$ ; b) $\frac{3}{4}$

Exercise 60 MediumIB · P2 TZ2 May 2018

The random variable $X$ follows a normal distribution with mean $μ = 50$ and variance $σ^{2} = 16$ .

a) Sketch an approximate graph of the probability density function of $X$ and shade the region that represents $P (μ - 2 σ < X < μ + σ)$ .

b) Find the value of $P (μ - 2 σ < X < μ + σ)$ .

c) Find the value of $k$ for which $P (μ - k σ < X < μ + k σ) = 0.5$ .

Show answer

Answer: b) $0.819$ ; c) $k \approx 0.674$

Show working and answer

a) Normal density curve with mean 50 and standard deviation 4, bell-shaped and symmetric about x = 50; the x-axis has marks at 42 (μ−2σ), 50 (μ) and 54 (μ+σ), and the vertical axis f(x) has no numerical scale. The area under the curve between x = 42 and x = 54 is shaded; the rest of the curve is not shaded

With $σ = 16 = 4$ : $μ - 2 σ = 42$ and $μ + σ = 54$ .

b) Using the GDC (normal cdf, limits 42 and 54, $μ = 50$ , $σ = 4$ ): $P (42 < X < 54) = P (- 2 < Z < 1) \approx 0.819.$

c) Standardising, $P (- k < Z < k) = 0.5$ , so $P (Z < k) = 0.75$ . Using the GDC (inverse normal, area $0.75$ ): $k \approx 0.674$ .

Answer: b) $0.819$ ; c) $k \approx 0.674$

Exercise 61 MediumIB · P2 TZ2 May 2016

A random variable $X$ follows a normal distribution with mean 3 and variance $2^{2}$ .

a) Find $P (0 \leq X \leq 2)$ .

b) Find $P (∣ X ∣ > 1)$ .

c) Given that $P (X > c) = 0.44$ , find the value of $c$ .

Show answer

Answer: a) $0.242$ ; b) $0.864$ ; c) $c \approx 3.30$

Show working and answer

Let $X \sim N (3; 2^{2})$ .

a) Using the GDC (normal cdf): $P (0 \leq X \leq 2) \approx 0.242$ .

b) $P (∣ X ∣ > 1) = 1 - P (- 1 \leq X \leq 1)$ . Using the GDC, $P (- 1 \leq X \leq 1) \approx 0.136$ , so $P (∣ X ∣ > 1) \approx 1 - 0.136 = 0.864.$

c) $P (X > c) = 0.44$ is equivalent to $P (X < c) = 0.56$ . Using the GDC (inverse normal, area $0.56$ , $μ = 3$ , $σ = 2$ ): $c \approx 3.30$ .

Answer: a) $0.242$ ; b) $0.864$ ; c) $c \approx 3.30$

Exercise 62 MediumIB · P1 TZ1 May 2005

Let $X$ be a normal random variable with mean 25 and variance 4. Find $P (∣ X - 25∣ < 3)$ .

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Answer: $P (∣ X - 25∣ < 3) \approx 0.866$

Show working and answer

Let $X \sim N (25; 4)$ , with $σ = 2$ . $P (∣ X - 25∣ < 3) = P (22 < X < 28) = P (- 1.5 < Z < 1.5)$ . Using the GDC (normal cdf): $P \approx 0.866.$

Answer: $P (∣ X - 25∣ < 3) \approx 0.866$

Exercise 63 MediumIB · P1 TZ0 November 2006

The weight of a certain type of vegetable follows a normal distribution with a mean of 450 grams and a standard deviation of 50 grams.

a) In a consignment containing 2,000 of these vegetables, calculate the expected number of vegetables that weigh more than 525 grams.

b) Find the value of the third quartile of the distribution.

Show answer

Answer: a) $134$ vegetables; b) $Q_{3} \approx 484$ g

Show working and answer

Let $X$ be the weight, $X \sim N (450, 5 0^{2})$ .

a) Using the GDC (normal cdf): $P (X > 525) = 0.0668$ . The expected number among 2000 vegetables is $2000 \cdot 0.0668 = 133.6 \approx 134.$

b) The third quartile satisfies $P (X < Q_{3}) = 0.75$ . Using the GDC (inverse normal): $Q_{3} = 483.7$ .

Answer: a) $134$ vegetables; b) $Q_{3} \approx 484$ g

Exercise 64 MediumIB · P1 TZ2 May 2007

The length of a particular species of lizard follows a normal distribution, with a mean length of 50 cm and a standard deviation of 4 cm. A lizard is chosen at random.

a) Find the probability that its length is greater than 45 cm.

b) Given that its length is greater than 45 cm, find the probability that its length is greater than 55 cm.

Show answer

Answer: a) $0.894$ ; b) $0.118$

Show working and answer

Let $L$ be the length, $L \sim N (50, 4^{2})$ .

a) Using the GDC (normal cdf): $P (L > 45) = 0.894$ .

b) Conditional probability. Since ${L > 55} \subset {L > 45}$ : $P (L > 55 ∣ L > 45) = \frac{P ( L > 55 )}{P ( L > 45 )} = \frac{0.1056}{0.8944} = 0.118.$

Answer: a) $0.894$ ; b) $0.118$

Exercise 65 HardIB · P2 TZ2 May 2011

The weights of the fish in a lake follow a normal distribution with a mean of 1.3 kg and a standard deviation of 0.2 kg.

a) Determine the probability that a fish we have just caught in the lake weighs less than 1.4 kg.

b) John catches 6 fish. Calculate the probability that at least 4 of these fish weigh more than 1.4 kg.

c) Determine the probability that a fish we have just caught in the lake weighs less than 1 kg, given that it weighs less than 1.4 kg.

Show answer

Answer: a) $0.691$ ; b) $0.077$ ; c) $0.097$

Show working and answer

Let $X \sim N (1.3; 0. 2^{2})$ .

a) Using the GDC (cumulative normal): $P (X < 1.4) = 0.6915 \approx 0.691$ .

b) The probability that a fish weighs more than $1.4$ kg is $p = 1 - 0.6915 = 0.3085$ . Let $Y$ be the number of fish, out of the 6, that weigh more than $1.4$ kg: $Y \sim B (6; 0.3085)$ . Using the GDC: $P (Y \geq 4) = 1 - P (Y \leq 3) = 0.0775 \approx 0.077.$

c) $P (X < 1 ∣ X < 1.4) = \frac{P ( X < 1 )}{P ( X < 1.4 )} = \frac{0.0668}{0.6915} \approx 0.0966 \approx 0.097.$

Answer: a) $0.691$ ; b) $0.077$ ; c) $0.097$

Exercise 66 HardIB · P2 TZ1 May 2017

The times taken by male runners to complete a marathon can be modelled by a normal distribution with mean 196 minutes and standard deviation 24 minutes.

a) Find the probability that a randomly selected runner completes the marathon in less than 3 hours.

It is known that 5% of male runners complete the marathon in less than $T_{1}$ minutes.

b) Calculate $T_{1}$ .

The times taken by female runners to complete the marathon can be modelled by a normal distribution with mean 210 minutes. It is found that 58% of female runners complete the marathon in between 185 and 235 minutes.

c) Find the standard deviation of the times of the female runners.

Show answer

Answer: a) $0.252$ ; b) $T_{1} \approx 156$ min ( $156.5$ ); c) $σ \approx 31.0$ min

Show working and answer

Let $X \sim N (196; 2 4^{2})$ (minutes).

a) Three hours is $180$ minutes. Using the GDC (normal cdf): $P (X < 180) \approx 0.252$ .

b) We look for $T_{1}$ with $P (X < T_{1}) = 0.05$ . Using the GDC (inverse normal, area $0.05$ ): $T_{1} \approx 156$ minutes (more exactly $156.5$ ).

c) Let $Y \sim N (210, σ^{2})$ . The values 185 and 235 are $25$ from the mean, so $P (185 < Y < 235) = P (∣ Y - 210∣ < 25) = 0.58$ . Then $P (Y < 235) = 0.5 + 0.29 = 0.79$ , and $\frac{25}{σ} = z_{0.79} \approx 0.8064 \Rightarrow σ \approx 31.0 minutes .$

Answer: a) $0.252$ ; b) $T_{1} \approx 156$ min ( $156.5$ ); c) $σ \approx 31.0$ min

Inverse normal and standardisation: finding values, μ and σ

Exercise 67 BasicIB · P1 May 2001

$Z$ is a normal random variable with mean 0 and variance 1. Calculate the value of $a$ for which $P (∣ Z ∣ \leq a) = 0.75$ .

Show answer

Answer: $a \approx 1.15$

Show working and answer

$P (∣ Z ∣ \leq a) = P (- a \leq Z \leq a) = 2Φ (a) - 1 = 0.75$ , so $Φ (a) = 0.875$ . Using the GDC (inverse normal, area $0.875$ ): $a \approx 1.15$ .

Answer: $a \approx 1.15$

Exercise 68 BasicIB · P1 TZ0 May 2006

The weights, in grams, of the loaves of bread sold by a supermarket are normally distributed with a mean of 200 g. The weights of 88% of the loaves are less than 220 g. Find the standard deviation.

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Answer: $σ \approx 17.0$ g

Show working and answer

Let $X \sim N (200; σ^{2})$ . $P (X < 220) = 0.88$ . Using the GDC (inverse normal, area $0.88$ ): $z \approx 1.175$ . So $\frac{220 - 200}{σ} = 1.175 \Rightarrow σ = \frac{20}{1.175} \approx 17.0 g .$

Answer: $σ \approx 17.0$ g

Exercise 69 BasicIB · P2 TZ1 May 2010

In a spectacles factory, it is known that the weight of the spectacles has a mean of 160 grams. It is also known that the interquartile range of the weights of the spectacles is 28 grams. Assuming that the weights of the spectacles are normally distributed, calculate the standard deviation of the weights of the spectacles.

Show answer

Answer: $σ = 20.8$ g

Show working and answer

The interquartile range of a normal distribution is $Q_{3} - Q_{1}$ . With $P (Z < z) = 0.75$ we have $z = 0.6745$ , so $Q_{3} = 160 + 0.6745 σ, Q_{1} = 160 - 0.6745 σ, Q_{3} - Q_{1} = 1.349 σ = 28.$ Therefore $σ = \frac{28}{1.349} = 20.8$ g.

Answer: $σ = 20.8$ g

Exercise 70 MediumIB · P1 May 1997

A zoologist knows that the lengths of a certain type of tropical snake are normally distributed with a mean length of $L$ metres and a standard deviation of 0.12 metres. If 20% of the snakes are longer than 0.70 metres, find the value of $L$ .

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Answer: $L \approx 0.599$ m

Show working and answer

Let $X \sim N (L, 0.1 2^{2})$ . We know that $P (X > 0.70) = 0.20$ , that is, $P (X < 0.70) = 0.80$ .

Using the GDC (inverse normal): $z = 0.8416$ leaves $80 %$ of the area to its left. Then $\frac{0.70 - L}{0.12} = 0.8416 \Rightarrow L = 0.70 - 0.12 \cdot 0.8416 = 0.599.$

Answer: $L \approx 0.599$ m

Exercise 71 MediumIB · P2 TZ2 November 2010

The weight loss, in kilograms, of people following the DELGAMAS slimming diet over a period of three months is modelled by a random variable $X$ . Experimental data showed that 67% of the people who used DELGAMAS lost up to 5 kilograms, while 12.4% lost at least seven kilograms. Assuming that $X$ follows a normal distribution, find the expected weight loss for a person who follows the DELGAMAS diet for three months.

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Answer: expected weight loss $μ \approx 3.77$ kg

Show working and answer

Let $X \sim N (μ, σ^{2})$ . Data: $P (X \leq 5) = 0.67$ and $P (X \geq 7) = 0.124$ , that is, $P (X < 7) = 0.876$ .

Using the GDC (inverse normal, $N (0, 1)$ ): $z_{0.67} = 0.4399$ and $z_{0.876} = 1.1552$ . Then $\frac{5 - μ}{σ} = 0.4399, \frac{7 - μ}{σ} = 1.1552.$ Subtracting: $\frac{2}{σ} = 0.7153 \Rightarrow σ = 2.796$ , and $μ = 5 - 0.4399 \cdot 2.796 = 3.77$ .

Answer: expected weight loss $μ \approx 3.77$ kg

Exercise 72 MediumIB · P2 TZ0 November 2013

Over a given year, the duration of direct flights from London to Singapore followed a normal distribution, with mean $μ$ and standard deviation $σ$ .

92% of the flights lasted less than 13 hours, while only 12% of the flights lasted less than 12 hours and 35 minutes.

Find $μ$ and $σ$ , giving your answers to the nearest minute.

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Answer: $μ \approx 766$ min (12 h 46 min); $σ \approx 10$ min

Show working and answer

Convert everything to minutes: $13$ h $= 780$ min and $12$ h $35$ min $= 755$ min. Data: $P (X < 780) = 0.92$ and $P (X < 755) = 0.12$ .

Using the GDC (inverse normal, $N (0, 1)$ ): $z_{0.92} = 1.4051$ and $z_{0.12} = - 1.1750$ : $\frac{780 - μ}{σ} = 1.4051, \frac{755 - μ}{σ} = - 1.1750.$ Subtracting: $\frac{25}{σ} = 2.5801 \Rightarrow σ = 9.69 \approx 10$ min. Then $μ = 780 - 1.4051 \cdot 9.69 = 766.4 \approx 766$ min.

Answer: $μ \approx 766$ min (12 h 46 min); $σ \approx 10$ min

Exercise 73 MediumIB · P2 TZ2 May 2010

The survival time of weeds in a field, after they have been sprayed with a herbicide, follows a normal distribution whose mean is 15 days.

a) If the probability of survival for more than 21 days is 0.2, find the standard deviation of the survival time. When another field is sprayed, the survival time of the weeds follows a normal distribution whose mean is 18 days.

b) If the standard deviation of the survival time does not change, find the probability of survival for more than 21 days.

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Answer: a) $σ \approx 7.13$ days; b) $0.337$

Show working and answer

Let $X \sim N (15, σ^{2})$ .

a) $P (X > 21) = 0.2$ is equivalent to $P (X < 21) = 0.8$ . Using the GDC (inverse normal): $z_{0.8} = 0.8416$ , so $\frac{21 - 15}{σ} = 0.8416 \Rightarrow σ = \frac{6}{0.8416} = 7.13 days .$

b) Now $X \sim N (18; 7.1 3^{2})$ . Using the GDC (cumulative normal): $P (X > 21) = P (Z > \frac{3}{7.129}) = P (Z > 0.4208) \approx 0.337.$

Answer: a) $σ \approx 7.13$ days; b) $0.337$

Exercise 74 MediumIB · P2 TZ1 May 2015

The finishing times in a marathon race follow a normal distribution with mean 210 minutes and standard deviation 22 minutes.

a) Find the probability that a runner finishes the race in less than three hours.

The fastest 90% of those who finish receive a certificate.

b) Find the time below which a competitor must complete the race in order to receive a certificate.

Show answer

Answer: a) $0.0863$ ; b) $238.2$ minutes

Show working and answer

Let $T \sim N (210, 2 2^{2})$ in minutes. Three hours is $180$ minutes.

a) Using the GDC (normal cdf): $P (T < 180) = P (Z < - 1.364) = 0.0863$ .

b) We look for $t$ with $P (T < t) = 0.90$ . Using the GDC (inverse normal): $t = 238.2$ minutes (about 3 h 58 min).

Answer: a) $0.0863$ ; b) $238.2$ minutes

Exercise 75 MediumIB · P2 TZ2 May 2014

The weights, in kg, of yearling bear cubs follow a normal distribution with mean $μ$ and standard deviation $σ$ .

a) Given that the weight corresponding to the third quartile is 21.3 kg and the weight corresponding to the first quartile is 17.1 kg, calculate the value of $μ$ and the value of $σ$ .

A random sample of 100 cubs is taken.

b) Find the expected number of cubs that weigh more than 22 kg.

Show answer

Answer: a) $μ = 19.2$ kg, $σ \approx 3.11$ kg; b) $\approx 18$ cubs ( $18.4$ )

Show working and answer

a) The first quartile leaves $25 %$ to its left and the third quartile leaves $75 %$ . Using the GDC (inverse normal): $z_{0.75} = 0.6745$ and $z_{0.25} = - 0.6745$ : $\frac{21.3 - μ}{σ} = 0.6745, \frac{17.1 - μ}{σ} = - 0.6745.$ Subtracting: $\frac{4.2}{σ} = 1.349 \Rightarrow σ = 3.11$ . Adding (or by symmetry): $μ = \frac{21.3 + 17.1}{2} = 19.2$ .

b) Using the GDC (normal cdf): $P (X > 22) = P (Z > \frac{22 - 19.2}{3.1135}) = P (Z > 0.8993) = 0.1842$ . The expected number is $100 \cdot 0.1842 = 18.4 \approx 18$ cubs.

Answer: a) $μ = 19.2$ kg, $σ \approx 3.11$ kg; b) $\approx 18$ cubs ( $18.4$ )

Exercise 76 MediumIB · P2 TZ0 November 2018

It is known that 56% of Infiglow batteries last less than 16 hours and that 94% last less than 17 hours. The lifetime of these batteries can be assumed to follow a normal distribution $N (μ, σ^{2})$ .

a) Find the value of $μ$ and the value of $σ$ .

b) Find the probability that an Infiglow battery chosen at random lasts at least 15 hours.

Show answer

Answer: a) $μ \approx 15.9$ ; $σ \approx 0.712$ ; b) $0.895$

Show working and answer

a) Let $X \sim N (μ, σ^{2})$ . $P (X < 16) = 0.56$ and $P (X < 17) = 0.94$ . Using the GDC (inverse normal): $z_{0.56} = 0.1510$ and $z_{0.94} = 1.5548$ : $\frac{16 - μ}{σ} = 0.1510, \frac{17 - μ}{σ} = 1.5548.$ Subtracting: $\frac{1}{σ} = 1.4038 \Rightarrow σ = 0.712$ . Then $μ = 16 - 0.1510 \cdot 0.7123 = 15.9$ .

b) Using the GDC (normal cdf) with $N (15.8925; 0.712 3^{2})$ : $P (X \geq 15) = P (Z \geq - 1.2528) = 0.895$ .

Answer: a) $μ \approx 15.9$ ; $σ \approx 0.712$ ; b) $0.895$

Exercise 77 MediumIB · P2 TZ0 November 2017

One in every five cups of coffee contains more than 120 mg of caffeine.

Three in every five cups contain more than 110 mg of caffeine.

The caffeine content of coffee is modelled by a normal distribution.

Calculate the mean and the standard deviation of the caffeine content of coffee.

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Answer: $μ \approx 112$ mg (more exactly $112.3$ mg); $σ \approx 9.13$ mg

Show working and answer

Let $X \sim N (μ, σ^{2})$ . The data are $P (X > 120) = 0.2$ and $P (X > 110) = 0.6$ , that is, $P (X < 110) = 0.4$ .

Using the GDC (inverse normal): $z_{0.8} \approx 0.8416$ and $z_{0.4} \approx - 0.2533$ . Standardising: $\frac{120 - μ}{σ} = 0.8416, \frac{110 - μ}{σ} = - 0.2533.$ Subtracting: $\frac{10}{σ} = 1.0949$ , so $σ \approx 9.13$ . Then $μ = 120 - 0.8416 \cdot 9.1327 \approx 112.3$ .

Answer: $μ \approx 112$ mg (more exactly $112.3$ mg); $σ \approx 9.13$ mg

Exercise 78 MediumIB · P2 TZ1 May 2016

The heights of students in a single year group at a school can be modelled by a normal distribution. It is known that 40% of the students measure less than 1.62 m and 25% are taller than 1.79 m. Calculate the mean and the standard deviation of the students' heights.

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Answer: $μ \approx 1.67$ m; $σ \approx 0.183$ m

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Let $X \sim N (μ, σ^{2})$ be the height. $P (X < 1.62) = 0.4$ and $P (X > 1.79) = 0.25$ , that is, $P (X < 1.79) = 0.75$ .

Using the GDC (inverse normal): $z_{0.4} \approx - 0.2533$ and $z_{0.75} \approx 0.6745$ . Then $\frac{1.62 - μ}{σ} = - 0.2533, \frac{1.79 - μ}{σ} = 0.6745.$ Subtracting: $\frac{0.17}{σ} = 0.9278$ , so $σ \approx 0.183$ . Then $μ = 1.62 + 0.2533 \cdot 0.1832 \approx 1.67$ .

Answer: $μ \approx 1.67$ m; $σ \approx 0.183$ m

Exercise 79 MediumIB · P2 TZ0 November 2016

A random variable $X$ is normally distributed with mean $μ$ and standard deviation $σ$ , such that $P (X < 30.31) = 0.1180$ and $P (X > 42.52) = 0.3060$ .

a) Find $μ$ and $σ$ .

b) Find $P (∣ X - μ ∣ < 1.2 σ)$ .

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Answer: a) $μ \approx 38.9$ ; $σ \approx 7.22$ ; b) $0.770$

Show working and answer

a) $P (X < 30.31) = 0.1180$ and $P (X < 42.52) = 1 - 0.3060 = 0.6940$ . Using the GDC (inverse normal): $z \approx - 1.1850$ and $z \approx 0.5072$ . So $\frac{30.31 - μ}{σ} = - 1.1850, \frac{42.52 - μ}{σ} = 0.5072.$ Subtracting: $\frac{12.21}{σ} = 1.6922$ , so $σ \approx 7.22$ ; and $μ = 30.31 + 1.1850 \cdot 7.215 \approx 38.9$ .

b) $P (∣ X - μ ∣ < 1.2 σ) = P (∣ Z ∣ < 1.2) = 2Φ (1.2) - 1$ . Using the GDC: $\approx 0.770$ .

Answer: a) $μ \approx 38.9$ ; $σ \approx 7.22$ ; b) $0.770$

Exercise 80 MediumIB · P1 May 1999

A factory has a machine designed to produce 1 kg bags of sugar. It has been found that the mean weight of sugar in the bags is 1.02 kg. Assuming that the weights of the bags are normally distributed, find the standard deviation if 1.7% of the bags weigh less than 1 kg. Give your answer correct to the nearest 0.1 gram.

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Answer: $σ \approx 9.4$ g

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Let $X \sim N (1.02; σ^{2})$ be the weight in kg. We have $P (X < 1) = 0.017$ . Using the GDC (inverse normal, area $0.017$ ): $z \approx - 2.120$ . So $\frac{1 - 1.02}{σ} = - 2.120 \Rightarrow σ = \frac{0.02}{2.120} \approx 0.00943 kg \approx 9.4 g .$

Answer: $σ \approx 9.4$ g

Exercise 81 MediumIB · P2 November 2003

A random variable $X$ is normally distributed with mean $μ$ and standard deviation $σ$ , such that $P (X > 50.32) = 0.119$ and $P (X < 43.56) = 0.305$ .

a) Calculate $μ$ and $σ$ .

b) Hence find $P (∣ X - μ ∣ < 5)$ .

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Answer: a) $μ \approx 45.6$ ; $σ \approx 4.00$ ; b) $0.789$

Show working and answer

a) $P (X < 50.32) = 1 - 0.119 = 0.881$ . Using the GDC (inverse normal): $z \approx 1.180$ and, for $0.305$ , $z \approx - 0.5101$ . So $\frac{50.32 - μ}{σ} = 1.180, \frac{43.56 - μ}{σ} = - 0.5101.$ Subtracting: $\frac{6.76}{σ} = 1.690$ , so $σ \approx 4.00$ , and $μ = 50.32 - 1.180 \cdot 4.00 \approx 45.6$ .

b) $P (∣ X - μ ∣ < 5) = P (40.6 < X < 50.6)$ . Using the GDC (normal cdf, $μ = 45.6$ , $σ = 4$ ): $\approx 0.789$ .

Answer: a) $μ \approx 45.6$ ; $σ \approx 4.00$ ; b) $0.789$

Exercise 82 MediumIB · P1 May 2003

The random variable $X$ has a normal distribution, and moreover

$P (X \leq 10) = 0.670 P (X \leq 12) = 0.937$

Find $E (X)$ .

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Answer: $E (X) = μ \approx 9.19$

Show working and answer

Using the GDC (inverse normal): $z_{0.670} \approx 0.4399$ and $z_{0.937} \approx 1.5301$ . So $\frac{10 - μ}{σ} = 0.4399, \frac{12 - μ}{σ} = 1.5301.$ Subtracting: $\frac{2}{σ} = 1.0902$ , so $σ \approx 1.835$ . Thus $μ = 10 - 0.4399 \cdot 1.835 \approx 9.19$ .

Answer: $E (X) = μ \approx 9.19$

Exercise 83 MediumIB · P1 TZ1 May 2004

The weights of adult male dogs of a certain breed can be assumed to be normally distributed with a mean of 25 kg and a standard deviation of 3 kg. Given that 30% of the weights lie between 25 kg and $x$ kg, where $x > 25$ , find the value of $x$ .

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Answer: $x \approx 27.5$ kg

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Let $X \sim N (25; 3^{2})$ . $P (25 < X < x) = 0.3$ and $P (X < 25) = 0.5$ , so $P (X < x) = 0.8$ . Using the GDC (inverse normal, area $0.8$ , $μ = 25$ , $σ = 3$ ): $x \approx 27.5$ .

Answer: $x \approx 27.5$ kg

Exercise 84 MediumIB · P1 TZ0 November 2005

The random variable $X$ is normally distributed with mean $μ$ and variance $σ^{2}$ . If $P (X > 6.2) = 0.9474$ and $P (X < 9.8) = 0.6368$ , calculate the value of $μ$ and the value of $σ$ .

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Answer: $μ \approx 9.16$ ; $σ \approx 1.83$

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$P (X > 6.2) = 0.9474$ is equivalent to $P (X < 6.2) = 0.0526$ . Using the GDC (inverse normal): $z \approx - 1.620$ for $0.0526$ , and $z \approx 0.3499$ for $0.6368$ . So $\frac{6.2 - μ}{σ} = - 1.620, \frac{9.8 - μ}{σ} = 0.3499.$ Subtracting: $\frac{3.6}{σ} = 1.970$ , so $σ \approx 1.83$ . And $μ = 9.8 - 0.3499 \cdot 1.827 \approx 9.16$ .

Answer: $μ \approx 9.16$ ; $σ \approx 1.83$

Exercise 85 MediumIB · P1 TZ0 November 2007

A furniture manufacturer makes tables. A table leg is considered too wide if its width is more than 10.5 cm, and too narrow if its width is less than 9.5 cm. Experience shows that 2% of the table legs manufactured are too wide and 4% of the table legs are too narrow. The widths of the table legs follow a normal distribution, with mean $μ$ cm and standard deviation $σ$ cm. Find the values of $μ$ and $σ$ .

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Answer: $μ = 9.96$ cm; $σ = 0.263$ cm

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Let $X \sim N (μ, σ^{2})$ be the width. The data are $P (X > 10.5) = 0.02$ and $P (X < 9.5) = 0.04$ .

Using the standard inverse normal: $P (Z < z) = 0.98 \Rightarrow z = 2.054$ and $P (Z < z) = 0.04 \Rightarrow z = - 1.751$ .
Standardising: $\frac{10.5 - μ}{σ} = 2.054, \frac{9.5 - μ}{σ} = - 1.751.$
Subtracting the equations $10.5 - μ = 2.054 σ$ and $9.5 - μ = - 1.751 σ$ : $1 = 3.805 σ \Rightarrow σ = 0.263.$
$μ = 10.5 - 2.054 \cdot 0.2629 = 9.96$ .

Answer: $μ = 9.96$ cm; $σ = 0.263$ cm

Exercise 86 MediumIB · P2 TZ1 May 2007

The time, $T$ minutes, that students at Amy's school spend each day sending text messages can be modelled by a normal distribution. 30% of the students spend less than 10 minutes per day, and 35% spend more than 15 minutes per day. Find the mean and the standard deviation of $T$ .

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Answer: $μ = 12.9$ min; $σ = 5.50$ min

Show working and answer

Let $T \sim N (μ, σ^{2})$ with $P (T < 10) = 0.30$ and $P (T > 15) = 0.35$ .

Inverse standard normal: $P (Z < z) = 0.30 \Rightarrow z = - 0.5244$ ; $P (Z < z) = 0.65 \Rightarrow z = 0.3853$ .
Equations: $\frac{10 - μ}{σ} = - 0.5244, \frac{15 - μ}{σ} = 0.3853.$
Subtracting: $\frac{5}{σ} = 0.9097 \Rightarrow σ = 5.50$ .
$μ = 10 + 0.5244 \cdot 5.496 = 12.9$ .

Answer: $μ = 12.9$ min; $σ = 5.50$ min

Exercise 87 MediumIB · P2 TZ1 May 2014

A student sits a national test and is told that the scores follow a normal distribution with mean 100. The student receives a score of 124 and is told that they are at the 68th percentile. Calculate the variance of the distribution.

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Answer: $σ^{2} \approx 2630$

Show working and answer

Let $X \sim N (100, σ^{2})$ . Being at the 68th percentile means $P (X < 124) = 0.68$ .

Inverse standard normal: $P (Z < z) = 0.68 \Rightarrow z = 0.4677$ .
Standardising: $\frac{124 - 100}{σ} = 0.4677 \Rightarrow σ = \frac{24}{0.4677} = 51.3$ .
Variance: $σ^{2} = 51. 3^{2} \approx 2630$ (with more figures, $2633$ ).

Answer: $σ^{2} \approx 2630$

Combined contextual problems

Exercise 88 MediumIB · P2 TZ2 May 2008

The distances travelled by students to get to class at Gauss school follow a normal distribution with a mean of 6 km and a standard deviation of 1.5 km.

a) (i) Find the probability that the distance travelled to Gauss school by a student chosen at random lies between 4.8 km and 7.5 km.

(ii) 15% of students travel less than $d$ km to get to Gauss school. Find the value of $d$ .

At Euler school, the distances travelled by students to get to class follow a normal distribution with a mean of $μ$ km and a standard deviation of $σ$ km.

b) If 10% of students travel more than 8 km and 5% of students travel less than 2 km, find the values of $μ$ and $σ$ .

Show answer

Answer: a) i) $0.629$ ; ii) $d \approx 4.45$ km; b) $μ \approx 5.37$ km, $σ \approx 2.05$ km

Show working and answer

For Gauss school, $X \sim N (6; 1. 5^{2})$ .

a) i) Using the GDC (cumulative normal): $P (4.8 < X < 7.5) = 0.8413 - 0.2119 = 0.629$ .

(ii) Using the GDC (inverse normal): the 15% percentile of $N (6; 1. 5^{2})$ is $d = 4.45$ km.

b) Let $Y \sim N (μ, σ^{2})$ . With $P (Y > 8) = 0.10$ and $P (Y < 2) = 0.05$ , the inverse normals give $z_{0.90} = 1.2816$ and $z_{0.05} = - 1.6449$ : $\frac{8 - μ}{σ} = 1.2816, \frac{2 - μ}{σ} = - 1.6449.$ Subtracting: $\frac{6}{σ} = 2.9265 \Rightarrow σ = 2.05$ . And $μ = 8 - 1.2816 \cdot 2.0503 = 5.37$ .

Answer: a) i) $0.629$ ; ii) $d \approx 4.45$ km; b) $μ \approx 5.37$ km, $σ \approx 2.05$ km

Exercise 89 MediumIB · P2 TZ2 May 2017

A machine produces packets of biscuits. The weights $X$ , in grams, of the packets of biscuits can be modelled by a normal distribution, where $X \sim N (μ, σ^{2})$ .

A packet of biscuits is considered to have an insufficient weight if it weighs less than 250 grams.

a) Given that $μ = 253$ and $σ = 1.5$ , find the probability that a randomly chosen packet of biscuits has an insufficient weight.

The manufacturer decides that the probability that a packet has an insufficient weight should be equal to 0.002. To achieve this, $μ$ is increased while $σ$ does not change.

b) Calculate the new value of $μ$ , correct to two decimal places.

The manufacturer is happy with the decision that the probability of an insufficient weight should be 0.002, but is unhappy with the way this has been achieved. Therefore, the machine is adjusted to reduce $σ$ and make $μ$ equal to 253 again.

c) Calculate the new value of $σ$ .

Show answer

Answer: a) $0.0228$ ; b) $μ \approx 254.32$ g; c) $σ \approx 1.04$ g

Show working and answer

a) Using the GDC (normal cdf, $μ = 253$ , $σ = 1.5$ ): $P (X < 250) = P (Z < - 2) \approx 0.0228$ .

b) Now $P (X < 250) = 0.002$ with $σ = 1.5$ . Using the GDC (inverse normal, area $0.002$ ): $z \approx - 2.878$ . Then $\frac{250 - μ}{1.5} = - 2.878 \Rightarrow μ = 250 + 1.5 \cdot 2.878 \approx 254.32.$

c) With $μ = 253$ : $\frac{250 - 253}{σ} = - 2.878$ , so $σ = \frac{3}{2.878} \approx 1.04$ .

Answer: a) $0.0228$ ; b) $μ \approx 254.32$ g; c) $σ \approx 1.04$ g

Exercise 90 MediumIB · P2 November 2002

a) The probability $P (A)$ that all the materials arrive on time at a building site under construction is 0.85. The probability that the building is completed on time is 0.60.

The probability that the materials arrive on time and that the building is completed on time is 0.55.

(i) Show that the events $A$ and $B$ are not independent.

(ii) All the materials arrive on time. Find the probability that the building is not completed on time.

b) A team of ten people was working on the building, including three electricians and two plumbers. The architect called a meeting with five of the team members, and chose at random the people who were to attend. Calculate the probability that exactly two electricians and one plumber are called to the meeting.

c) The number of hours per week worked by the members of the team is normally distributed, with a mean of 42 hours. 10% of the team works 48 or more hours per week. Find the probability that both plumbers have worked more than 40 hours in a given week.

Show answer

Answer: a) ii) $\frac{6}{17} \approx 0.353$ ; b) $\frac{5}{21} \approx 0.238$ ; c) $0.443$

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a) i) $P (A) \cdot P (B) = 0.85 \cdot 0.60 = 0.51 \neq = 0.55 = P (A \cap B)$ , so $A$ and $B$ are not independent.

(ii) $P (B^{'} ∣ A) = 1 - P (B ∣ A) = 1 - \frac{0.55}{0.85} = \frac{6}{17} \approx 0.353$ .

b) Five of the 10 people are chosen: 3 electricians, 2 plumbers and 5 of another type. We must choose 2 electricians, 1 plumber and 2 of the other 5: $P = \frac{( 2 3 ) ( 1 2 ) ( 2 5 )}{( 5 10 )} = \frac{3 \cdot 2 \cdot 10}{252} = \frac{5}{21} \approx 0.238.$

c) Let $H \sim N (42, σ^{2})$ with $P (H \geq 48) = 0.1$ . Using the GDC (inverse normal, area $0.9$ ): $z \approx 1.2816$ , so $σ = \frac{6}{1.2816} \approx 4.68$ . Then $P (H > 40) \approx 0.665$ . Assuming the two plumbers work independently: $P = 0.66 5^{2} \approx 0.443.$

Answer: a) ii) $\frac{6}{17} \approx 0.353$ ; b) $\frac{5}{21} \approx 0.238$ ; c) $0.443$

Exercise 91 MediumIB · P2 TZ2 May 2007

The time taken by buses to make the journey between two given cities follows a normal distribution, with a mean of 35 minutes and a standard deviation of 7 minutes.

a) Find the probability that a bus chosen at random takes less than 40 minutes to make the journey.

b) 90% of buses take less than $t$ minutes to make the journey. Find the value of $t$ .

c) For a random sample of 10 buses, the duration of the journey between the two cities is recorded. Find the probability that exactly 6 of these buses take less than 40 minutes to make the journey.

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Answer: a) $0.762$ ; b) $t = 44.0$ min; c) $0.131$

Show working and answer

Let $T$ be the time, $T \sim N (35, 7^{2})$ .

a) Using the GDC (normal cdf): $P (T < 40) = 0.762$ .

b) We seek $t$ with $P (T < t) = 0.9$ . Using the GDC (inverse normal): $t = 44.0$ minutes.

c) Each bus takes less than 40 min with probability $p = 0.7625$ (unrounded). The number of buses $Y$ out of 10 is $Y \sim B (10; 0.7625)$ : $P (Y = 6) = (6 10) p^{6} (1 - p)^{4} = 0.131.$

Answer: a) $0.762$ ; b) $t = 44.0$ min; c) $0.131$

Exercise 92 MediumIB · P2 TZ1 May 2009

Bob measured the heights of 63 students. After analysis, he conjectured that the height, $H$ , of the students could be modelled by a normal distribution with mean 166.5 cm and standard deviation 5 cm.

a) On the basis of this assumption, estimate the number of students whose height is at least 170 cm.

Later Bob noticed that the tape measure he had used to measure the heights was faulty, since it started at the 5 cm mark and not at the zero mark.

b) What are the correct values of the mean and the variance of the distribution of the heights of these students?

Show answer

Answer: a) $15$ students; b) mean $= 161.5$ cm; variance $= 25$ cm $^{2}$

Show working and answer

a) $H \sim N (166.5; 5^{2})$ . Using the GDC (normal cdf): $P (H \geq 170) = 0.242$ . Out of 63 students: $63 \cdot 0.242 = 15.2 \approx 15 students .$

b) The tape starts at the 5 cm mark, so each recorded measurement is 5 cm greater than the true height. If $H$ is the recorded height, the true height is $H - 5$ :

Correct mean: $166.5 - 5 = 161.5$ cm.
Subtracting a constant does not change the spread: the variance is still $5^{2} = 25$ cm $^{2}$ .

Answer: a) $15$ students; b) mean $= 161.5$ cm; variance $= 25$ cm $^{2}$

Exercise 93 MediumIB · P1 TZ2 May 2015

Emily walks to school every day. The time it takes her can be modelled by a normal distribution with a mean of 11 minutes and a standard deviation of 3 minutes. She is late if her journey takes more than 15 minutes.

a) Find the probability that she is late next Monday.

b) Find the probability that she is late at least once during the following week (from Monday to Friday).

Show answer

Answer: a) $0.0912$ ; b) $0.380$

Show working and answer

Let $T \sim N (11, 3^{2})$ be the time of the journey.

a) Using the GDC (normal cumulative distribution): $P (T > 15) = 0.0912$ .

b) Each day she is late with probability $p = 0.0912$ . Let $Y$ be the number of late days in 5 days, so $Y \sim B (5; 0.0912)$ : $P (Y \geq 1) = 1 - P (Y = 0) = 1 - (1 - 0.0912)^{5} = 0.380.$

Answer: a) $0.0912$ ; b) $0.380$

Exercise 94 MediumIB · P2 TZ1 May 2021

A company produces bags of sugar whose weights can be modelled by a normal distribution with mean 1,000 and standard deviation 3.5. A bag of sugar is rejected for sale if its weight is less than 995 grams.

a) Calculate the probability that a randomly selected bag is rejected.

b) Estimate the number of rejected bags in a random sample of 100 bags.

c) Given that a bag is not rejected, calculate the probability that it weighs more than 1,005 grams.

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Answer: a) $0.0766$ ; b) $\approx 8$ bags; c) $0.0829$

Show working and answer

Let $X \sim N (1000; 3. 5^{2})$ be the weight.

a) Using the GDC (normal cumulative distribution): $P (X < 995) = 0.0766$ .

b) Expected number out of 100 bags: $100 \cdot 0.0766 = 7.66 \approx 8$ bags.

c) Conditional probability. Since ${X > 1005} \subset {X \geq 995}$ : $P (X > 1005 ∣ X \geq 995) = \frac{P ( X > 1005 )}{1 - 0.0766} = \frac{0.0766}{0.9234} = 0.0829.$

Answer: a) $0.0766$ ; b) $\approx 8$ bags; c) $0.0829$

Exercise 95 MediumIB · P2 TZ2 May 2022

Rachel and Sophia are taking part in a javelin throwing competition. The distances ( $R$ metres) that Rachel throws follow a normal distribution with mean 56.5 and standard deviation equal to 3. The distances ( $S$ metres) that Sophia throws follow a normal distribution with mean 57.5 and standard deviation equal to 1.8. In the first round of the competition, each competitor has five throws. To qualify for the next round of the competition, the competitor must achieve at least one throw of 60 metres or more in the first round.

Find the probability that only one of the two, Rachel or Sophia, qualifies for the next round of the competition.

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Answer: $0.493$

Show working and answer

For each competitor, the probability that a throw is at least 60 m, using the GDC (normal cumulative distribution):

Rachel: $R \sim N (56.5; 3^{2})$ , $p_{R} = P (R \geq 60) = 0.1217$ .
Sophia: $S \sim N (57.5; 1. 8^{2})$ , $p_{S} = P (S \geq 60) = 0.0824$ .

A competitor qualifies if they achieve at least one throw of 60 m or more in five attempts (binomial with $n = 5$ ): $q_{R} = 1 - (1 - 0.1217)^{5} = 0.4773, q_{S} = 1 - (1 - 0.0824)^{5} = 0.3496.$

Exactly one of the two qualifies (independent events): $q_{R} (1 - q_{S}) + q_{S} (1 - q_{R}) = 0.4773 \cdot 0.6504 + 0.3496 \cdot 0.5227 = 0.310 + 0.183 = 0.493.$

Answer: $0.493$

Exercise 96 HardIB · P2 TZ0 November 2008

a) A biscuit box is considered to have insufficient weight if it weighs less than 228 grams. The weights of these biscuit boxes are known to follow a normal distribution, with a mean of 231 grams and a standard deviation of 1.5 grams. What is the probability that a box has insufficient weight?

b) The manufacturer decides that the probability that a box has insufficient weight should be reduced to a value of 0.002.

(i) Bill suggests increasing the mean and not changing the standard deviation. Find the value of the new mean.

(ii) Sarah suggests reducing the standard deviation and not changing the mean. Find the value of the new standard deviation.

c) After the probability that a box has insufficient weight has been reduced to 0.002, a group of customers buys 100 boxes of biscuits. Find the probability that at least two of the boxes have insufficient weight.

Show answer

Answer: a) $0.0228$ ; b) i) $μ \approx 232.3$ g; ii) $σ \approx 1.04$ g; c) $0.0174$

Show working and answer

a) $X \sim N (231; 1. 5^{2})$ . Using the GDC (cumulative normal): $P (X < 228) = P (Z < - 2) = 0.0228$ .

b) We require $P (X < 228) = 0.002$ . Using the GDC (inverse normal): $z_{0.002} = - 2.8782$ .

(i) With $σ = 1.5$ : $\frac{228 - μ}{1.5} = - 2.8782 \Rightarrow μ = 228 + 2.8782 \cdot 1.5 = 232.3$ g.

(ii) With $μ = 231$ : $\frac{228 - 231}{σ} = - 2.8782 \Rightarrow σ = \frac{3}{2.8782} = 1.04$ g.

c) Let $Y$ be the number of boxes with insufficient weight: $Y \sim B (100; 0.002)$ . Using the GDC: $P (Y \geq 2) = 1 - P (Y \leq 1) = 1 - 0.9826 = 0.0174.$

Answer: a) $0.0228$ ; b) i) $μ \approx 232.3$ g; ii) $σ \approx 1.04$ g; c) $0.0174$

Exercise 97 HardIB · P2 TZ2 May 2009

A study carried out at a bottling plant has shown that the volume of drink in the bottles of mineral water filled by Machine A follows a normal distribution with mean 998 ml and standard deviation 2.5 ml.

a) Verify that the probability that a bottle chosen at random from those filled by Machine A contains more than 1,000 ml of mineral water is 0.212.

b) A random sample of 5 bottles is taken from Machine A. Find the probability that exactly 3 of them contain more than 1,000 ml of mineral water each.

c) Find the minimum number of bottles that must be included in the sample so that the probability that at least one of the bottles taken from Machine A contains more than 1,000 ml of mineral water is greater than 0.99.

d) It has been shown that, for Machine B, the probability that a bottle contains less than 996 ml of mineral water is 0.1151. The probability that a bottle contains more than 1,000 ml is 0.3446. Find the mean and the standard deviation of the volume of mineral water contained in the bottles filled by Machine B.

Show answer

Answer: a) $0.212$ ; b) $0.0591$ ; c) $20$ bottles; d) $μ = 999$ ml, $σ = 2.5$ ml

Show working and answer

Let $X \sim N (998; 2. 5^{2})$ for Machine A.

a) $P (X > 1000) = P (Z > \frac{1000 - 998}{2.5}) = P (Z > 0.8) = 1 - 0.7881 = 0.2119 \approx 0.212$ .

b) Let $Y$ be the number of bottles, out of 5, with more than 1000 ml: $Y \sim B (5; 0.2119)$ . Using the GDC (binomial): $P (Y = 3) = 0.0591$ .

c) With $n$ bottles: $P (at least one) = 1 - (1 - 0.2119)^{n} = 1 - 0.788 1^{n} > 0.99$ , that is, $0.788 1^{n} < 0.01$ : $n > \frac{ln 0.01}{ln 0.7881} = 19.3 \Rightarrow n = 20.$ (Check: with $n = 19$ we get $0.989$ and with $n = 20$ we get $0.991$ .)

d) Let $W \sim N (μ, σ^{2})$ . Using the GDC (inverse normal): $P (W < 996) = 0.1151 \Rightarrow z = - 1.2$ ; $P (W > 1000) = 0.3446 \Rightarrow P (W < 1000) = 0.6554 \Rightarrow z = 0.4$ . Then $\frac{996 - μ}{σ} = - 1.2, \frac{1000 - μ}{σ} = 0.4.$ Subtracting: $\frac{4}{σ} = 1.6 \Rightarrow σ = 2.5$ , and $μ = 1000 - 0.4 \cdot 2.5 = 999$ .

Answer: a) $0.212$ ; b) $0.0591$ ; c) $20$ bottles; d) $μ = 999$ ml, $σ = 2.5$ ml

Exercise 98 HardIB · P2 TZ2 May 2012

At a market stall, apples, pears and plums are sold.

a) The weights of the apples follow a normal distribution with mean 200 grams and standard deviation 25 grams.

(i) Given that there are 450 apples at the stall, what is the expected number of apples with a weight greater than 225 grams?

(ii) Given that 70% of the apples weigh less than $m$ grams, find the value of $m$ .

b) The weights of the pears follow a normal distribution with mean $μ$ grams and standard deviation $σ$ grams. Given that 8% of these pears weigh more than 270 grams and 15% weigh less than 250 grams, find $μ$ and $σ$ .

c) The weights of the plums follow a normal distribution with mean 80 grams and standard deviation 4 grams. Five plums are picked at random. What is the probability that exactly 3 of them weigh more than 82 grams?

Show answer

Answer: a) $\approx 71$ apples ( $71.39$ ); ii) $m \approx 213$ g ( $213.1$ ); b) $μ \approx 258.5$ g, $σ \approx 8.19$ g; c) $0.140$

Show working and answer

a) i) $X \sim N (200, 2 5^{2})$ . $P (X > 225) = P (Z > 1) = 0.1587$ , so the expected number is $450 \cdot 0.1587 = 71.39 \approx 71$ apples.

(ii) Using the GDC (inverse normal): $P (X < m) = 0.70$ gives $m = 213.1$ g.

b) Let $Y \sim N (μ, σ^{2})$ : $P (Y > 270) = 0.08$ (that is, $P (Y < 270) = 0.92$ , $z = 1.4051$ ) and $P (Y < 250) = 0.15$ ( $z = - 1.0364$ ): $\frac{270 - μ}{σ} = 1.4051, \frac{250 - μ}{σ} = - 1.0364.$ Subtracting: $\frac{20}{σ} = 2.4415 \Rightarrow σ = 8.19$ g, and $μ = 270 - 1.4051 \cdot 8.1917 = 258.5$ g.

c) For one plum, $P (C > 82) = P (Z > 0.5) = 0.3085$ . The number of plums, out of 5, weighing more than $82$ g is $B (5; 0.3085)$ ; using the GDC (binomial): $P (exactly 3) = 0.1404 \approx 0.140.$

Answer: a) $\approx 71$ apples ( $71.39$ ); ii) $m \approx 213$ g ( $213.1$ ); b) $μ \approx 258.5$ g, $σ \approx 8.19$ g; c) $0.140$

Exercise 99 HardIB · P2 TZ2 May 2005

A company buys 44% of its stock of screws from manufacturer A and the rest from manufacturer B. The diameter of the screws produced by each manufacturer follows a normal distribution with standard deviation 0.16 mm. The mean diameter of the screws from manufacturer A is 1.56 mm. 24.2% of the screws from manufacturer B have a diameter less than 1.52 mm.

a) Find the mean diameter of the screws produced by manufacturer B.

A screw is chosen at random from the company's stock.

b) Verify that the probability that the diameter is less than 1.52 mm is 0.312, correct to three significant figures.

c) The diameter of the screw is found to be less than 1.52 mm. Find the probability that the screw came from manufacturer B.

d) Manufacturer B produces 8,000 screws a day. It makes a profit of $ 1.50 for each screw sold, provided that its diameter is between 1.52 mm and 1.83 mm.

Screws with a diameter less than 1.52 mm are discarded, with a loss of $ 0.85 per screw. Screws with a diameter greater than 1.83 mm are sold with a reduced profit of $ 0.50 per screw. Find the expected profit of manufacturer B.

Show answer

Answer: a) $μ_{B} \approx 1.632$ mm; b) $0.312$ ; c) $0.434$ ; d) expected profit $\approx 6587$ $

Show working and answer

a) For manufacturer B, $P (D_{B} < 1.52) = 0.242$ . Using the GDC (inverse normal): $z = - 0.6999$ . Then $\frac{1.52 - μ _{B}}{0.16} = - 0.6999 \Rightarrow μ_{B} = 1.52 + 0.6999 \cdot 0.16 = 1.632 mm .$

b) For manufacturer A: $P (D_{A} < 1.52) = P (Z < \frac{1.52 - 1.56}{0.16}) = P (Z < - 0.25) = 0.4013$ . By the law of total probability: $P (D < 1.52) = 0.44 \cdot 0.4013 + 0.56 \cdot 0.242 = 0.1766 + 0.1355 = 0.3121 \approx 0.312.$

c) By Bayes' theorem: $P (B ∣ D < 1.52) = \frac{0.56 \cdot 0.242}{0.3121} = \frac{0.1355}{0.3121} \approx 0.434.$

d) For B: $D_{B} \sim N (1.632; 0.1 6^{2})$ . Using the GDC (normal cdf):

$P (D_{B} < 1.52) = 0.242$ (loss of $0.85$ $ per screw),
$P (D_{B} > 1.83) = 0.1079$ (profit of $0.50$ $),
$P (1.52 < D_{B} < 1.83) = 1 - 0.242 - 0.1079 = 0.6501$ (profit of $1.50$ $).

Expected profit per screw: $1.50 \cdot 0.6501 - 0.85 \cdot 0.242 + 0.50 \cdot 0.1079 = 0.8234$ $. For $8000$ screws: $8000 \cdot 0.8234 \approx 6587$ $.

Answer: a) $μ_{B} \approx 1.632$ mm; b) $0.312$ ; c) $0.434$ ; d) expected profit $\approx 6587$ $

Exercise 100 HardIB · P2 TZ2 May 2019

Timmy owns a shop. His daily income from the sale of his items can be modelled by a normal distribution, with a mean daily income of $820 and a standard deviation of $230. To make a profit, Timmy's daily income must be greater than $1,000.

a) Calculate the probability that, on a randomly chosen day, Timmy makes a profit.

The shop opens 24 days a month.

b) Calculate the probability that, in a randomly chosen month, Timmy makes a profit on between 5 and 10 days (inclusive).

Show answer

Answer: a) $0.217$ ; b) $0.613$

Show working and answer

a) Let $X \sim N (820, 23 0^{2})$ . Using the GDC (normal cdf): $P (X > 1000) = P (Z > \frac{180}{230}) = P (Z > 0.7826) = 0.217.$

b) Let $Y$ be the number of days with a profit in a month of 24 days: $Y \sim B (24; 0.2169)$ . Using the GDC (binomial): $P (5 \leq Y \leq 10) = P (Y \leq 10) - P (Y \leq 4) = 0.613.$

Answer: a) $0.217$ ; b) $0.613$

Exercise 101 HardIB · P2 May 2000

A machine produces bags of salt whose weights are normally distributed, with a mean of 110 g and a standard deviation of 1.142 g. If the weight of a bag is less than 108 g, the bag is rejected. With this setting, 4% of the bags are rejected.

The machine's settings are altered and it is found that 7% of the bags are rejected.

a) i) If the mean has not changed, find the new standard deviation correct to three decimal places.

The machine is adjusted to operate with the new value of the standard deviation.

(ii) Calculate the value, correct to two decimal places, at which the mean must be set so that only 4% of the bags are rejected.

b) With the new settings from part (a), it is found that 80% of the bags of salt have a weight between $A$ g and $B$ g, where $A$ and $B$ are symmetric about the mean. Calculate the values of $A$ and $B$ , giving your answer correct to two decimal places.

Show answer

Answer: a) i) $σ \approx 1.355$ g; ii) $μ \approx 110.37$ g; b) $A \approx 108.63$ g, $B \approx 112.11$ g

Show working and answer

Let $X \sim N (μ, σ^{2})$ . It is rejected if $X < 108$ .

a) i) With $μ = 110$ and $P (X < 108) = 0.07$ : using the GDC (inverse normal, area $0.07$ ), $z \approx - 1.4758$ . So $\frac{108 - 110}{σ} = - 1.4758 \Rightarrow σ = \frac{2}{1.4758} \approx 1.355 g .$

(ii) Now $σ = 1.355$ and $P (X < 108) = 0.04$ : $z \approx - 1.7507$ , so $μ = 108 + 1.7507 \cdot 1.355 \approx 110.37 g .$

b) With $μ = 110.37$ and $σ = 1.355$ : $P (A < X < B) = 0.8$ with $A, B$ symmetric implies $P (X < B) = 0.9$ , that is, $z \approx 1.2816$ . The distance from the mean is $1.2816 \cdot 1.355 \approx 1.74$ . Thus, $A = 110.37 - 1.74 \approx 108.63, B = 110.37 + 1.74 \approx 112.11.$

Answer: a) i) $σ \approx 1.355$ g; ii) $μ \approx 110.37$ g; b) $A \approx 108.63$ g, $B \approx 112.11$ g

Exercise 102 HardIB · P2 November 2000

(i) A satellite relies on solar cells in order to operate, as long as at least one of the cells is working. The cells fail independently of each other, and the probability that an individual cell fails within one year is 0.8.

a) For a satellite with ten solar cells, find the probability that all ten cells fail within one year.

b) For a satellite with ten solar cells, find the probability that the satellite is still working at the end of one year.

c) For a satellite with $n$ solar cells, write down the probability that the satellite is still working at the end of one year. Hence, find the smallest number of solar cells needed for the probability that the satellite is still working at the end of one year to be at least $0.95$ .

(ii) The lifetime of a particular component of a solar cell is $Y$ years, where $Y$ is a continuous random variable with probability density function

$f (y) = {0, 0.5 e^{- \frac{y}{2}}, when y < 0 when y \geq 0$

a) Determine the probability, to four significant figures, that a given component fails within a semester.

Each solar cell has three components which work independently, and the cell will continue working if at least two of the components continue working.

b) Calculate the probability that a solar cell fails within a semester.

Show answer

Answer: i) a) $0.107$ ; b) $0.893$ ; c) $1 - 0. 8^{n}$ , $n = 14$ cells; ii) a) $0.2212$ ; b) $0.125$

Show working and answer

(i) Let $F$ be the number of cells that fail: $F \sim B (n; 0.8)$ .

a) $P (all 10 fail) = 0. 8^{10} \approx 0.107$ .

b) The satellite works if at least one cell works: $1 - 0. 8^{10} \approx 0.893$ .

c) With $n$ cells, $P (works) = 1 - 0. 8^{n}$ . We require $1 - 0. 8^{n} \geq 0.95$ , that is, $0. 8^{n} \leq 0.05$ : $n \geq \frac{ln 0.05}{ln 0.8} \approx 13.4.$ The smallest integer is $n = 14$ (with 13 we obtain $0.945 < 0.95$ ).

(ii) a) The component fails within a semester if $Y < 0.5$ : $P (Y < 0.5) = \int_{0}^{0.5} 0.5 e^{- y /2} d y = [- e^{- y /2}]_{0}^{0.5} = 1 - e^{- 0.25} \approx 0.2212.$

b) Let $p = 0.2212$ be the probability that a component fails. The cell fails if at least two of the three components fail. With $N \sim B (3; p)$ : $P (N \geq 2) = 3 p^{2} (1 - p) + p^{3} \approx 0.125.$

Answer: i) a) $0.107$ ; b) $0.893$ ; c) $1 - 0. 8^{n}$ , $n = 14$ cells; ii) a) $0.2212$ ; b) $0.125$

Exercise 103 HardIB · P2 TZ0 November 2004

Ian and Kart have been chosen to represent their countries in the Olympic discus throw. The distance achieved by athletes in the throw is assumed to be normally distributed. The mean distance achieved by Ian last year was 60.33 m, with a standard deviation of 1.95 m.

a) Last year, 80% of Ian's throws exceeded $x$ metres. Find $x$ , giving your answer correct to two decimal places.

b) Last year, 80% of Kart's throws exceeded 56.52 m. If the mean distance of his throws was 59.39 m, find the standard deviation of his throws, giving your answer correct to two decimal places.

c) This year, Kart's throws have a mean of 59.50 m and a standard deviation of 3.00 m. Ian's throws still have a mean of 60.33 m and a standard deviation of 1.95 m. In the competition, an athlete must have at least one throw of 65 m or more in the first round to qualify for the final. Each athlete is allowed three throws in the first round.

(i) Determine which of the two athletes has the greater chance of qualifying for the final on their first throw.

(ii) Find the probability that both athletes qualify for the final.

Show answer

Answer: a) $x \approx 58.69$ m; b) $σ \approx 3.41$ m; c) i) Kart ( $0.0334$ compared with $0.00831$ for Ian); ii) $0.0024$

Show working and answer

a) Let $X \sim N (60.33; 1.9 5^{2})$ . $P (X > x) = 0.8$ is equivalent to $P (X < x) = 0.2$ . Using the GDC (inverse normal, area $0.2$ ): $x \approx 58.69$ m.

b) $P (Y > 56.52) = 0.8$ with $μ = 59.39$ , that is, $P (Y < 56.52) = 0.2$ and $z \approx - 0.8416$ : $\frac{56.52 - 59.39}{σ} = - 0.8416 \Rightarrow σ = \frac{2.87}{0.8416} \approx 3.41 m .$

c) i) Using the GDC (normal cdf):

Kart, $N (59.5; 3^{2})$ : $P (X \geq 65) \approx 0.0334$ .
Ian, $N (60.33; 1.9 5^{2})$ : $P (X \geq 65) \approx 0.00831$ .

On his first throw, Kart has the greater chance.

(ii) With three throws, $P (qualify) = 1 - (1 - p)^{3}$ :

Kart: $1 - (1 - 0.0334)^{3} \approx 0.0968$ .
Ian: $1 - (1 - 0.00831)^{3} \approx 0.0247$ .

The athletes are independent, so $0.0968 \cdot 0.0247 \approx 0.00239 \approx 0.0024$ .

Answer: a) $x \approx 58.69$ m; b) $σ \approx 3.41$ m; c) i) Kart ( $0.0334$ compared with $0.00831$ for Ian); ii) $0.0024$

Exercise 104 HardIB · P2 TZ1 May 2008

A company produces computer microchips, whose lifetime follows a normal distribution with a mean of 90 months and a standard deviation of 3.7 months.

a) If a microchip is guaranteed for 84 months, find the probability that it will fail before the guarantee ends.

b) The probability that a microchip does not fail before the end of the guarantee is required to be 99%. For how many months should it be guaranteed?

c) A rival company produces microchips for which the probability of failure after 84 months is 0.88. Given that their lifetime also follows a normal distribution with standard deviation 3.7 months, find the mean.

Show answer

Answer: a) $0.0524$ ; b) $81.4$ months; c) $μ = 88.3$ months

Show working and answer

Let $X$ be the lifetime of the microchip, $X \sim N (90; 3. 7^{2})$ .

a) Using the GDC (normal cdf): $P (X < 84) = 0.0524$ .

b) We seek $t$ with $P (X > t) = 0.99$ , that is, $P (X < t) = 0.01$ . Using the GDC (inverse normal): $t = 81.4$ months.

c) Now $Y \sim N (μ, 3. 7^{2})$ with $P (Y > 84) = 0.88$ , that is, $P (Y < 84) = 0.12$ . The standard inverse normal gives $z = - 1.175$ , so $\frac{84 - μ}{3.7} = - 1.175 \Rightarrow μ = 84 + 1.175 \cdot 3.7 = 88.3.$

Answer: a) $0.0524$ ; b) $81.4$ months; c) $μ = 88.3$ months

Exercise 105 HardIB · P2 TZ2 May 2015

The farmer Suzie grows turnips. The weights of the turnips she harvests follow a normal distribution with mean 122 g and standard deviation 14.7 g.

a) i) Calculate the percentage of the turnips that weigh between 110 g and 130 g.

(ii) Suzie has 100 turnips ready to take to market. Find the expected number of turnips that weigh more than 130 g.

(iii) Find the probability that at least 30 of these 100 turnips weigh more than 130 g.

The farmer Ray also grows turnips. The weights of the turnips he grows follow a normal distribution with mean 144 g. Ray only takes to market those turnips that weigh more than 130 g. Over a certain period, Ray observes that he has to reject 1 in every 15 turnips for weighing less than they should.

b) i) Find the standard deviation of the weights of Ray's turnips.

(ii) Ray has 200 turnips ready to take to market. Find the expected number of turnips that weigh more than 150 g.

Show answer

Answer: a) i) $50.0%$ ( $49.97%$ ); ii) $29.3$ turnips; iii) $0.478$ ; b) i) $σ = 9.33$ g; ii) $55.7$ turnips

Show working and answer

Let $W \sim N (122; 14. 7^{2})$ be the weight of one of Suzie's turnips.

a) i) Using the GDC (normal cumulative distribution): $P (110 < W < 130) = 0.4997$ , that is, $50.0%$ (more precisely $49.97%$ ).

(ii) $P (W > 130) = 0.2931$ . Expected number out of 100 turnips: $100 \cdot 0.2931 = 29.3$ turnips.

(iii) Let $Y$ be the number of turnips weighing more than 130 g, so $Y \sim B (100; 0.2931)$ . Using the GDC (binomial cumulative distribution): $P (Y \geq 30) = 1 - P (Y \leq 29) = 0.478.$

b) i) Let $V \sim N (144, σ^{2})$ be the weight of Ray's turnips. He rejects those weighing less than 130 g, 1 in every 15: $P (V < 130) = \frac{1}{15}$ . Inverse standard normal: $z = - 1.501$ . $\frac{130 - 144}{σ} = - 1.501 \Rightarrow σ = 9.33 g .$

(ii) The 200 turnips ready for market already weigh more than 130 g, so we must use conditional probability: $P (V > 150 ∣ V > 130) = \frac{P ( V > 150 )}{P ( V > 130 )} = \frac{0.2600}{0.9333} = 0.2786.$ Expected number: $200 \cdot 0.2786 = 55.7$ turnips.

Answer: a) i) $50.0%$ ( $49.97%$ ); ii) $29.3$ turnips; iii) $0.478$ ; b) i) $σ = 9.33$ g; ii) $55.7$ turnips

Exercise 106 HardIB · P2 TZ0 November 2015

A survey is carried out in a large office building. It is found that 30% of the office workers weigh less than 62 kg, and 25% of the office workers weigh more than 98 kg. The weights of the office workers can be modelled by a normal distribution with mean $μ$ and standard deviation $σ$ .

a) i) Determine a system of two linear equations satisfied by $μ$ and $σ$ .

(ii) Find the value of $μ$ and the value of $σ$ .

b) Find the probability that an office worker weighs more than 100 kg.

The building has lifts that take the office workers to their offices. Given that a particular lift contains 10 office workers,

c) Find the probability that there are at least four office workers who weigh more than 100 kg.

Given that there are 10 office workers in the lift and that at least one of them weighs more than 100 kg,

d) Find the probability that there are fewer than four office workers who weigh more than 100 kg.

Show answer

Answer: a) ii) $μ = 77.7$ kg; $σ = 30.0$ kg; b) $0.229$ ; c) $0.178$ ; d) $0.808$

Show working and answer

Let $W \sim N (μ, σ^{2})$ be the weight, with $P (W < 62) = 0.30$ and $P (W > 98) = 0.25$ .

a) i) Inverse standard normal: $P (Z < z) = 0.30 \Rightarrow z = - 0.5244$ ; $P (Z < z) = 0.75 \Rightarrow z = 0.6745$ . So: ${μ - 0.5244 σ = 62 μ + 0.6745 σ = 98$

(ii) Subtracting: $1.1989 σ = 36 \Rightarrow σ = 30.0$ ; then $μ = 62 + 0.5244 \cdot 30.03 = 77.7$ .

b) Using the GDC (normal cumulative distribution, unrounded values): $P (W > 100) = 0.229$ .

c) Let $Y$ be the number of the 10 office workers who weigh more than 100 kg: $Y \sim B (10; 0.2293)$ . Using the GDC: $P (Y \geq 4) = 1 - P (Y \leq 3) = 0.178.$

d) Conditional probability: $P (Y < 4 ∣ Y \geq 1) = \frac{P ( 1 \leq Y \leq 3 )}{P ( Y \geq 1 )} = \frac{0.8220 - 0.0739}{1 - 0.0739} = 0.808.$

Answer: a) ii) $μ = 77.7$ kg; $σ = 30.0$ kg; b) $0.229$ ; c) $0.178$ ; d) $0.808$

Exercise 107 HardIB · P2 TZ2 May 2021

The flight times ( $T$ minutes) between two given cities follow a normal distribution with mean 75 minutes and standard deviation equal to $σ$ minutes.

a) Given that 2% of the flight times are greater than 82 minutes, find the value of $σ$ .

b) Find the probability that a flight chosen at random has a flight time of more than 80 minutes.

c) Given that a particular flight between these two cities lasted more than 80 minutes, find the probability that it lasted less than 82 minutes.

On a particular day, there are 64 flights scheduled between these two cities.

d) Find the expected number of flights that will have a flight time of more than 80 minutes.

e) Find the probability that more than 6 of the flights scheduled on that particular day have a flight time of more than 80 minutes.

Show answer

Answer: a) $σ = 3.41$ min; b) $0.0712$ ; c) $0.719$ ; d) $4.56$ flights; e) $0.169$

Show working and answer

Let $T \sim N (75, σ^{2})$ .

a) $P (T > 82) = 0.02 \Rightarrow P (Z < z) = 0.98 \Rightarrow z = 2.054$ . Then $\frac{82 - 75}{σ} = 2.054 \Rightarrow σ = 3.41 min .$

b) Using the GDC (normal cumulative distribution, $σ = 3.408$ unrounded): $P (T > 80) = 0.0712$ .

c) Conditional probability: $P (T < 82 ∣ T > 80) = \frac{P ( 80 < T < 82 )}{P ( T > 80 )} .$ Using the GDC: $P (80 < T < 82) = 0.98 - 0.9288 = 0.0512$ , and therefore $P (T < 82 ∣ T > 80) = \frac{0.0512}{0.0712} = 0.719.$

d) Each flight exceeds 80 min with probability $0.0712$ . Out of 64 flights: $64 \cdot 0.0712 = 4.56$ flights.

e) Let $Y$ be the number of flights lasting more than 80 min, so $Y \sim B (64; 0.0712)$ . Using the GDC (binomial cumulative distribution): $P (Y > 6) = 1 - P (Y \leq 6) = 0.169.$

Answer: a) $σ = 3.41$ min; b) $0.0712$ ; c) $0.719$ ; d) $4.56$ flights; e) $0.169$

Exercise 108 HardIB · P2 TZ1 May 2022

A bakery makes two types of muffins: chocolate muffins and banana muffins. The weights, $C$ grams, of the chocolate muffins are normally distributed with a mean of 62 g and a standard deviation of 2.9 g.

a) Find the probability that a randomly selected chocolate muffin weighs less than 61 g.

b) In a random selection of 12 chocolate muffins, find the probability that exactly 5 of them weigh less than 61 g.

The weights, $B$ grams, of the banana muffins are normally distributed with a mean of 68 g and a standard deviation of 3.4 g. Each day, 60% of the muffins made are chocolate. On a particular day, one muffin is selected at random from all those made in the bakery.

c) i) Find the probability that the randomly selected muffin weighs less than 61 g.

(ii) Given that the randomly selected muffin weighs less than 61 g, calculate the probability that it is a chocolate muffin.

The machine that makes the chocolate muffins is adjusted so that the mean weight of the chocolate muffins stays the same but their standard deviation changes to $σ$ g. The machine that makes the banana muffins is not adjusted. The probability that the weight of a randomly selected muffin from these machines is less than 61 g is now 0.157.

d) Calculate the value of $σ$ .

Show answer

Answer: a) $0.365$ ; b) $0.214$ ; c) i) $0.227$ ; ii) $0.965$ ; d) $σ = 1.47$ g

Show working and answer

Let $C \sim N (62; 2. 9^{2})$ be the weight of a chocolate muffin.

a) Using the GDC (normal cumulative distribution): $P (C < 61) = 0.365$ .

b) Let $Y$ be the number of the 12 muffins that weigh less than 61 g, so $Y \sim B (12; 0.3651)$ : $P (Y = 5) = (5 12) (0.3651)^{5} (0.6349)^{7} = 0.214.$

c) i) For the banana muffins, $B \sim N (68; 3. 4^{2})$ : $P (B < 61) = 0.0198$ . By the law of total probability: $P (weight < 61) = 0.6 \cdot 0.3651 + 0.4 \cdot 0.0198 = 0.227.$

(ii) Bayes' theorem: $P (chocolate ∣ weight < 61) = \frac{0.6 \cdot 0.3651}{0.2270} = 0.965.$

d) Now $C \sim N (62, σ^{2})$ and the total probability of weighing less than 61 g is $0.157$ : $0.6 P (C < 61) + 0.4 \cdot 0.0198 = 0.157 \Rightarrow P (C < 61) = \frac{0.157 - 0.00790}{0.6} = 0.2485.$ Inverse standard normal: $z = - 0.6785$ , and $\frac{61 - 62}{σ} = - 0.6785$ gives $σ = 1.47$ g. (Using the GDC, you can also solve the equation $0.6 P (C < 61) + 0.4 \cdot 0.0198 = 0.157$ for $σ$ directly.)

Answer: a) $0.365$ ; b) $0.214$ ; c) i) $0.227$ ; ii) $0.965$ ; d) $σ = 1.47$ g

Exercise 109 HardIB · P2 TZ0 November 2022

In a large company, the time that employees work ( $T$ hours per week) follows a normal distribution with mean 42 and standard deviation equal to 10.7.

a) Find the probability that an employee chosen at random works more than 40 hours per week.

b) A group of four employees is chosen at random. These employees are asked, one at a time, whether they work more than 40 hours per week. Find the probability that the fourth employee is the only one in the group who works more than 40 hours per week.

c) There is a large group of employees who work more than 40 hours per week.

(i) An employee is chosen at random from this large group. Find the probability that this employee works less than 55 hours per week.

(ii) Ten employees are chosen at random from this large group. Find the probability that exactly five of them work less than 55 hours per week.

It is known that $P (a \leq T \leq b) = 0.904$ and that $P (T > b) = 2 P (T < a)$ , where $a$ and $b$ are numbers of hours worked per week. Employees who work less than $a$ hours per week are considered part-time workers.

d) Find the maximum time (in hours per week) that an employee can work and still be considered a part-time worker.

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Answer: a) $0.574$ ; b) $0.0443$ ; c) i) $0.805$ ; ii) $0.0242$ ; d) $22.2$ hours per week

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Let $T \sim N (42; 10. 7^{2})$ .

a) Using the GDC (normal cumulative distribution): $P (T > 40) = 0.574$ .

b) With $p = 0.5741$ : the first three do not work more than 40 hours and the fourth does: $(1 - p)^{3} p = 0.425 9^{3} \cdot 0.5741 = 0.0443.$

c) i) Conditional probability: $P (T < 55 ∣ T > 40) = \frac{P ( 40 < T < 55 )}{P ( T > 40 )} = \frac{0.4619}{0.5741} = 0.805.$

(ii) Let $Y$ be the number of the 10 employees who work less than 55 hours, so $Y \sim B (10; 0.8046)$ : $P (Y = 5) = (5 10) (0.8046)^{5} (0.1954)^{5} = 0.0242.$

d) Let $q = P (T < a)$ . Then $P (T > b) = 2 q$ , and $P (a \leq T \leq b) = 1 - q - 2 q = 0.904 \Rightarrow q = \frac{0.096}{3} = 0.032.$ Using the GDC (inverse normal): $a = 22.2$ hours per week.

Answer: a) $0.574$ ; b) $0.0443$ ; c) i) $0.805$ ; ii) $0.0242$ ; d) $22.2$ hours per week

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