IB Analysis and Approaches HL · Continuous random variables and the normal distribution · Section 3 of 4

Normal calculations and the inverse normal

3 worked examples · By Álvaro López

Key idea

In Paper 2, almost every question on the normal distribution is solved with two GDC tools: the normal cumulative distribution (from a value to a probability) and the inverse normal (from a probability to a value). The task is to translate the question into a probability, draw a sketch and use the right tool.

From a value to a probability

For $X \sim N (μ, σ^{2})$ , the probability $P (a < X < b)$ is the area under the bell between $a$ and $b$ . The GDC calculates it with the normal cumulative distribution function, to which you give the lower bound, the upper bound, the mean $μ$ and the standard deviation $σ$ .

The three basic types of area are:

Question	Sketch	How to calculate
$P (X < a)$	left tail	from $μ - 10 σ$ to $a$
$P (X > a)$	right tail	from $a$ to $μ + 10 σ$
$P (a < X < b)$	central band	from $a$ to $b$

As the normal is continuous, $P (X \leq a) = P (X < a)$ . And as the total area is 1, it is always true that $P (X > a) = 1 - P (X < a)$ .

Three small normal curves in a row, each with its mean μ: in the first, the left tail up to a is shaded and labelled P(X<a); in the second, the right tail from a is shaded and labelled P(X>a); in the third, the strip between a and b around the mean is shaded and labelled P(a<X<b). The rest of each curve is left unshaded

From a probability to a value: the inverse normal

Sometimes the probability is known and the value is asked for: "the heaviest 10% weigh more than…", "the time that is exceeded with probability 0.98". The GDC's inverse normal function returns the value $a$ such that $P (X < a) = p$ , for given $μ$ and $σ$ .

A very common mistake is to give the calculator the area of the right tail without converting it. If the top 5% exceed a value, the area to the left of that value is $0.95$ .

Context problems and expected number

In real problems you must define the variable, write down its distribution, sketch and calculate. When several elements of a sample are asked for, the normal is combined with the binomial: the probability $p$ that one element meets the condition comes from the normal, and then the number $N$ of elements that meet it among $n$ chosen at random follows $N \sim B (n, p)$ . In a population of $M$ elements, the expected number is $M \cdot p$ .

Worked examples

Example 1 Basic

Journey time

The time $T$ (in minutes) that a person takes to get to work follows a normal distribution $N (35, 4^{2})$ . Using the GDC, find:

a) the probability that it takes less than 30 minutes;

b) the probability that it takes more than 42 minutes;

c) the probability that it takes between 32 and 38 minutes.

Solution

We have $μ = 35$ and $σ = 4$ .

a) $P (T < 30) = 0.106$ (area of the left tail).

b) $P (T > 42) = 0.0401$ (area of the right tail; it can be calculated from 42 to a very high bound).

c) $P (32 < T < 38) = 0.547$ (strip between the two values).

Answer: a) $0.106$ ; b) $0.0401$ ; c) $0.547$ .

Example 2 Medium

Apple sizes

The mass $W$ (in grams) of the apples from a farm follows a normal distribution $N (180, 1 5^{2})$ .

a) Find the probability that an apple weighs more than 200 g.

b) 90% of the apples weigh less than $w$ grams. Find $w$ .

c) The heaviest apples, the top 5%, are sold as "extra". Find the minimum mass of an "extra" apple.

Solution

a) $P (W > 200) = 0.0912$ .

b) We need $w$ such that $P (W < w) = 0.9$ . Using the inverse normal ( $μ = 180$ , $σ = 15$ , area $0.9$ ): $w = 199.22 \dots \approx 199 g$

c) The top 5% is a right tail: $P (W > m) = 0.05$ , that is, $P (W < m) = 0.95$ . Using the inverse normal with area $0.95$ : $m = 204.67 \dots \approx 205 g$

Answer: a) $0.0912$ ; b) $199$ g; c) $205$ g.

Example 3 Hard

Battery quality control

The lifetime $L$ (in hours) of a battery follows a normal distribution $N (48, 5^{2})$ . A battery is considered defective if it lasts less than 40 hours.

a) Find the probability that a randomly chosen battery is defective.

b) The manufacturer wants to guarantee a lifetime $t$ such that 98% of the batteries last more than $t$ hours. Find $t$ .

c) 25 batteries are chosen at random. Find the probability that at least 2 are defective.

d) In a batch of 400 batteries, find the expected number of defective ones.

Solution

a) $p = P (L < 40) = 0.0548$ .

b) We need $P (L > t) = 0.98$ , that is, $P (L < t) = 0.02$ . Using the inverse normal with area $0.02$ (the guaranteed value lies below the mean): $t = 37.73 \dots \approx 37.7 hours$

c) Let $N$ be the number of defective ones among 25. The batteries are independent and each one is defective with probability $p = 0.0548$ : $N \sim B (25; 0.0548)$ . Then $P (N \geq 2) = 1 - P (N \leq 1) = 1 - [P (N = 0) + P (N = 1)] = 0.401$

d) Expected number: $400 \cdot 0.0548 = 21.9$ batteries.

Answer: a) $0.0548$ ; b) $37.7$ h; c) $0.401$ ; d) $21.9$ .

In the exam: store the full value of $p$ in the calculator to use it in the binomial, without rounding it first. Write $L \sim N (48, 5^{2})$ and $N \sim B (25; p)$ before calculating: they earn the marks for defining the variable and the model.

Common mistakes

Summary

For $X \sim N (μ, σ^{2})$ , $P (a < X < b)$ is obtained with the GDC's normal cumulative distribution, entering $σ$ (not $σ^{2}$ ).
Always draw the sketch; $P (X > a) = 1 - P (X < a)$ and $P (X \leq a) = P (X < a)$ .
The inverse normal gives the value with a given area to its left: convert right tails first.
A central band of area $p$ leaves $\frac{1 - p}{2}$ in each tail.
In samples: the normal probability is the $p$ of a binomial $B (n, p)$ ; the expected number in $M$ elements is $M \cdot p$ .
Answers to 3 significant figures; do not round intermediate values.

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