IB Analysis and Approaches HL · Continuous random variables and the normal distribution · Section 2 of 4

The normal distribution

3 worked examples · By Álvaro López

Key idea

The normal distribution is the most important continuous random variable: it describes heights, measurement errors, process times and many natural data. It is determined by two numbers, $μ$ and $σ$ . You need to know the shape of its curve, know what each parameter changes, and use symmetry and the 68-95-99.7 rule to estimate probabilities without a calculator.

The bell curve

Many quantities cluster around a central value and spread out from it symmetrically: few measurements very low, many close to the mean, few very high. The model that describes this behaviour is the normal distribution.

Pay attention to the notation: in the IB, $N (μ, σ^{2})$ uses the variance. If you are given $X \sim N (50, 16)$ , the standard deviation is $σ = 4$ , not 16.

You do not need to integrate this function (it has no elementary antiderivative): probabilities are obtained with the GDC. What you do need to know is what the curve tells you.

Bell curve of the normal distribution with marks μ−3σ, μ−2σ, μ−σ, μ, μ+σ, μ+2σ and μ+3σ: the zone between μ−σ and μ+σ is shaded and labelled approx. 68%; the strips between μ−2σ and μ−σ and between μ+σ and μ+2σ, each approx. 13.5%; the tails beyond ±2σ are left unshaded

What each parameter changes

The mean $μ$ moves the bell left or right without changing its shape: it shifts the whole curve. The standard deviation $σ$ controls the width. With a small $σ$ the data cluster together and the bell is tall and narrow; with a large $σ$ they spread out and the bell is low and wide. Since the total area is always 1, a wider bell must be lower.

The maximum value of the density is $f (μ) = \frac{1}{σ 2 π}$ , which depends only on $σ$ .

The 68-95-99.7 rule

There are three areas worth remembering, because they let you estimate probabilities in Paper 1 and check GDC results.

Symmetry gives more areas: if 68% lies within $\pm σ$ , then 32% lies outside, and half of that (16%) on each side. For example, $P (X > μ + σ) \approx 0.16$ and $P (X > μ + 2 σ) \approx 0.025$ .

Worked examples

Example 1 Basic

Tomato masses

The mass $X$ (in grams) of the tomatoes from a cooperative follows a normal distribution $N (200, 1 0^{2})$ . Use the 68-95-99.7 rule to estimate:

a) the probability that a tomato has a mass between 190 g and 210 g;

b) the probability that the mass is between 180 g and 220 g;

c) the probability that a tomato weighs more than 220 g.

Solution

Here $μ = 200$ and $σ = 10$ .

a) $190 = μ - σ$ and $210 = μ + σ$ : $P (190 < X < 210) \approx 0.68$ .

b) $180 = μ - 2 σ$ and $220 = μ + 2 σ$ : $P (180 < X < 220) \approx 0.95$ .

c) Outside $μ \pm 2 σ$ remains $1 - 0.95 = 0.05$ ; by symmetry half of it is above 220: $P (X > 220) \approx 0.025$ .

Answer: a) $0.68$ ; b) $0.95$ ; c) $0.025$ .

Example 2 Medium

Using symmetry

The time $X$ (in minutes) that an operator takes to assemble a part follows a normal distribution with mean 50. It is known that $P (X < 42) = 0.2$ .

a) Find $P (X > 58)$ .

b) Find $P (42 < X < 58)$ .

c) Find $P (X < 58)$ .

Solution

The mean is 50, and both 42 and 58 are 8 units away from it ( $50 \mp 8$ ), so they are symmetric about the mean.

a) By symmetry of the curve, the area to the right of 58 equals the area to the left of 42: $P (X > 58) = P (X < 42) = 0.2$ .

b) The remaining area, outside that interval, is $0.2 + 0.2 = 0.4$ : $P (42 < X < 58) = 1 - 0.4 = 0.6$

c) $P (X < 58) = 1 - P (X > 58) = 1 - 0.2 = 0.8$ .

Answer: a) $0.2$ ; b) $0.6$ ; c) $0.8$ .

Example 3 Hard

Two machines with the same mean

Two machines fill bottles of oil. The volume (in ml) from machine A follows $X \sim N (500, 5^{2})$ and that from machine B follows $Y \sim N (500, 1 0^{2})$ . A bottle is rejected if it contains more than 510 ml.

a) Find the maximum value of the density function of $X$ and that of $Y$ . Explain which machine is more precise and what relationship there is between the heights of the bell curves.

b) Using the GDC, find the probability that a bottle from each machine is rejected.

c) Using the GDC, for each machine, find the probability that the volume is within one standard deviation of the mean, and explain why the same value is obtained.

Solution

a) The maximum is at the mean and equals $\frac{1}{σ 2 π}$ : $f_{X} (500) = \frac{1}{5 2 π} \approx 0.0798 f_{Y} (500) = \frac{1}{10 2 π} \approx 0.0399$ The bell of A is twice as tall (and half as wide) because its $σ$ is half. Machine A is more precise: its volumes cluster more closely around 500 ml.

b) For A, $510 = μ + 2 σ$ : by the rule, $P (X > 510) \approx 0.025$ (with the GDC, $0.0228$ ). For B, $510 = μ + σ$ : $P (Y > 510) \approx 0.16$ (with the GDC, $0.159$ ).

Machine B rejects almost seven times as many bottles.

c) In both cases, $P (μ - σ < X < μ + σ) \approx 0.683$ . Measured in standard deviations, the shape of the curve is the same for any normal; only the scale of the horizontal axis changes.

Answer: a) $0.0798$ and $0.0399$ ; b) $0.0228$ and $0.159$ ; c) $0.683$ in both.

In the exam: reason with $μ$ and $σ$ (for example, "510 is 2 standard deviations above the mean") and not only with the number the calculator gives. To compare two normals, what matters is how many $σ$ separate the value from the mean.

Common mistakes

Summary

$X \sim N (μ, σ^{2})$ : symmetric bell curve about $x = μ$ ; mean, median and mode coincide at $μ$ . The second number is the variance.
Density: $f (x) = \frac{1}{σ 2 π} e^{- (x - μ)^{2} / (2 σ^{2})}$ ; maximum $\frac{1}{σ 2 π}$ ; points of inflexion at $μ \pm σ$ .
$μ$ shifts the bell; $σ$ widens it (lower) or narrows it (taller). The total area is always 1.
Rule: $\approx 68 %$ within $\pm σ$ , $\approx 95 %$ within $\pm 2 σ$ , $\approx 99.7 %$ within $\pm 3 σ$ .
Symmetry gives $P (X < μ) = \frac{1}{2}$ and $P (X < μ - a) = P (X > μ + a)$ .

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