๐Ÿ”” The Normal Distribution

A Level Maths Year 2 ยท Statistics

Level 3 · Ages 17–18

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1 A Continuous Distribution

The binomial distribution from Year 1 (see The Binomial Distribution) is discrete: it counts things. Heights, masses, times and lengths are continuous โ€” they can take any value in a range. The most important model for continuous data is the normal distribution.

Notation
$X \sim \mathrm{N}(\mu,\ \sigma^2)$
mean $\mu$  ·  variance $\sigma^2$  ·  standard deviation $\sigma$
The second number is the variance. $X \sim \mathrm{N}(50,\ 16)$ has standard deviation $\sqrt{16} = 4$, not $16$. Entering $16$ as the standard deviation in your calculator is the single most common error on this topic.
For a continuous variable, probability is area under the curve. The total area is $1$, and the probability of any single exact value is zero: $\mathrm{P}(X = a) = 0$. So $\mathrm{P}(X < a) = \mathrm{P}(X \leqslant a)$ โ€” the inequality sign does not matter.
2 Properties of the Normal Curve
ฮผ โˆ’ 2ฯƒ ฮผ โˆ’ ฯƒ ฮผ ฮผ + ฯƒ ฮผ + 2ฯƒ โ‰ˆ 68% point of inflection Shaded: P(ฮผ โˆ’ ฯƒ < X < ฮผ + ฯƒ) โ‰ˆ 0.683
IntervalProbability (computed)Rule of thumb
$\mu \pm \sigma$$0.6827$about $68\%$
$\mu \pm 2\sigma$$0.9545$about $95\%$
$\mu \pm 3\sigma$$0.9973$almost all
Checking whether a normal model is plausible: data should be roughly symmetrical, unimodal, and have about $95\%$ of values within $2\sigma$ of the mean. A quantity that must be positive is a poor fit if $\mu - 3\sigma$ is negative.
3 Standardising and Finding Probabilities

Every normal distribution is a stretched and shifted copy of the standard normal distribution $Z \sim \mathrm{N}(0,\ 1)$. Converting $X$ to $Z$ is called standardising.

Standardising
$Z = \dfrac{X - \mu}{\sigma} \sim \mathrm{N}(0,\ 1)$     $\mathrm{P}(Z < z) = \Phi(z)$
$z$ measures how many standard deviations a value is from the mean. Your calculator's normal cumulative distribution function finds probabilities for any $\mathrm{N}(\mu,\sigma^2)$ directly, but standardising is essential for problems with an unknown $\mu$ or $\sigma$.
Worked Example 1 โ€” Basic probabilities

$X \sim \mathrm{N}(50,\ 16)$. Find (a) $\mathrm{P}(X < 56)$, (b) $\mathrm{P}(45 < X < 53)$, (c) $\mathrm{P}(X > 58)$.

โ‘ $\sigma = \sqrt{16} = 4$.
โ‘ก(a) $z = \dfrac{56 - 50}{4} = 1.5$, so $\mathrm{P}(X < 56) = \Phi(1.5) = 0.9332$.
โ‘ข(b) $z$-values $\dfrac{45-50}{4} = -1.25$ and $\dfrac{53-50}{4} = 0.75$; $\Phi(0.75) - \Phi(-1.25) = 0.6677$.
โ‘ฃ(c) $z = \dfrac{58-50}{4} = 2$; $\mathrm{P}(X > 58) = 1 - \Phi(2) = 0.0228$ (3 s.f.).
Sketch first. A quick bell curve with the region shaded tells you whether your answer should be more or less than $0.5$ โ€” an excellent check.
Worked Example 2 โ€” Normal and binomial together

The masses of bags of flour are $\mathrm{N}(1005,\ 4^2)$ grams. A bag is underweight if its mass is less than $1000$ g. Find the probability that a box of $10$ bags contains at most one underweight bag.

โ‘ $\mathrm{P}(M < 1000) = \Phi\!\left(\dfrac{1000 - 1005}{4}\right) = \Phi(-1.25) = 0.10565$
โ‘กLet $U$ = number underweight in the box: $U \sim \mathrm{B}(10,\ 0.10565)$, assuming bags are independent.
โ‘ข$\mathrm{P}(U \leqslant 1) = 0.89435^{10} + 10(0.10565)(0.89435)^9 = 0.714$ (3 s.f.)
4 Inverse Normal Problems

Sometimes you know the probability and want the value. Use the inverse normal function on your calculator โ€” but check whether it wants the area to the left (most do).

$p$$0.9$$0.95$$0.975$$0.99$$0.995$
$z$ with $\Phi(z) = p$$1.2816$$1.6449$$1.9600$$2.3263$$2.5758$
Worked Example 3 โ€” Finding a value from a probability

Adult heights are $H \sim \mathrm{N}(170,\ 8^2)$ cm. (a) Find $h$ such that $\mathrm{P}(H < h) = 0.9$. (b) Find the interval, symmetrical about the mean, containing the middle $95\%$ of heights.

โ‘ (a) $\Phi(z) = 0.9 \Rightarrow z = 1.2816$.
โ‘ก$h = \mu + z\sigma = 170 + 1.2816 \times 8 = 180.3$ cm.
โ‘ข(b) The middle $95\%$ leaves $2.5\%$ in each tail, so $z = \pm 1.96$.
โ‘ฃ$170 \pm 1.96 \times 8 = 170 \pm 15.68$, i.e. $154.3 < H < 185.7$ cm.
5 Finding an Unknown $\mu$ or $\sigma$
The method
Turn the probability into a $z$-value, then solve $\dfrac{x - \mu}{\sigma} = z$
Worked Example 4 โ€” Unknown standard deviation

$X \sim \mathrm{N}(30,\ \sigma^2)$ and $\mathrm{P}(X > 35) = 0.1$. Find $\sigma$.

โ‘ $\mathrm{P}(X < 35) = 0.9$, so $\dfrac{35 - 30}{\sigma} = 1.2816$.
โ‘ก$\sigma = \dfrac{5}{1.2816} = 3.90$ (3 s.f.)
Worked Example 5 โ€” Unknown mean

$X \sim \mathrm{N}(\mu,\ 2.5^2)$ and $\mathrm{P}(X < 20) = 0.05$. Find $\mu$.

โ‘ $\Phi(z) = 0.05 \Rightarrow z = -1.6449$. The $z$-value is negative because $20$ is in the lower tail.
โ‘ก$\dfrac{20 - \mu}{2.5} = -1.6449 \Rightarrow 20 - \mu = -4.112$
โ‘ข$\mu = 24.1$ (3 s.f.)
Worked Example 6 โ€” Both unknown: simultaneous equations

$X \sim \mathrm{N}(\mu,\ \sigma^2)$ with $\mathrm{P}(X < 40) = 0.2$ and $\mathrm{P}(X > 65) = 0.1$. Find $\mu$ and $\sigma$.

โ‘ $\Phi(z) = 0.2 \Rightarrow z = -0.8416$:   $40 - \mu = -0.8416\sigma$
โ‘ก$\Phi(z) = 0.9 \Rightarrow z = 1.2816$:   $65 - \mu = 1.2816\sigma$
โ‘ขSubtract: $25 = 2.1232\sigma \Rightarrow \sigma = 11.77$
โ‘ฃ$\mu = 40 + 0.8416 \times 11.77 = 49.9$ (3 s.f.)
Keep at least four decimal places in the $z$-values and the full calculator value of $\sigma$ until the end; rounding early can shift the final answer.
Sign of $z$. If the value is below the mean, its $z$-value must be negative. A quick sketch prevents a wrong sign, which otherwise gives a nonsensical $\sigma < 0$ or a mean on the wrong side.
6 Quick Reference

Notation

$X \sim \mathrm{N}(\mu,\sigma^2)$ โ€” second parameter is the variance.

Shape

Symmetric bell; mean = median = mode.

Inflection

Points of inflection at $x = \mu \pm \sigma$.

Standardise

$Z = \dfrac{X - \mu}{\sigma} \sim \mathrm{N}(0,1)$

68โ€“95โ€“99.7

Within $1$, $2$, $3$ s.d. of the mean.

Continuous

$\mathrm{P}(X = a) = 0$, so $<$ and $\leqslant$ give the same answer.

Inverse normal

$x = \mu + z\sigma$ where $\Phi(z) = p$.

Unknowns

Form $\dfrac{x-\mu}{\sigma} = z$ for each fact; solve.

Key $z$-values

$1.2816$, $1.6449$, $1.9600$, $2.3263$, $2.5758$

7 Practice Questions

Ten questions, from routine to exam-style. Give answers to $3$ significant figures unless told otherwise.

Question 1

$Z \sim \mathrm{N}(0,1)$. Find (a) $\mathrm{P}(Z < 1.2)$, (b) $\mathrm{P}(Z > -0.8)$, (c) $\mathrm{P}(-1 < Z < 1.5)$.

โ–ถ Show solution

(a) $\Phi(1.2) = 0.885$

(b) By symmetry $\mathrm{P}(Z > -0.8) = \Phi(0.8) = 0.788$

(c) $\Phi(1.5) - \Phi(-1) = 0.9332 - 0.1587 = 0.775$

Question 2

$X \sim \mathrm{N}(20,\ 9)$. Find (a) $\mathrm{P}(X > 24)$, (b) $\mathrm{P}(17 < X < 25)$.

โ–ถ Show solution

$\sigma = 3$.

(a) $z = \dfrac{4}{3}$; $1 - \Phi(1.333) = 0.0912$

(b) $z$ from $-1$ to $\dfrac{5}{3}$: $\Phi(1.667) - \Phi(-1) = 0.794$

Question 3

$X \sim \mathrm{N}(\mu,\sigma^2)$. Explain why $\mathrm{P}(X \leqslant a) = \mathrm{P}(X < a)$, and why $\mathrm{P}(X < \mu) = 0.5$.

โ–ถ Show solution

$X$ is continuous, so the probability of any single exact value is zero (a line has no area). Hence including or excluding $a$ makes no difference.

The curve is symmetrical about $\mu$, so exactly half the area lies on each side.

Question 4

IQ scores are modelled by $\mathrm{N}(100,\ 15^2)$. Find the score exceeded by only $5\%$ of the population.

โ–ถ Show solution

Need $\mathrm{P}(X < a) = 0.95$, so $z = 1.6449$.

$a = 100 + 1.6449 \times 15 = 124.7$ (to 1 d.p.)

Question 5

The curve of a normal distribution has points of inflection at $x = 12$ and $x = 18$. Find $\mu$ and $\sigma$, and hence $\mathrm{P}(X > 20)$.

โ–ถ Show solution

Inflections are at $\mu \pm \sigma$, so $\mu = \dfrac{12 + 18}{2} = 15$ and $\sigma = 3$.

$\mathrm{P}(X > 20) = 1 - \Phi\!\left(\dfrac{5}{3}\right) = 0.0478$

Question 6

$X \sim \mathrm{N}(\mu,\ 16)$ and $\mathrm{P}(X < 10) = 0.3$. Find $\mu$.

โ–ถ Show solution

$\sigma = 4$. $\Phi(z) = 0.3 \Rightarrow z = -0.5244$.

$\dfrac{10 - \mu}{4} = -0.5244 \Rightarrow \mu = 10 + 2.098 = 12.1$

Question 7

$X \sim \mathrm{N}(50,\ \sigma^2)$ and $\mathrm{P}(45 < X < 55) = 0.8$. Find $\sigma$.

โ–ถ Show solution

By symmetry, $0.1$ lies in each tail, so $\mathrm{P}(X < 55) = 0.9$.

$\dfrac{55 - 50}{\sigma} = 1.2816 \Rightarrow \sigma = 3.90$

Question 8

$X \sim \mathrm{N}(\mu,\ \sigma^2)$ with $\mathrm{P}(X > 80) = 0.15$ and $\mathrm{P}(X < 50) = 0.25$. Find $\mu$ and $\sigma$.

โ–ถ Show solution

$\Phi(z) = 0.85 \Rightarrow z = 1.0364$:   $80 - \mu = 1.0364\sigma$

$\Phi(z) = 0.25 \Rightarrow z = -0.6745$:   $50 - \mu = -0.6745\sigma$

Subtract: $30 = 1.7109\sigma \Rightarrow \sigma = 17.5$

$\mu = 80 - 1.0364 \times 17.53 = 61.8$

Question 9

$X \sim \mathrm{N}(64,\ 2.5^2)$. Find the interval, symmetrical about the mean, that contains the middle $90\%$ of the distribution.

โ–ถ Show solution

$5\%$ in each tail: $z = \pm 1.6449$.

$64 \pm 1.6449 \times 2.5 = 64 \pm 4.11$, so $59.9 < X < 68.1$.

Question 10

The lifetimes of a brand of battery are modelled by $L \sim \mathrm{N}(220,\ 15^2)$ hours.

(a) Find $\mathrm{P}(L > 250)$. (b) Find $\mathrm{P}(200 < L < 240)$. (c) The manufacturer wants only $2\%$ of batteries to fail before the guaranteed time $t$. Find $t$. (d) Find the probability that all four batteries in a pack last more than $200$ hours. (e) A battery has already lasted $230$ hours. Find the probability that it lasts more than $250$ hours.

โ–ถ Show solution

(a) $z = 2$: $1 - \Phi(2) = 0.0228$

(b) $z = \pm\dfrac{4}{3}$: $\Phi(1.333) - \Phi(-1.333) = 0.818$

(c) $\Phi(z) = 0.02 \Rightarrow z = -2.0537$; $t = 220 - 2.0537 \times 15 = 189$ hours (3 s.f.)

(d) $\mathrm{P}(L > 200) = \Phi(1.333) = 0.9088$; assuming independence, $0.9088^4 = 0.682$

(e) Conditional probability: $\mathrm{P}(L > 250 \mid L > 230) = \dfrac{\mathrm{P}(L > 250)}{\mathrm{P}(L > 230)}$, since $L > 250$ already implies $L > 230$.

$\mathrm{P}(L > 230) = 1 - \Phi(0.667) = 0.2525$, so the answer is $\dfrac{0.02275}{0.2525} = 0.0901$.

The Normal Distribution · A Level Maths Year 2 · Created with MathJax