By the end of the course you have two distributions to choose from. The DfE content (area N) asks you to select an appropriate probability distribution for a context, with appropriate reasoning — so the choice itself earns marks.
| Binomial $\mathrm{B}(n,\,p)$ | Normal $\mathrm{N}(\mu,\,\sigma^2)$ | |
|---|---|---|
| Type | Discrete — counts $0, 1, \ldots, n$ | Continuous — measurements |
| Conditions | Fixed number of trials $n$; each trial success/failure; trials independent; constant $p$ | Symmetrical, bell-shaped data clustered about the mean |
| Mean | $np$ | $\mu$ |
| Variance | $np(1-p)$ | $\sigma^2$ |
| Typical context | Number of defective items in a sample; number of correct guesses | Heights, masses, times, measurement errors |
Suggest a suitable distribution for each variable, or explain why neither is suitable.
(a) The number of faulty bulbs in a box of $24$. (b) The volume of drink in a $330$ ml can. (c) The number of rolls of a die until the first six. (d) The time a customer waits in a queue.
When $n$ is large, binomial probabilities are tedious to calculate and the bar chart of $\mathrm{B}(n,\,p)$ looks very like a normal curve. We can then approximate it by a normal distribution with the same mean and variance.
$X \approx Y$ where $Y \sim \mathrm{N}\big(np,\ np(1-p)\big)$
Rule of thumb: $np > 5$ and $nq > 5$, where $q = 1 - p$
A binomial variable takes whole-number values; a normal variable is continuous. Each integer $k$ is represented by the interval from $k - 0.5$ to $k + 0.5$. Think of each binomial bar as having width $1$, centred on $k$.
| Binomial $X$ | Integers included | Normal $Y$ |
|---|---|---|
| $\mathrm{P}(X \leqslant 20)$ | $\ldots, 19, 20$ | $\mathrm{P}(Y < 20.5)$ |
| $\mathrm{P}(X < 20)$ | $\ldots, 18, 19$ | $\mathrm{P}(Y < 19.5)$ |
| $\mathrm{P}(X \geqslant 20)$ | $20, 21, \ldots$ | $\mathrm{P}(Y > 19.5)$ |
| $\mathrm{P}(X > 20)$ | $21, 22, \ldots$ | $\mathrm{P}(Y > 20.5)$ |
| $\mathrm{P}(X = 20)$ | $20$ | $\mathrm{P}(19.5 < Y < 20.5)$ |
$X \sim \mathrm{B}(60,\ 0.4)$. Use a normal approximation to find (a) $\mathrm{P}(X \leqslant 20)$, (b) $\mathrm{P}(X = 25)$, (c) $\mathrm{P}(20 < X < 30)$.
$30\%$ of customers at a café order a cake. Use a suitable approximation to find the probability that, of $200$ customers, at least $70$ order a cake.
A model is only useful if it describes reality. Ways to judge a model:
- Compare frequencies: calculate the frequencies the model predicts and set them beside the observed ones.
- Compare mean and variance: for a binomial model the variance $np(1-p)$ must be less than the mean $np$. If the data's variance is much larger, the binomial is a poor fit.
- Normal checks: is the data roughly symmetrical? Is about $68\%$ within $1$ s.d. and $95\%$ within $2$ s.d. of the mean? Does the model give a meaningful probability to impossible values (e.g. negative times)?
- Question the assumptions: are trials really independent? Is $p$ really constant?
A supplier claims $70\%$ of its seeds germinate. A gardener plants $6$ seeds in each of $80$ trays and counts how many germinate.
| Germinated | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Trays | 0 | 1 | 3 | 9 | 22 | 28 | 17 |
Compare the data with $\mathrm{B}(6,\ 0.7)$ and with a binomial model using the observed proportion.
The times customers spend waiting on a helpline have mean $4.2$ minutes and standard deviation $3.8$ minutes. Comment on modelling the waiting time by $\mathrm{N}(4.2,\ 3.8^2)$.
Binomial
Discrete count; fixed $n$, independent trials, constant $p$.
Normal
Continuous, symmetrical, bell-shaped.
Approximation
$\mathrm{B}(n,p) \approx \mathrm{N}\big(np,\ np(1-p)\big)$
Conditions
$n$ large, $p$ near $\tfrac12$; $np > 5$ and $nq > 5$.
Continuity correction
Integer $k$ becomes the interval $k \pm 0.5$.
$\leqslant k$ / $\geqslant k$
$Y < k + 0.5$ / $Y > k - 0.5$
$< k$ / $> k$
$Y < k - 0.5$ / $Y > k + 0.5$
Binomial check
Variance should be less than mean; compare expected and observed frequencies.
Normal check
Symmetry; $68\%$ / $95\%$ rule; no impossible values.
Ten questions, from routine to exam-style. Give answers to $3$ significant figures unless told otherwise.
State, with a reason, whether a binomial or normal model is more suitable for: (a) the number of left-handed students in a class of $30$; (b) the lengths of leaves from an oak tree.
▶ Show solution
(a) Binomial $\mathrm{B}(30,\,p)$: a count from a fixed number of people, each left-handed or not, assumed independent with constant $p$.
(b) Normal: length is a continuous measurement likely to be symmetrical about a mean.
For which of these is a normal approximation suitable? (a) $\mathrm{B}(40,\ 0.1)$ (b) $\mathrm{B}(40,\ 0.45)$ (c) $\mathrm{B}(500,\ 0.02)$
▶ Show solution
(a) $np = 4 < 5$: not suitable — the distribution is too skewed.
(b) $np = 18$, $nq = 22$: suitable.
(c) $np = 10$, $nq = 490$: the rule of thumb is satisfied, so it is acceptable, though $p$ is far from $\tfrac12$ and the approximation will be less accurate in the tails.
$X \sim \mathrm{B}(n,\,p)$ is approximated by $Y$. Write the continuity-corrected version of (a) $\mathrm{P}(X < 15)$, (b) $\mathrm{P}(X \geqslant 15)$, (c) $\mathrm{P}(X = 15)$, (d) $\mathrm{P}(12 \leqslant X < 18)$.
▶ Show solution
(a) $X \leqslant 14$: $\mathrm{P}(Y < 14.5)$
(b) $X = 15, 16, \ldots$: $\mathrm{P}(Y > 14.5)$
(c) $\mathrm{P}(14.5 < Y < 15.5)$
(d) $X = 12, \ldots, 17$: $\mathrm{P}(11.5 < Y < 17.5)$
A fair coin is tossed $50$ times. Use a normal approximation to find the probability of at least $30$ heads.
▶ Show solution
$X \sim \mathrm{B}(50,\ 0.5)$; $\mu = 25$, $\sigma^2 = 12.5$, $\sigma = 3.536$.
$\mathrm{P}(X \geqslant 30) \approx \mathrm{P}(Y > 29.5) = 1 - \Phi(1.273) = 0.102$
(Exact: $0.101$.)
$X \sim \mathrm{B}(100,\ 0.35)$. Use a normal approximation to find $\mathrm{P}(30 \leqslant X \leqslant 40)$.
▶ Show solution
$\mu = 35$, $\sigma^2 = 22.75$, $\sigma = 4.770$.
$\mathrm{P}(29.5 < Y < 40.5) = \Phi(1.153) - \Phi(-1.153) = 0.751$
(Exact: $0.751$.)
$X \sim \mathrm{B}(80,\ 0.6)$. Use a normal approximation to estimate $\mathrm{P}(X = 50)$, and compare with the exact value.
▶ Show solution
$\mu = 48$, $\sigma^2 = 19.2$, $\sigma = 4.382$.
$\mathrm{P}(49.5 < Y < 50.5) = \Phi(0.571) - \Phi(0.342) = 0.0819$
Exact: $\binom{80}{50}(0.6)^{50}(0.4)^{30} = 0.0827$ — the approximation is good.
The number of cracked eggs in each of $50$ boxes of $10$ eggs is recorded.
| Cracked | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Boxes | 22 | 8 | 5 | 4 | 3 | 3 | 2 | 1 | 1 | 1 | 0 |
The mean is $1.86$ and the variance is $5.60$. Explain why $\mathrm{B}(10,\,p)$ is not a good model.
▶ Show solution
Estimate $p = 1.86 / 10 = 0.186$. A binomial with this $p$ has variance $10 \times 0.186 \times 0.814 = 1.51$, far less than the observed $5.60$.
Also $\mathrm{B}(10,\ 0.186)$ predicts about $50 \times 0.814^{10} = 6.4$ boxes with no cracks, against $22$ observed.
The eggs in a box are not independent: if a box is dropped, several eggs crack together. So the binomial assumption of independent trials fails.
The masses of $200$ apples have mean $152$ g and standard deviation $11$ g. $136$ of the apples are between $141$ g and $163$ g, and $191$ are between $130$ g and $174$ g. Does a normal model seem reasonable?
▶ Show solution
$141$ to $163$ is $\mu \pm \sigma$: $\dfrac{136}{200} = 68\%$, matching the expected $68.3\%$.
$130$ to $174$ is $\mu \pm 2\sigma$: $\dfrac{191}{200} = 95.5\%$, matching the expected $95.4\%$.
A normal model $\mathrm{N}(152,\ 11^2)$ seems reasonable (a histogram would confirm symmetry).
An airline sells $200$ tickets for a flight with $190$ seats. Each passenger independently turns up with probability $0.92$. (a) Use a normal approximation to find the probability that more than $190$ passengers turn up. (b) The exact binomial probability is $0.0374$. Explain why the approximation is not very accurate here.
▶ Show solution
(a) $X \sim \mathrm{B}(200,\ 0.92)$: $np = 184$, $nq = 16$, $\sigma^2 = 14.72$, $\sigma = 3.837$.
$\mathrm{P}(X > 190) \approx \mathrm{P}(Y > 190.5) = 1 - \Phi(1.694) = 0.0451$
(b) $p = 0.92$ is far from $\tfrac12$, so the binomial is negatively skewed (it cannot exceed $200$, only $16$ above the mean). The symmetrical normal curve overestimates the upper tail.
A multiple-choice test has $40$ questions, each with $4$ options. A student guesses every answer at random. Let $X$ be the number correct.
(a) State the distribution of $X$ and give its mean and variance. (b) Explain why a normal approximation is suitable. (c) Use it to estimate $\mathrm{P}(X \geqslant 15)$. (d) The examiner wants the probability of passing by guessing to be less than $5\%$. Using the approximation, find the smallest suitable pass mark, and check it with the exact binomial distribution.
▶ Show solution
(a) $X \sim \mathrm{B}(40,\ 0.25)$; mean $10$, variance $40 \times 0.25 \times 0.75 = 7.5$.
(b) $np = 10 > 5$ and $nq = 30 > 5$.
(c) $Y \sim \mathrm{N}(10,\ 7.5)$, $\sigma = 2.739$. $\mathrm{P}(Y > 14.5) = 1 - \Phi(1.643) = 0.0502$.
(d) Need $\mathrm{P}(X \geqslant k) < 0.05$. From (c), $k = 15$ gives $0.0502$ — just too big.
$k = 16$: $\mathrm{P}(Y > 15.5) = 1 - \Phi(2.008) = 0.0223 < 0.05$ ✓. The pass mark should be $16$.
Exact check: $\mathrm{P}(X \geqslant 15) = 0.0544$ and $\mathrm{P}(X \geqslant 16) = 0.0262$, so $16$ is confirmed.