2.04StatisticsStretch

Probability Distributions

Q: How do I know whether to use binomial or normal?

Binomial for counting successes in a fixed number of independent trials; normal for continuous measurements that are roughly symmetric about a mean.

Q: Why is P(X = a) = 0 for a normal distribution?

Probability is area under the curve, and a single point has no width, so no area.

Q: What is the mean of a binomial distribution?

np, the expected number of successes. Its variance is np(1 - p).

Q: When can the normal approximate the binomial?

When n is large and p is not close to 0 or 1, so the binomial is roughly symmetric. Use a continuity correction.

A $\textbf{probability distribution}$ lists every value a random quantity can take together with how likely each is. Two distributions do most of the work at A-Level: the $\textbf{binomial}$ , which counts successes in a fixed number of independent trials, and the $\textbf{normal}$ , the bell-shaped curve that models heights, measurement errors and countless other continuous quantities.

50 min Probability

The big picture

A distribution is a model, and every model rests on assumptions. The binomial model $\mathrm{B}(n, p)$ needs a fixed number of trials, two outcomes, a constant probability of success and independent trials — and exam questions routinely ask you to state these in context or criticise them. The normal model $\mathrm{N}(\mu, \sigma^{2})$ describes symmetric continuous data, where probabilities are areas under a curve and single values have probability zero. Working fluently with both — calculator functions for probabilities, standardising to $Z$ , and finding unknown parameters from given probabilities — is what makes hypothesis testing possible in the next lesson.

What you'll be able to do

Work with discrete random variables and their probability distributions
State the conditions for a binomial model and calculate binomial probabilities
Understand the normal distribution, its shape and parameters, and find probabilities
Standardise using $Z = \dfrac{X - \mu}{\sigma}$ and find unknown values or parameters

Discrete random variables

A $\textbf{random variable}$ $X$ takes numerical values determined by chance. For a discrete random variable, list each value $x$ with $\mathrm{P}(X = x)$ . The probabilities must sum to 1.

The distribution may be given as a table or a formula, such as $\mathrm{P}(X = x) = kx$ for $x = 1, 2, 3, 4$ . Use $\sum \mathrm{P}(X = x) = 1$ to find an unknown constant.

A $\textbf{discrete uniform}$ distribution gives every value the same probability, like the score on a fair die.

Example 1

\mathrm{P}(X = x) = kx

for

x = 1, 2, 3, 4

. Find

k

and

\mathrm{P}(X \ge 3)

k(1 + 2 + 3 + 4) = 1

, so

k = 0.1

\mathrm{P}(X \ge 3) = 0.3 + 0.4

Answer

k = 0.1

;

\mathrm{P}(X \ge 3) = 0.7

The binomial distribution

$X \sim \mathrm{B}(n, p)$ counts the number of successes in $n$ trials when: there is a $\textbf{fixed number}$ of trials; each trial has $\textbf{two outcomes}$ ; the probability of success $p$ is $\textbf{constant}$ ; and the trials are $\textbf{independent}$ .

The probability of exactly $r$ successes is $\binom{n}{r}p^{r}(1 - p)^{n-r}$ — choose which trials succeed, then multiply their probabilities.

Cumulative probabilities $\mathrm{P}(X \le r)$ come from the calculator. Translate inequalities carefully: $\mathrm{P}(X \ge 5) = 1 - \mathrm{P}(X \le 4)$ and $\mathrm{P}(X < 5) = \mathrm{P}(X \le 4)$ .

\mathrm{P}(X = r) = \binom{n}{r}p^{r}(1 - p)^{n - r}

Binomial probability.

Example 1Each seed germinates with probability 0.3, independently. For 12 seeds, find

\mathrm{P}(X = 4)

and

\mathrm{P}(X \ge 5)

X \sim \mathrm{B}(12, 0.3)

\mathrm{P}(X = 4) = \binom{12}{4}(0.3)^{4}(0.7)^{8} = 0.2311

\mathrm{P}(X \ge 5) = 1 - \mathrm{P}(X \le 4) = 1 - 0.7237

Answer

0.231

and

0.276

(3 s.f.)

Tip — When asked to criticise a binomial model, pick the condition most likely to fail in the context — often independence (seeds in the same tray share conditions) or constant $p$ .

The normal distribution

$X \sim \mathrm{N}(\mu, \sigma^{2})$ is continuous, symmetric about the mean $\mu$ , and bell-shaped, with points of inflection at $\mu \pm \sigma$ . About 68% of values lie within one standard deviation of the mean, 95% within two, and 99.7% within three.

Probabilities are areas under the curve. Since a single value has zero area, $\mathrm{P}(X \le a) = \mathrm{P}(X < a)$ .

Use the calculator’s normal cumulative distribution function directly, or standardise: $Z = \dfrac{X - \mu}{\sigma} \sim \mathrm{N}(0, 1)$ .

Z = \dfrac{X - \mu}{\sigma} \sim \mathrm{N}(0, 1^{2})

Standardising.

Example 1Heights of adult men are modelled as

\mathrm{N}(170, 8^{2})

cm. Find

\mathrm{P}(X > 180)

and

\mathrm{P}(160 < X < 175)

\mathrm{P}(X > 180) = \mathrm{P}(Z > 1.25) = 0.1056

\mathrm{P}(160 < X < 175) = \mathrm{P}(-1.25 < Z < 0.625) = 0.6284

Answer

0.106

and

0.628

(3 s.f.)

Why this works

The second parameter in $\mathrm{N}(\mu, \sigma^{2})$ is the $\textit{variance}$ . $\mathrm{N}(170, 64)$ has standard deviation 8, not 64 — a mix-up that costs marks every year.

Working backwards

To find a value $a$ with a given probability, use the inverse normal function: if $\mathrm{P}(X < a) = 0.9$ , then $a = \mu + 1.2816\sigma$ .

To find an unknown $\mu$ or $\sigma$ , convert the probability to a $z$ -value, write $\dfrac{a - \mu}{\sigma} = z$ , and solve. Two such conditions give simultaneous equations for both parameters.

A binomial distribution with large $n$ may have a shape that can be approximated by a normal model. For OCR H240 you need to recognise this modelling connection; calculations using the normal approximation to the binomial, including continuity-correction calculations, are excluded. Use exact binomial calculator functions for binomial probabilities and tests.

Example 1The masses of apples are normally distributed with standard deviation 12 g, and 20% weigh more than 150 g. Find the mean.

\mathrm{P}(X > 150) = 0.2

means

\mathrm{P}(Z > z) = 0.2

, so

z = 0.8416

\dfrac{150 - \mu}{12} = 0.8416

, so

\mu = 150 - 10.10

Answer

\mu \approx 139.9

Tip — Sketch a normal curve and shade the region every time. It shows immediately whether the $z$ -value should be positive or negative.

Think like an examiner

State binomial conditions in the context of the question, not generically.
Show the distribution, e.g. " $X \sim \mathrm{B}(12, 0.3)$ ", before calculating.
Convert $\ge$ , $>$ and $<$ correctly to cumulative probabilities for discrete distributions.
Use a sketch and standardisation when finding unknown normal parameters.

Common misconceptions

Writing

\mathrm{P}(X \ge 5) = 1 - \mathrm{P}(X \le 5)

for a binomial.

\mathrm{P}(X \ge 5) = 1 - \mathrm{P}(X \le 4)

Treating

\sigma^{2}

\mathrm{N}(\mu, \sigma^{2})

as the standard deviation.

Square-root the variance to get

\sigma

Using a binomial model when trials are clearly dependent.

Check and state all four conditions in context.

Giving a positive

z

for a value below the mean.

Values below the mean standardise to negative

z

Distributions

\sum \mathrm{P}(X = x) = 1

Discrete distributions.

\mathrm{P}(X = r) = \binom{n}{r}p^{r}q^{n-r}

Binomial,

q = 1 - p

Z = \tfrac{X - \mu}{\sigma}

Standard normal.

E(X) = np

Binomial mean.

Stretch yourself

A machine fills bags with a mass $X$ grams, where $X \sim \mathrm{N}(\mu, \sigma^{2})$ . 5% of bags weigh less than 495 g and 10% weigh more than 510 g. Find $\mu$ and $\sigma$ .

Hint — Convert each percentage to a $z$ -value (one negative, one positive), write two equations, and subtract.

Reveal solution

Solution

\mathrm{P}(X < 495) = 0.05

gives

\dfrac{495 - \mu}{\sigma} = -1.6449

, so

495 = \mu - 1.6449\sigma

\mathrm{P}(X > 510) = 0.10

gives

\dfrac{510 - \mu}{\sigma} = 1.2816

, so

510 = \mu + 1.2816\sigma

. Subtracting:

15 = 2.9265\sigma

, so

\sigma = 5.126

. Then

\mu = 510 - 1.2816 \times 5.126 = 503.4

. So

\mu \approx 503.4

g and

\sigma \approx 5.13

g. Check:

\mathrm{P}(X < 495) = \mathrm{P}\left(Z < \tfrac{495 - 503.43}{5.126}\right) = \mathrm{P}(Z < -1.645) \approx 0.05

✓.

\blacksquare

Questions students ask

How do I know whether to use binomial or normal?

Binomial for counting successes in a fixed number of independent trials; normal for continuous measurements that are roughly symmetric about a mean.

Why is

\mathrm{P}(X = a) = 0

for a normal distribution?

Probability is area under the curve, and a single point has no width, so no area.

What is the mean of a binomial distribution?

np

, the expected number of successes. Its variance is

np(1 - p)

When can the normal approximate the binomial?

When

n

is large and

p

is not close to 0 or 1, so the binomial is roughly symmetric. Use a continuity correction.

Key takeaways

Discrete distributions list values with probabilities summing to 1.
$\mathrm{B}(n, p)$ : fixed trials, two outcomes, constant $p$ , independence.
Translate binomial inequalities carefully using $\mathrm{P}(X \le r)$ .
$\mathrm{N}(\mu, \sigma^{2})$ is continuous and symmetric; probabilities are areas.
Standardise with $Z = (X - \mu)/\sigma$ to find unknown values or parameters.