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 $probability distribution$ 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 $binomial$ , which counts successes in a fixed number of independent trials, and the $normal$ , 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 $B (n, p)$ 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 $N (μ, σ^{2})$ 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 $Z = \frac{X - μ}{σ}$ and find unknown values or parameters

Discrete random variables

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

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

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

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

, so

k = 0.1

P (X \geq 3) = 0.3 + 0.4

Answer

k = 0.1

;

P (X \geq 3) = 0.7

The binomial distribution

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

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

Cumulative probabilities $P (X \leq r)$ come from the calculator. Translate inequalities carefully: $P (X \geq 5) = 1 - P (X \leq 4)$ and $P (X < 5) = P (X \leq 4)$ .

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

Binomial probability.

X \sim B (12, 0.3)

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

P (X \geq 5) = 1 - P (X \leq 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

$X \sim N (μ, σ^{2})$ is continuous, symmetric about the mean $μ$ , and bell-shaped, with points of inflection at $μ \pm σ$ . 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, $P (X \leq a) = P (X < a)$ .

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

Z = \frac{X - μ}{σ} \sim N (0, 1^{2})

Standardising.

P (X > 180) = P (Z > 1.25) = 0.1056

P (160 < X < 175) = P (- 1.25 < Z < 0.625) = 0.6284

Answer

0.106

and

0.628

(3 s.f.)

Why this works

The second parameter in $N (μ, σ^{2})$ is the $variance$ . $N (170, 64)$ 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 $P (X < a) = 0.9$ , then $a = μ + 1.2816 σ$ .

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

The normal distribution can approximate $B (n, p)$ when $n$ is large and $p$ is close to 0.5, using $μ = n p$ and $σ^{2} = n p (1 - p)$ with a continuity correction.

P (X > 150) = 0.2

means

P (Z > z) = 0.2

, so

z = 0.8416

\frac{150 - μ}{12} = 0.8416

, so

μ = 150 - 10.10

Answer

μ \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. " $X \sim B (12, 0.3)$ ", before calculating.
Convert $\geq$ , $>$ and $<$ correctly to cumulative probabilities for discrete distributions.
Use a sketch and standardisation when finding unknown normal parameters.

Common misconceptions

Writing

P (X \geq 5) = 1 - P (X \leq 5)

for a binomial.

P (X \geq 5) = 1 - P (X \leq 4)

Treating

σ^{2}

N (μ, σ^{2})

as the standard deviation.

Square-root the variance to get

σ

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 P (X = x) = 1

Discrete distributions.

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

Binomial,

q = 1 - p

Z = \frac{X - μ}{σ}

Standard normal.

E (X) = n p

Binomial mean.

Stretch yourself

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

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

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

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?

n p

, the expected number of successes. Its variance is

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

How this fits the course

Build on

Probability Laws

Binomial Expansion

Leads to

Hypothesis Testing

Test yourself

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Next step

Practise this topic

Real past-paper questions on Probability Distributions, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

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Probability Distributions

Discrete random variables

The binomial distribution

The normal distribution

Working backwards

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Related topics

Probability Distributions

Discrete random variables

The binomial distribution

The normal distribution

Working backwards

Practise this topic

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