Mini-Lesson

Statistical distributions

2 missions. Guess, learn, then check. Your place is saved.

Mission 1Guess all 10 four-option questions. Chance of getting 5 or more?
Mission 2Heights, test scores, bag masses: why the same bell shape?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Guess all 10 four-option questions. Chance of getting 5 or more?

Make a guess: Guess all 10 four-option questions. Chance of getting 5 or more?

Statistical distributions · binomial

Discrete random variables

A random variable $X$ X takes numerical values by chance. It is discrete if you can list its values.
Its probability distribution gives $P (X = x)$ P(X = x) for every value $x$ x, as a table or a formula. The probabilities must add to 1:

\sum P (X = x) = 1

A discrete uniform distribution gives every value the same probability, e.g. the score on a fair die: $P (X = x) = \frac{1}{6}$ P(X = x) = 1/6 for $x = 1, 2, \dots, 6$ x = 1, 2, …, 6.
Capital $X$ X is the variable; lower-case $x$ x is a value it takes.

Calculate

?The discrete random variable

X

has probability distribution

P (X = x) = k (x + 2)

for

x = 1, 2, 3, 4

. Find

P (X \geq 3)

. Give your answer to 3 decimal places.

Statistical distributions · binomial

The binomial distribution

$X \sim B (n, p)$ X ~ B(n, p) counts the successes in $n$ n trials when:
there is a fixed number of trials $n$ n; each trial has two outcomes (success or failure); trials are independent; the probability of success $p$ p is constant.
Then, with $q = 1 - p$ q = 1 − p:

P (X = x) = (x n) p^{x} (1 - p)^{n - x}, x = 0, 1, \dots, n

On a tree diagram every path with $x$ x successes has probability $p^{x} q^{n - x}$ p^x q^(n − x), and there are $(x n)$ C(n, x) such paths: these are the terms of the expansion of $(q + p)^{n}$ (q + p)ⁿ.
In context, state the conditions as assumptions, e.g. "seeds germinate independently of each other".

Worked example Statistical distributions · binomial

Using the formula and the calculator

Question

X \sim B (12, 0.3)

. Find

P (X = 4)

and

P (X \leq 4)

, to 4 decimal places.

1
Use the formula: $P (X = 4) = (4 12) (0.3)^{4} (0.7)^{8}$ P(X = 4) = C(12, 4)(0.3)⁴(0.7)⁸
2
Evaluate: $495 \times 0.0081 \times 0.05764801 = 0.2311$ 495 × 0.0081 × 0.05764801 = 0.2311
3
For $P (X \leq 4)$ P(X ≤ 4) use the calculator's binomial cumulative function with $n = 12$ n = 12, $p = 0.3$ p = 0.3, $x = 4$ x = 4: $0.7237$ 0.7237

Answer

P (X = 4) = 0.2311

and

P (X \leq 4) = 0.7237

.

⚠️ Watch out

"binomial PD" gives one value

P (X = x)

; "binomial CD" gives

P (X \leq x)

. Check which you need.

Sort it

?Can

X

be modelled by a binomial distribution?

Tap a statement, then tap the group it belongs in. Then press Check.

Statistical distributions · binomial

Cumulative probabilities

Your calculator gives $P (X \leq x)$ P(X ≤ x) directly. Rewrite every other inequality in that form:
$P (X < x) = P (X \leq x - 1)$ P(X < x) = P(X ≤ x − 1)
$P (X \geq x) = 1 - P (X \leq x - 1)$ P(X ≥ x) = 1 − P(X ≤ x − 1) and $P (X > x) = 1 - P (X \leq x)$ P(X > x) = 1 − P(X ≤ x)
$P (a \leq X \leq b) = P (X \leq b) - P (X \leq a - 1)$ P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a − 1)
For example, if $X \sim B (8, 0.4)$ X ~ B(8, 0.4), then $P (X \geq 3) = 1 - P (X \leq 2) = 1 - 0.3154 = 0.6846$ P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − 0.3154 = 0.6846.
You expect about $n p$ np successes, e.g. $20 \times 0.3 = 6$ 20 × 0.3 = 6 for $B (20, 0.3)$ B(20, 0.3). In a 2-tail test, compare the result with $n p$ np to choose the tail.

Quick check

?

X \sim B (15, 0.4)

. Find

P (X \geq 6)

.

Calculate

?Seeds of a certain type germinate with probability 0.85. Tom plants 20 seeds. Assuming the seeds germinate independently, find the probability that at least 15 and at most 18 of them germinate. Give your answer to 3 significant figures.

Mission check

?Guess all 10 four-option questions. Chance of getting 5 or more?

Mission 2

Heights, test scores, bag masses: why the same bell shape?

Make a guess: Heights, test scores, bag masses: why the same bell shape?

Statistical distributions · the normal distribution

The normal distribution as a model (A-level only)

A normal distribution models a continuous variable with a symmetric bell-shaped curve. Write $X \sim N (μ, σ^{2})$ X ~ N(μ, σ²): the second number is the variance.
Mean = median = mode $= μ$ = μ. The points of inflection are at $x = μ \pm σ$ x = μ ± σ.
Probability is area under the curve; the total area is 1. $P (X = a) = 0$ P(X = a) = 0, so $P (X < a) = P (X \leq a)$ P(X < a) = P(X ≤ a).
About 68% (two-thirds) of values lie within $μ \pm σ$ μ ± σ, about 95% within $μ \pm 2 σ$ μ ± 2σ, almost all within $μ \pm 3 σ$ μ ± 3σ.

Quick check

?

X \sim N (50, 16)

. Where are the points of inflection on the curve of

X

?

Statistical distributions · the normal distribution

Standardising and working backwards

Any normal variable can be turned into the standard normal $Z \sim N (0, 1)$ Z ~ N(0, 1):

Z = \frac{X - μ}{σ}

$z$ z counts standard deviations from the mean. $P (Z < z)$ P(Z < z) is written $Φ (z)$ Phi(z).
Your calculator's normal cdf gives probabilities directly. Its inverse normal finds the value with a given area to its left.
If $μ$ μ or $σ$ σ is unknown, standardise and use an inverse normal $z$ z-value: $x = μ + z σ$ x = μ + zσ.

Worked example Statistical distributions · the normal distribution

A probability between two values

Question

The masses of bags of sugar,

X

grams, are modelled by

X \sim N (506, 4^{2})

. Find

P (500 < X < 510)

.

1
Standardise each end: $z = \frac{500 - 506}{4} = - 1.5$ z = (500 − 506)/4 = −1.5 and $z = \frac{510 - 506}{4} = 1$ z = (510 − 506)/4 = 1
2
Write as areas to the left: $P (- 1.5 < Z < 1) = Φ (1) - Φ (- 1.5)$ P(−1.5 < Z < 1) = Phi(1) − Phi(−1.5)
3
Use the calculator: $0.84134 - 0.06681$ 0.84134 − 0.06681

Answer

P (500 < X < 510) = 0.7745

(4 d.p.)

⚠️ Watch out

with a calculator's normal cdf you can enter 500, 510,

μ = 506

and

σ = 4

directly. Entering the variance 16 as

σ

gives a wrong answer.

Worked example Statistical distributions · the normal distribution

Finding $μ$ μ and $σ$ σ together

Question

Times to finish a puzzle,

X

minutes, are normally distributed. 20% of people take less than 30 minutes and 10% take more than 60 minutes. Find

μ

and

σ

.

1
Inverse normal for each area to the left: $P (Z < - 0.8416) = 0.2$ P(Z < −0.8416) = 0.2 and $P (Z < 1.2816) = 0.9$ P(Z < 1.2816) = 0.9
2
Write $x = μ + z σ$ x = μ + zσ twice: $μ - 0.8416 σ = 30$ μ − 0.8416σ = 30 and $μ + 1.2816 σ = 60$ μ + 1.2816σ = 60
3
Subtract to eliminate $μ$ μ: $2.1232 σ = 30$ 2.1232σ = 30, so $σ = 14.13$ σ = 14.13
4
Substitute back: $μ = 30 + 0.8416 \times 14.13 = 41.89$ μ = 30 + 0.8416 × 14.13 = 41.89

Answer

μ = 41.9

minutes and

σ = 14.1

minutes (3 s.f.)

⚠️ Watch out

"10% take more than 60" means 90% take less, so use

z = 1.2816

, not

- 1.2816

.

Calculate

?A commuter's journey time,

T

minutes, is modelled by

T \sim N (34, 5^{2})

. She is late if the journey takes more than 40 minutes. Find the expected number of late days in 20 independent working days. Give your answer to 3 significant figures.

Statistical distributions · the normal distribution

Binomial or normal?

Binomial $B (n, p)$ B(n, p): a count of successes in a fixed number of independent trials with constant $p$ p. Normal: a continuous measurement, roughly symmetric about its mean.
A binomial count is a total of $n$ n small yes/no effects. For large $n$ n with $p$ p near 0.5 its bar chart is close to a bell: $B (n, p) \approx N (n p, n p (1 - p))$ B(n, p) ≈ N(np, np(1−p)).
A normal model fits badly if the data are strongly skewed or impossible values get real probability.
For a calculation, use a continuity correction: $P (X \leq 30)$ P(X ≤ 30) becomes $P (Y < 30.5)$ P(Y < 30.5).

Calculate

?

X \sim B (80, 0.45)

. Use a suitable normal approximation, with a continuity correction, to estimate

P (X \leq 30)

. Give your answer to 3 significant figures.

Sort it

?Which model fits each variable best?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?Adult heights, test scores and bag masses all make the same bell shape. Why?

Recap

Big ideas

A discrete probability distribution lists $P (X = x)$ P(X = x) for each value; the probabilities add to 1.
$X \sim B (n, p)$ X ~ B(n, p) needs a fixed $n$ n, two outcomes, independent trials and constant $p$ p; $P (X = x) = (x n) p^{x} (1 - p)^{n - x}$ P(X = x) = C(n, x)p^x(1 − p)^(n − x).
Turn every inequality into $P (X \leq x)$ P(X ≤ x) form: $P (X \geq x) = 1 - P (X \leq x - 1)$ P(X ≥ x) = 1 − P(X ≤ x − 1).
$X \sim N (μ, σ^{2})$ X ~ N(μ, σ²): a symmetric bell about $μ$ μ, inflection points at $μ \pm σ$ μ ± σ, probability = area. About 68%, 95% and 99.7% of values lie within 1, 2 and 3 standard deviations.
Standardise with $Z = \frac{X - μ}{σ}$ Z = (X − μ)/σ; use the inverse normal and $x = μ + z σ$ x = μ + zσ to find a value, $μ$ μ or $σ$ σ.
Binomial for counts, normal for continuous measurements; $B (n, p)$ B(n, p) looks like $N (n p, n p (1 - p))$ N(np, np(1−p)) for large $n$ n and $p$ p near 0.5.
Approximating a binomial by a normal needs a continuity correction.

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Mission 1Guess all 10 four-option questions. Chance of getting 5 or more?
Mission 2Heights, test scores, bag masses: why the same bell shape?

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Specification points and sources for this lesson are listed at the bottom of the page.

Statistical distributions

Guess all 10 four-option questions. Chance of getting 5 or more?

Discrete random variables

The binomial distribution

Using the formula and the calculator

Cumulative probabilities

Heights, test scores, bag masses: why the same bell shape?

The normal distribution as a model (A-level only)

Standardising and working backwards

A probability between two values

Finding μμ and σσ together

Binomial or normal?

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Finding $μ$ μ and $σ$ σ together