The Poisson and normal distributions · Probability & distributions

The binomial distribution counts successes in a fixed number of trials. Two other distributions appear in Further Maths: the Poisson for rare events, and the normal for measurements that cluster round an average.

The Poisson distribution

When events happen at random at an average rate $\lambda$ (per day, per page, per batch), the number that happen is Poisson:

The Poisson distributionλ is the mean number of events

It is also a good approximation to the binomial when $n$ is large and $p$ is small: use $\lambda = np$ . That saves working out large powers.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q12

In a bottle manufacturing company, it was observed that $5\%$ of the bottles manufactured were defective. In a random sample of 150 bottles manufactured, find the probability that: $[\text{Take } e = 2.7183]$

exactly 3;

between 3 and 6;

at most 4, bottles are defective.

Which distribution?
- Use the Poisson approximation with ${\lambda = np = 150 \times 0.05 = 7.5}$ .
- ${e^{-7.5} \approx 0.000553}$ .
Think first. n = 150 is large and p = 0.05 is small. What is λ?
Exactly 3
- ${P(3) = e^{-7.5} \times \frac{7.5^3}{3!}}$ .
- ${= 0.000553 \times 70.3125}$ .
- ${\approx 0.0389}$ .
Think first. e^(−λ) λ³ ÷ 3!
Between 3 and 6
- ${P(4) = e^{-7.5} \times \frac{7.5^4}{4!} \approx 0.0729}$ .
- ${P(5) = e^{-7.5} \times \frac{7.5^5}{5!} \approx 0.1094}$ .
- Add: ${0.1823}$ .
Think first. Between 3 and 6 means 4 or 5.
At most 4
- ${P(0) + P(1) + P(2) + P(3) + P(4)}$ .
- The five terms: ${0.0006}$ , ${0.0041}$ , ${0.0156}$ , ${0.0389}$ , ${0.0729}$ .
- Add them: ${0.1321}$ .

More: the Poisson distribution

NECO 2023 · Paper 1 · Q42One out of a thousand bulbs manufactured by a factory is found to be defective. If 2,000 bulbs were produced, find the probability …

The normal distribution

Heights, marks and weights often follow a normal distribution: a symmetric bell-shaped curve centred on the mean $\mu$ , with spread measured by the standard deviation $\sigma$ . The probability of a value between $x_1$ and $x_2$ is the area under the curve between them.

The normal curveChange x to z, then read the area from the table

Every normal distribution becomes the standard one (mean 0, standard deviation 1) when you change $x$ to $z = \dfrac{x - \mu}{\sigma}$ . Tables give $\Phi(z)$ , the area to the left of $z$ , so

P(x_1 < X < x_2) = \Phi(z_2) - \Phi(z_1)

Areas under the normal curveSet the mean, the spread and two values

0.5z₁ = (64 − 60) ÷ 82z₂ = (76 − 60) ÷ 80.2858P(64 < X < 76)

mean μ = 60standard deviation σ = 8x₁ = 64x₂ = 76

Change each value to a z-score, the number of standard deviations from the mean. Then the probability is the area between z₁ and z₂: Φ(2) − Φ(0.5) = 0.9772 − 0.6915 = 0.2858.

Worked example · NECO 2023

NECO 2023 · Paper 2 · Q15 (b)

The scores of 5000 students follow a normal distribution with mean 72 and variance 25. Determine the number of students who obtained scores between 80 and 84.

Mean and standard deviation
- ${\mu = 72}$ , and ${\sigma = \sqrt{25} = 5}$ .
Think first. The variance is 25. What is σ?
The z-scores
- ${z_1 = \frac{80 - 72}{5} = 1.6}$ .
- ${z_2 = \frac{84 - 72}{5} = 2.4}$ .
The probability
- ${\Phi(2.4) - \Phi(1.6) = 0.9918 - 0.9452}$ .
- ${= 0.0466}$ .
Think first. Read Φ(2.4) and Φ(1.6) from the table.
The number of students
- ${0.0466 \times 5000 = 233}$ students.

Your turn

WAEC 2023 · Paper 2 · Q12

Probability & distributions

(a)
A farmer finds out that for every 100 oranges he harvests, 15 are bad. If he selects 20 oranges from his farm, what is the probability that three of them are bad?
(b)
In an international market, the demand for mobile phones has a Poisson distribution with mean of $1.8$ . Find the probability that in a randomly chosen period, the demand is at most two.

Worked solution (try it first)

(a)

$p = \frac{15}{100} = 0.15$ and $n = 20$ : $P(3) = \binom{20}{3}(0.15)^3(0.85)^{17}$
$\approx 0.2428$ .

(b)

$\lambda = 1.8$ : $P(X \le 2) = e^{-1.8}\left(1 + 1.8 + \dfrac{1.8^2}{2}\right)$
$= 4.42e^{-1.8}$
$\approx 0.7306$ .

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