Poisson to Normal Approximation

A2 · S2 · 11 min

When the mean of a Poisson distribution is large, adding up Poisson terms by hand becomes impractical: $P(X \le 40)$ for $X \sim \text{Po}(48)$ has forty-one terms. Fortunately, a Poisson distribution with a large mean is almost perfectly bell-shaped, so it can be replaced by a normal distribution with the same mean and variance. This approximation, with its continuity correction, is examined on almost every Paper 6, often inside a longer question or a hypothesis test.

Why the shape becomes normal

For small $\lambda$ the Poisson distribution is crowded against zero and skewed to the right. As $\lambda$ increases, zero becomes far away from the mean (measured in standard deviations, it is $\lambda/\sqrt{\lambda} = \sqrt{\lambda}$ standard deviations away), the skew fades and the bars trace out a symmetric bell.

There is a second way to see it. A $\text{Po}(20)$ count is the total of twenty independent $\text{Po}(1)$ counts, one per unit of the interval (see sums of Poisson variables). Totals of many independent pieces tend to be normally distributed. That is the same idea as the central limit theorem.

The graph shows $\text{Po}(20)$ as bars of width $1$ , with the curve of $N(20, 20)$ on top. The shaded area under the curve up to $15.5$ approximates the total area of the bars for $r = 0, 1, \dots, 15$ , that is $P(X \le 15)$ .

y = 20^floor(x + 0.5) exp(-20) / fact(floor(x + 0.5)) y = exp(-(x - 20)^2 / 40) / sqrt(40 pi) fill 5 15.5 y = exp(-(x - 20)^2 / 40) / sqrt(40 pi)

The approximation

A normal approximation must match the mean and the variance of the distribution it replaces. A Poisson distribution has both equal to $\lambda$ .

Key result

If $X \sim \text{Po}(\lambda)$ and $\lambda$ is large, then approximately

X \sim N(\lambda, \lambda).

The syllabus condition is $\lambda > 15$ (approximately). A continuity correction must be used.

Notice that the second parameter of the normal is the variance, $\lambda$ . The standard deviation used when standardising is $\sqrt{\lambda}$ .

The continuity correction

A Poisson variable takes only whole-number values; a normal variable is continuous. Each whole number $r$ is represented by the bar from $r - 0.5$ to $r + 0.5$ , so every boundary moves by a half.

Key result

With $X \sim \text{Po}(\lambda)$ and $Y \sim N(\lambda, \lambda)$ :

Poisson	Normal
$P(X \le 20)$	$P(Y < 20.5)$
$P(X < 20) = P(X \le 19)$	$P(Y < 19.5)$
$P(X \ge 20)$	$P(Y > 19.5)$
$P(X > 20) = P(X \ge 21)$	$P(Y > 20.5)$
$P(X = 20)$	$P(19.5 < Y < 20.5)$
$P(15 \le X \le 25)$	$P(14.5 < Y < 25.5)$

The safest way to get it right every time: first rewrite the Poisson event using only $\le$ and $\ge$ (so "fewer than 20" becomes $X \le 19$ ), then move the boundary half a unit outwards, to include the whole of the end bar.

Normal approximation to a Poisson

State the Poisson distribution with the correct mean for the interval: $X \sim \text{Po}(\lambda)$ .
Check $\lambda > 15$ and state the approximation: $X \approx N(\lambda, \lambda)$ .
Rewrite the event with $\le$ or $\ge$ , then apply the continuity correction.
Standardise: $z = \dfrac{\text{corrected value} - \lambda}{\sqrt{\lambda}}$ , to 3 decimal places.
Use the normal table, with a sketch if the region is not obvious.

Three probabilities

The number of cars arriving at a car park in a 10-minute period has the distribution $\text{Po}(25)$ . Use a suitable approximation to find the probability that, in a 10-minute period,

(a) at most $20$ cars arrive,

(b) more than $30$ cars arrive,

Solution

$X \sim \text{Po}(25)$ . Since $25 > 15$ , use $Y \sim N(25, 25)$ , so $\sigma = 5$ .

(a)

P(X \le 20) \approx P(Y < 20.5) = P\left(Z < \frac{20.5 - 25}{5}\right) = P(Z < -0.9) = 1 - \Phi(0.9) = 1 - 0.8159 = 0.184

(b) More than $30$ is $X \ge 31$ :

P(X \ge 31) \approx P(Y > 30.5) = P\left(Z > \frac{30.5 - 25}{5}\right) = P(Z > 1.1) = 1 - 0.8643 = 0.136

P(21.5 < Y < 27.5) = P(-0.7 < Z < 0.5) = \Phi(0.5) - (1 - \Phi(0.7)) = 0.6915 - 0.2420 = 0.4495 = 0.450

The exact Poisson answers are $0.185$ , $0.137$ and $0.453$ , so the approximation is good.

Rescaling first

A help desk receives calls at random at an average rate of $3.5$ per hour. Use a suitable approximation to find the probability that more than $35$ calls are received in an 8-hour working day.

Solution

For 8 hours, $\lambda = 8 \times 3.5 = 28$ . Let $X \sim \text{Po}(28)$ .

$\lambda = 28 > 15$ , so $X \approx Y \sim N(28, 28)$ .

More than $35$ means $X \ge 36$ :

P(X \ge 36) \approx P(Y > 35.5) = P\left(Z > \frac{35.5 - 28}{\sqrt{28}}\right) = P(Z > 1.417)

= 1 - \Phi(1.417) = 1 - 0.9218 = 0.0782

Working backwards

Two kinds of inverse question appear.

Finding a boundary. "Find the smallest stock so that the probability of running out is less than 5%." Set up the corrected inequality and use the critical value from the table.

Finding $\lambda$ . If a probability is given and $\lambda$ is unknown, standardising gives an equation in which $\lambda$ appears both as the mean and inside the square root. Let $u = \sqrt{\lambda}$ to turn it into a quadratic.

Smallest stock level

A garage sells tyres at random at an average rate of $6$ per day. It is open $5$ days a week and receives one delivery a week. Use a suitable approximation to find the smallest number of tyres the garage should have in stock at the start of a week so that the probability that demand exceeds stock during the week is less than $0.05$ .

Solution

Weekly demand $X \sim \text{Po}(30)$ , and $30 > 15$ , so $X \approx Y \sim N(30, 30)$ .

With stock $n$ , we need $P(X > n) < 0.05$ . Now $X > n$ means $X \ge n + 1$ , which is $Y > n + 0.5$ after the continuity correction:

P\left(Z > \frac{n + 0.5 - 30}{\sqrt{30}}\right) < 0.05 \;\Rightarrow\; \frac{n + 0.5 - 30}{\sqrt{30}} > 1.645

n > 30 - 0.5 + 1.645\sqrt{30} = 38.51

The smallest stock is $39$ tyres.

Finding an unknown mean

The number of emails received by an office in a day has the distribution $\text{Po}(\lambda)$ , where $\lambda > 15$ . Using a normal approximation, the probability of receiving more than $30$ emails in a day is $0.1$ . Find $\lambda$ .

Solution

$P(X > 30) = P(X \ge 31) \approx P(Y > 30.5)$ with $Y \sim N(\lambda, \lambda)$ .

$P(Y > 30.5) = 0.1$ , so $30.5$ is above the mean by $1.282$ standard deviations:

\frac{30.5 - \lambda}{\sqrt{\lambda}} = 1.282

Let $u = \sqrt{\lambda}$ : $30.5 - u^2 = 1.282u$ , so $u^2 + 1.282u - 30.5 = 0$ .

u = \frac{-1.282 + \sqrt{1.282^2 + 4(30.5)}}{2} = \frac{-1.282 + 11.120}{2} = 4.919

The negative root is rejected because $u = \sqrt{\lambda} > 0$ . So $\lambda = 4.919^2 = 24.2$ (3 s.f.), which does satisfy $\lambda > 15$ .

Common mistakes

Using $\lambda$ as the standard deviation. $N(25, 25)$ has $\sigma = 5$ . Divide by $\sqrt{\lambda}$ .
Missing or wrong-way continuity corrections. "More than 30" is $X \ge 31$ , corrected to $Y > 30.5$ , not $Y > 31.5$ or $Y > 30$ .
Approximating when $\lambda$ is small. For $\lambda \le 15$ calculate the Poisson probability exactly.
Forgetting to rescale $\lambda$ before checking the condition. A rate of $3.5$ per hour is too small, but $28$ per day is fine; check the condition on the mean for the interval in the question.
Keeping the negative root. In "find $\lambda$ " questions, $\sqrt{\lambda}$ must be positive.
Rounding $z$ too early. Give $z$ to 3 decimal places and use the "add" columns of the table.

Exam tip

State both distributions: " $X \sim \text{Po}(28)$ , approximated by $N(28, 28)$ ". The mark scheme has a mark for the correct normal parameters and a separate mark for the continuity correction.
Show the standardisation in full, with the corrected value visible: $\dfrac{35.5 - 28}{\sqrt{28}}$ . A wrong continuity correction then loses one mark rather than several.
"Justify the use of a normal approximation": " $\lambda = 28 > 15$ ".
Approximations are always approximate. If you check your answer by working out the exact Poisson probability, do not write the exact value as your answer when an approximation was asked for.
In "smallest stock" or "least $n$ " questions, finish with a whole number and make the direction of rounding match the inequality.

Summary

For $X \sim \text{Po}(\lambda)$ with $\lambda > 15$ : $X \approx N(\lambda, \lambda)$ .
Standardise with $\sqrt{\lambda}$ : $z = \dfrac{x - \lambda}{\sqrt{\lambda}}$ .
Always use a continuity correction: rewrite with $\le$ or $\ge$ , then extend by $0.5$ outward.
Rescale $\lambda$ to the interval first, then check $\lambda > 15$ .
Inverse problems: use a critical $z$ from the table; for unknown $\lambda$ , solve a quadratic in $\sqrt{\lambda}$ .

Practice questions

Question

$X \sim \text{Po}(18)$ . Use a suitable approximation to find (a) $P(X \ge 20)$ , (b) $P(X < 15)$ .
Explain why a normal approximation should not be used for $X \sim \text{Po}(4)$ , and find $P(X \ge 2)$ exactly.
Vehicles pass a point on a road at random at an average rate of $1.2$ per minute. Use a suitable approximation to find the probability that between $30$ and $40$ vehicles inclusive pass in a 30-minute period.
$X \sim \text{Po}(50)$ . Use a normal approximation to estimate $P(X = 50)$ , and compare with the exact value $0.0563$ .
$X \sim \text{Po}(\lambda)$ . Using a normal approximation, $P(X < 20) = 0.05$ . Find $\lambda$ .
A football team scores goals at random at an average rate of $2.5$ per match. Find the probability that the team scores more than $100$ goals in a season of $38$ matches.
Requests arrive at a web server at random at an average rate of $0.8$ per minute. Find the probability that fewer than $40$ requests arrive in an hour, and state one assumption needed for your calculation to be valid.
A radioactive source emits particles at random at an average rate of $6$ per 10 seconds. (a) Find the probability that at least $40$ particles are emitted in a one-minute period. (b) Five separate one-minute periods are chosen. Find the probability that at least $40$ particles are emitted in exactly two of them.

Answers

$Y \sim N(18, 18)$ , $\sqrt{18} = 4.2426$ . (a) $P(X \ge 20) \approx P(Y > 19.5) = P(Z > 0.354) = 1 - 0.6383 = 0.362$ . (b) $P(X < 15) = P(X \le 14) \approx P(Y < 14.5) = P(Z < -0.825) = 1 - 0.7953 = 0.205$ .
$\lambda = 4$ is not greater than $15$ ; the distribution is skewed, so a normal curve is a poor fit. $P(X \ge 2) = 1 - e^{-4}(1 + 4) = 1 - 5e^{-4} = 0.908$ .
$\lambda = 30 \times 1.2 = 36$ , $Y \sim N(36, 36)$ , $\sigma = 6$ . $P(30 \le X \le 40) \approx P(29.5 < Y < 40.5) = P(-1.083 < Z < 0.75)$ $= \Phi(0.75) - (1 - \Phi(1.083)) = 0.7734 - (1 - 0.8606) = 0.7734 - 0.1394 = 0.634$ .
$P(49.5 < Y < 50.5)$ with $Y \sim N(50, 50)$ : $z = \pm\dfrac{0.5}{\sqrt{50}} = \pm 0.071$ . $= 2\Phi(0.071) - 1 = 2(0.5283) - 1 = 0.0566$ . The exact value is $0.0563$ , so the approximation is very close (error about $0.5\%$ ).
$P(X \le 19) \approx P(Y < 19.5) = 0.05$ , so $\dfrac{19.5 - \lambda}{\sqrt{\lambda}} = -1.645$ . With $u = \sqrt{\lambda}$ : $u^2 - 1.645u - 19.5 = 0$ , $u = \dfrac{1.645 + \sqrt{2.706 + 78}}{2} = 5.3144$ . $\lambda = 5.3144^2 = 28.2$ (3 s.f.).
$\lambda = 38 \times 2.5 = 95$ , $Y \sim N(95, 95)$ . $P(X > 100) = P(X \ge 101) \approx P(Y > 100.5) = P\left(Z > \dfrac{5.5}{\sqrt{95}}\right) = P(Z > 0.564) = 1 - 0.7136 = 0.286$ .
$\lambda = 60 \times 0.8 = 48$ , $Y \sim N(48, 48)$ . $P(X < 40) = P(X \le 39) \approx P(Y < 39.5) = P\left(Z < \dfrac{-8.5}{\sqrt{48}}\right) = P(Z < -1.227) = 1 - 0.8901 = 0.110$ . Assumption: requests arrive independently of each other at a constant average rate throughout the hour.
(a) Per minute, $\lambda = 36$ ; $Y \sim N(36, 36)$ . $P(X \ge 40) \approx P(Y > 39.5) = P(Z > 0.583) = 1 - 0.7201 = 0.2799$ , so $0.280$ (3 s.f.). (b) $N \sim B(5, 0.2799)$ : $P(N = 2) = \binom{5}{2}(0.2799)^2(0.7201)^3 = 10 \times 0.078344 \times 0.37341 = 0.293$ .

Why the shape becomes normal

The approximation

The continuity correction

Working backwards

Practice questions

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