AS & A Level · Practice questions · Approximations: binomial to Poisson, Poisson to normal

Pick the shortcut, then justify it.

Six original Cambridge-style Paper 6 questions on approximating the binomial by the Poisson and the Poisson by the normal, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Check the conditions, choose, then correct.

Binomial, rare successes: \(n > 50\), \(np < 5\): use \(\text{Po}(np)\), no continuity correction.
\(p\) close to 1: count the failures instead.
Poisson, large mean: \(\lambda > 15\): use \(N(\lambda, \lambda)\), standard deviation \(\sqrt{\lambda}\).
Continuity correction: list the integers included, then go half a unit beyond the last one.
Justify: quote the numerical conditions in your answer.

Recall

[2 marks]

State the conditions under which \(B(n, p)\) may be approximated by a Poisson distribution, and the parameter of that Poisson distribution.

\(n\) large and \(p\) small: approximately \(n > 50\) and \(np < 5\). ✓
The Poisson parameter is \(\lambda = np\). ✓

Calculation

[4 marks]

\(X \sim B(400, 0.008)\). Use a suitable approximation to find (a) \(\text{P}(X = 3)\) (b) \(\text{P}(X \le 1)\).

\(n = 400 > 50\), \(np = 3.2 < 5\), so \(X \approx \text{Po}(3.2)\). ✓
(a) \(e^{-3.2}\dfrac{3.2^{3}}{6} = 0.223\) (3 s.f.). ✓
(b) \(e^{-3.2}(1 + 3.2) = 4.2e^{-3.2} = 0.171\) (3 s.f.). ✓

Calculation

[4 marks]

\(X \sim \text{Po}(40)\). Use a normal approximation to find \(\text{P}(35 \le X < 45)\).

\(Y \sim N(40, 40)\). \(35 \le X < 45\) means \(35 \le X \le 44\), so \(\text{P}(34.5 < Y < 44.5)\). ✓
\(z_1 = \dfrac{34.5 - 40}{\sqrt{40}} = -0.870\), \(z_2 = \dfrac{44.5 - 40}{\sqrt{40}} = 0.712\). ✓
\(\Phi(0.712) - \Phi(-0.870) = 0.7618 - (1 - 0.8078) = 0.570\) (3 s.f.). ✓

Analysis

[4 marks]

\(X \sim B(90, 0.025)\).

(a) Explain why a normal approximation is not suitable but a Poisson approximation is.

(b) Use the Poisson approximation to find \(\text{P}(X \ge 2)\).

(a) \(np = 2.25 < 5\), so the binomial is too skewed for a normal approximation; \(n = 90 > 50\) and \(np < 5\), so \(\text{Po}(2.25)\) is suitable. ✓
(b) \(\text{P}(X \ge 2) = 1 - e^{-2.25}(1 + 2.25) = 1 - 3.25 \times 0.1054 = 0.657\) (3 s.f.). ✓

Calculation

[3 marks]

A hospital's records show that 99% of patients do not react badly to a vaccine. 300 patients are vaccinated. Find, using a suitable approximation, the probability that more than 3 patients react badly.

\(R \sim B(300, 0.01)\); \(n > 50\), \(np = 3 < 5\), so \(R \approx \text{Po}(3)\). ✓
\(\text{P}(R > 3) = 1 - e^{-3}\left(1 + 3 + 4.5 + 4.5\right) = 1 - 13e^{-3}\). ✓
\(= 1 - 0.6472 = 0.353\) (3 s.f.). ✓

Analysis

[6 marks]

Calls arrive at a call centre at random at a mean rate of 1.5 per minute.

(a) Find the probability that at most 2 calls arrive in a given minute.

(b) Use a suitable approximation to find the probability that more than 100 calls arrive in a given hour, justifying your choice.

(a) \(X \sim \text{Po}(1.5)\): \(e^{-1.5}(1 + 1.5 + 1.125) = 3.625e^{-1.5} = 0.809\) (3 s.f.). ✓
(b) In an hour \(T \sim \text{Po}(90)\). Since \(\lambda = 90 > 15\), use \(Y \sim N(90, 90)\). ✓
More than 100 means 101 or more: \(\text{P}(Y > 100.5)\). ✓
\(z = \dfrac{100.5 - 90}{\sqrt{90}} = 1.107\), so \(1 - \Phi(1.107) = 1 - 0.8659 = 0.134\) (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.