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AS & A Level · Practice questions · The normal approximation to the binomial

Correct by a half.

Six original Cambridge-style questions on the normal approximation to the binomial distribution, at Paper 5 standard.

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, approximate, correct, standardise.

01
Recall
[2 marks]

\(X \sim \text{B}(80, 0.35)\). Show that a normal approximation is appropriate and state the approximating distribution.

  • \(np = 28 > 5\), \(nq = 52 > 5\). ✓
  • \(\text{N}(28, 18.2)\), since \(npq = 80 \times 0.35 \times 0.65 = 18.2\). ✓
02
Calculation
[3 marks]

\(X \sim \text{B}(80, 0.35)\). Use a normal approximation to find \(\text{P}(X > 32)\).

  • \(X > 32\) means 33 or more: \(\text{P}(Y > 32.5)\), \(Y \sim \text{N}(28, 18.2)\). ✓
  • \(z = \dfrac{32.5 - 28}{\sqrt{18.2}} = \dfrac{4.5}{4.266} = 1.055\). ✓
  • \(1 - \Phi(1.055) = 1 - 0.8543 = 0.146\) (3 s.f.). ✓
03
Calculation
[4 marks]

\(X \sim \text{B}(80, 0.35)\). Use a normal approximation to find \(\text{P}(25 \le X \le 30)\).

  • \(\text{P}(24.5 < Y < 30.5)\). ✓
  • \(z_1 = \dfrac{24.5 - 28}{4.266} = -0.820\), \(z_2 = \dfrac{30.5 - 28}{4.266} = 0.586\). ✓
  • \(\Phi(0.586) - [1 - \Phi(0.820)] = 0.7211 - 0.2061 = 0.515\) (3 s.f.). ✓
04
Calculation
[4 marks]

A fair die is thrown 120 times. Use a suitable approximation to find the probability that at least 25 sixes are obtained.

  • \(X \sim \text{B}(120, \tfrac{1}{6})\): \(np = 20\), \(nq = 100\), both over 5. ✓
  • \(Y \sim \text{N}\left(20, \tfrac{50}{3}\right)\), \(\sigma = 4.082\). ✓
  • \(\text{P}(Y > 24.5)\): \(z = \dfrac{4.5}{4.082} = 1.102\). ✓
  • \(1 - \Phi(1.102) = 1 - 0.8648 = 0.135\) (3 s.f.). ✓
05
Analysis
[4 marks]

30% of customers prefer brand A. A random sample of 200 customers is taken. Use a suitable approximation to find the probability that between 50 and 70 customers inclusive prefer brand A.

  • \(X \sim \text{B}(200, 0.3)\): \(np = 60\), \(nq = 140\). \(Y \sim \text{N}(60, 42)\), \(\sigma = 6.481\). ✓
  • \(\text{P}(49.5 < Y < 70.5)\): \(z = \pm\dfrac{10.5}{6.481} = \pm 1.620\). ✓
  • \(2\Phi(1.620) - 1 = 2(0.9474) - 1 = 0.895\) (3 s.f.). ✓
06
Analysis
[6 marks]

An airline sells 110 tickets for a flight with 104 seats. Each ticket holder turns up independently with probability 0.9.

(a) Use a suitable approximation to find the probability that more passengers turn up than there are seats. (b) The airline does this on 5 separate flights. Find the probability that at least one of the flights has more passengers than seats.

  • (a) \(X \sim \text{B}(110, 0.9)\): \(np = 99\), \(nq = 11\), both over 5. \(Y \sim \text{N}(99, 9.9)\), \(\sigma = 3.146\). ✓
  • More than 104 means 105 or more: \(\text{P}(Y > 104.5)\), \(z = \dfrac{5.5}{3.146} = 1.748\). ✓
  • \(1 - \Phi(1.748) = 1 - 0.9598 = 0.0402\) (3 s.f.). ✓
  • (b) \(1 - (1 - 0.0402)^5 = 1 - 0.9598^5 = 0.186\) (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.