The normal approximation and choosing a distribution - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 17 exam-style questions on The normal approximation and choosing a distribution for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

A greengrocer sells apples in bags of 10. Each apple is bruised with probability 0.12, independently of the other apples. Jonah models the number of bruised apples in a bag, X, by the normal distribution \(N(1.2, 1.056)\).

Give two reasons why a normal distribution is not a suitable model for X, and state a more suitable distribution for X.

Model answer

Reason 1: X is a discrete count taking only the values 0, 1, ..., 10, but the normal distribution is continuous.
Reason 2: n = 10 is small and p = 0.12 is far from 0.5, so the distribution of X is skewed; the normal model \(N(1.2, 1.056)\) would give probability 0.121 to negative values.
A more suitable model is \(X \sim B(10, 0.12)\).

Explanation

A normal approximation to \(B(n, p)\) needs n large and p close to 0.5; here n = 10 and p = 0.12, so the distribution is heavily skewed towards 0 and only 11 whole-number values are possible. The binomial distribution \(B(10, 0.12)\) models the count exactly, since the apples are independent with a constant probability. Answers about independence do not score, because the question states it.

Mark scheme

1 mark
B1 X is discrete (it takes only the whole-number values 0 to 10), but a normal distribution is continuous
1 mark for each different reason, max 2. Accept "X can only take a few values".
1 mark
B1 p = 0.12 is not close to 0.5 and n = 10 is small, so the distribution of X is skewed, not symmetrical (the normal model gives a large probability, about 0.12, to negative values)
Accept any one of: p not close to 0.5; n not large; the distribution is skewed; the normal model gives probability to impossible negative values. "n is small" alone is enough for this mark. Do not accept "the apples may not be independent" (the question states that they are).
1 mark
B1 \(X \sim B(10, 0.12)\)
Needs "binomial" with n = 10 and p = 0.12. Do not accept "calculator speak" without the word binomial or B.

Question 2

4 marks

At a supermarket, 64% of customers use the self-checkout tills. A random sample of 90 customers is taken.

Using a normal approximation, find the probability that more than 60 of these customers use the self-checkout tills. Give your answer to 3 significant figures.

Model answer

Let X be the number using the self-checkout: \(X \sim B(90, 0.64)\)
Mean \(= 57.6\), variance \(= 57.6 \times 0.36 = 20.736\), so \(X \approx Y \sim N(57.6, 20.736)\)
\(P(X > 60) = P(X \geqslant 61) \approx P(Y > 60.5)\)
\(= P\left(Z > \dfrac{60.5 - 57.6}{\sqrt{20.736}}\right) = P(Z > 0.6368) = 0.26211...\)
Probability = 0.262 (3 s.f.)

Explanation

"More than 60" means 61 or more, so the continuity correction gives \(P(Y > 60.5)\) with \(Y \sim N(57.6, 20.736)\). The mean and variance come from np and np(1 − p). Common errors are using 59.5, which includes 60, and entering the variance as the standard deviation.

Mark scheme

1 mark
M1 Attempts a normal approximation with mean \(90 \times 0.64 = 57.6\); or a correct final answer of 0.262 with no working
An answer of 0.262 with no working scores 4 marks. The exact binomial value 0.264 (0.26424...) scores 0 marks.
1 mark
A1 Correct standard deviation 4.55 (awrt 4.55, i.e. [4.545, 4.555)) or variance 20.736, correctly labelled
\(N(57.6, 20.736)\).
1 mark
M1 Attempts a continuity correction: \(P(Y > 60.5)\) (allow 59.5 or 61.5 as an attempt)
Accept the correction seen in a standardisation. No continuity correction (60, giving 0.299) scores M0.
1 mark
A1 0.262 (awrt 0.262, i.e. [0.2615, 0.2625))
Needs 60.5. Using 59.5 gives 0.338: A0.

Question 3

5 marks

In a large town, 48% of households recycle their food waste. A random sample of 100 households is taken, and X is the number of these households that recycle their food waste.

Explain why a continuity correction is needed when a normal distribution is used to approximate P(X = 50). Use a normal approximation to estimate P(X = 50), and find the difference between your estimate and the exact binomial probability. Give your answers to 3 significant figures.

Model answer

X is discrete, but the approximating normal variable Y is continuous, and \(P(Y = 50) = 0\). So the value 50 is represented by the interval 49.5 to 50.5.
\(X \sim B(100, 0.48)\), mean 48, variance \(48 \times 0.52 = 24.96\), so \(Y \sim N(48, 24.96)\)
\(P(X = 50) \approx P(49.5 < Y < 50.5) = 0.073600...\) = 0.0736 (3 s.f.)
Exact: \(P(X = 50) = 0.073465...\) = 0.0735 (3 s.f.)
Difference \(= 0.073600 - 0.073465 = 0.000135\) (3 s.f.)

Explanation

A continuous distribution gives probability 0 to any single value, so a whole number such as 50 is represented by the interval of width 1 around it, from 49.5 to 50.5. The approximation is very close to the exact binomial value here because n is large and p is close to 0.5. Round only at the end: rounding both probabilities to 3 s.f. before subtracting loses the difference.

Mark scheme

1 mark
B1 X is discrete but the normal distribution is continuous: for a continuous variable the probability of any single value is 0, so the value 50 is represented by the interval from 49.5 to 50.5. Correct numerical answers with no working score 4 marks (all except this mark)
Answers of 0.0736, 0.0735 and 0.000135 with no working score 4 marks. Accept "the binomial is discrete and the normal is continuous" or "P(Y = 50) = 0 for a continuous variable".
1 mark
M1 Uses \(N(48, 24.96)\) and finds \(P(49.5 < Y < 50.5)\)
Needs both the mean 48 and the variance 24.96 (or standard deviation 4.996) and an interval of width 1 around 50; accept the limits seen in a standardisation.
1 mark
A1 0.0736 (awrt 0.0736, i.e. [0.07355, 0.07365))
Exact value 0.073600...
1 mark
B1 Exact binomial probability \(P(X = 50) = 0.0735\) (awrt 0.0735, i.e. [0.07345, 0.07355))
From \(B(100, 0.48)\) on a calculator (0.073465...).
1 mark
B1 Difference 0.000135 (awrt 0.000135, i.e. [0.0001345, 0.0001355); accept 0.00013 or 0.00014 from values given to 4 s.f.)
Accept \(1.35 \times 10^{-4}\). 0.0001 from values rounded to 3 s.f. scores B0.

Question 4

7 marks

A research company telephones people at random to take part in a survey. Each person telephoned agrees to take part with probability 0.4, independently of the others. The company needs at least 30 people to take part.

Using a normal approximation, find the least number of people that the company should telephone so that the probability that at least 30 of them take part is at least 0.95.

Model answer

Let X be the number who take part: \(X \sim B(n, 0.4)\), approximated by \(Y \sim N(0.4n, 0.24n)\).
Need \(P(X \geqslant 30) \geqslant 0.95\), i.e. \(P(Y > 29.5) \geqslant 0.95\)
\(\dfrac{29.5 - 0.4n}{\sqrt{0.24n}} \leqslant -1.6449\)
\(0.4n - 29.5 \geqslant 1.6449\sqrt{0.24}\sqrt{n} = 0.80581\sqrt{n}\)
Let \(u = \sqrt{n}\): \(0.4u^2 - 0.80581u - 29.5 = 0\)
\(u = \dfrac{0.80581 + \sqrt{0.80581^2 + 4 \times 0.4 \times 29.5}}{0.8} = 9.6539\)
\(n = 9.6539^2 = 93.2\)
So the least number is n = 94 (check: n = 93 gives 0.948, n = 94 gives 0.956).

Explanation

The approximation has mean 0.4n and variance 0.24n, so the standardised condition gives an equation in \(\sqrt{n}\), which is a quadratic; its positive root gives n = 93.2, and n must be rounded up to 94. The continuity correction turns "at least 30" into 29.5. Common errors are forgetting the square root in the standard deviation and rounding 93.2 down.

Mark scheme

1 mark
B1 For n people telephoned, the approximation is \(N(0.4n, 0.24n)\). An answer of 94 with no working scores 0 marks
An answer of 94 with no working scores 0 marks: the exact binomial distribution also gives 94 (n = 93 gives 0.94998 and n = 94 gives 0.9575), so the use of the normal approximation must be seen. The minimum working is \(N(0.4n, 0.24n)\) with the continuity-corrected 29.5 and 1.6449 in an equation or inequality for n, or approximation trials for n = 93 and n = 94 (0.948 and 0.956). Exact binomial trials with no use of a normal approximation score 0 marks.
1 mark
M1 Uses a continuity correction: \(P(Y > 29.5) \geqslant 0.95\) (or \(P(Y \geqslant 29.5)\))
Allow 30.5 as an attempt.
1 mark
B1 Uses \(z = -1.6449\) (or 1.6449 with compatible signs)
Accept 1.645.
1 mark
M1 Standardises: \(\dfrac{29.5 - 0.4n}{\sqrt{0.24n}} = -1.6449\) (or \(\leqslant -1.6449\))
Must have \(\sqrt{0.24n}\) as the standard deviation.
1 mark
M1 (dM1) Rearranges into a quadratic in \(\sqrt{n}\), e.g. \(0.4n - 0.8058\sqrt{n} - 29.5 = 0\), and solves it
Depends on the previous M1. Alternative method: trials of n with the approximation showing that n = 93 gives 0.948 (< 0.95) and n = 94 gives 0.956 (≥ 0.95) earn this mark and the next A1.
1 mark
A1 \(\sqrt{n} = 9.654\) (awrt 9.65, i.e. [9.645, 9.655)) or \(n = 93.2\) (awrt 93.2, i.e. [93.15, 93.25))
Ignore the negative root of the quadratic.
1 mark
A1 n = 94
Depends on the dM1. cao. 93 scores A0 (it gives a probability below 0.95).

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