Exam-style questionsNormal Approximation to the Binomial questions with mark schemes
Easiest first. Try each question on paper, then open its mark scheme: M marks are for method, A marks for accurate answers that follow from the method, B marks for independent results.
Question 1easy5 marksYear 13
The random variable \(X \sim \mathrm{B}(80,\ 0.4)\).
(a) Give two reasons why a normal distribution can be used to approximate \(X\). (2)
(b) Use a normal approximation to find \(P(X \le 28)\). (3)
Mark scheme and answer
(a) \(n\) is large B1
\(p\) is close to 0.5 B1
(b) \(Y \sim \mathrm{N}(32,\ 19.2)\) B1
\(P(Y\lt 28.5)\) M1
\(P\left(Z\lt \dfrac{28.5-32}{\sqrt{19.2}}\right) = 0.2122\) A1
Answer: (a) n large, p close to 0.5; (b) 0.2122 (exact binomial 0.2131)
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 2easy3 marksYear 13
A fair coin is tossed 400 times. \(H\) is the number of heads.
(a) Using a normal approximation, find \(P(H \ge 215)\). (3)
Mark scheme and answer
(a) \(\mathrm{N}(200,\ 100)\) B1
\(P(Y \gt 214.5) = P(Z \gt 1.45)\) M1
\(= 0.07353\) A1
Answer: 0.07353
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 3easy3 marksYear 13
3% of the light bulbs made by a factory are faulty. A box contains 50 bulbs. \(F\) is the number of faulty bulbs in the box.
(a) Explain why a normal approximation to the distribution of \(F\) is not suitable. (1)
(b) Find \(P(F \le 2)\). (2)
Mark scheme and answer
(a) \(np = 1.5\) is small / \(p\) far from 0.5 so the distribution is highly skewed B1
(b) \(\mathrm{B}(50,\ 0.03)\), \(P(F\le2)\) M1
\(= 0.8108\) A1
Answer: (a) np = 1.5 too small / skewed; (b) 0.8108
Board notes: All boards expect candidates to recognise when the normal approximation is inappropriate (np and n(1-p) small / p far from 0.5) and to use the exact binomial distribution instead.
Question 4medium6 marksYear 13
A quarter of the visitors to a museum buy something from the gift shop. On one morning 120 visitors, who can be regarded as a random sample, enter the museum. Let \(X\) be the number who buy something.
(a) Using a suitable approximation, find \(P(25 \le X \lt 35)\). (4)
(b) Given that the exact binomial value is \(0.7075\), find the percentage error in your approximation. (2)
Mark scheme and answer
(a) \(\mathrm{N}(30,\ 22.5)\) B1
\(P(24.5 \lt Y \lt 34.5)\) M1M1
(M1 standardising, M1 both continuity corrections)
\(= 0.7055\) A1
(b) \(\dfrac{|0.7055 - 0.7075|}{0.7075}\times100\) M1
\(= 0.28\%\) A1
Answer: (a) 0.7055; (b) 0.28%
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 5medium4 marksYear 13
A seed company claims that 85% of its tomato seeds germinate. A gardener plants 200 of these seeds. Assume the claim is true and seeds germinate independently.
(a) Use a normal approximation to find the probability that more than 175 seeds germinate. (3)
(b) Explain why, although \(p = 0.85\) is not close to 0.5, the approximation is reasonable here. (1)
Mark scheme and answer
(a) \(\mathrm{N}(170,\ 25.5)\) B1
\(P(Y\gt 175.5)\) M1
\(= 0.1380\) A1
(b) \(n\) is large enough that \(n(1-p) = 30 \gt 5\) (distribution is not too skewed) B1
Answer: (a) 0.1380 (exact 0.1368); (b) n(1−p) = 30 > 5
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 6medium5 marksYear 13
The random variable \(X \sim \mathrm{B}(n,\ p)\) is approximated by \(Y \sim \mathrm{N}(36,\ 21.6)\).
(a) Find the value of \(n\) and the value of \(p\). (3)
(b) Using the approximation, find \(P(X = 40)\). (2)
Mark scheme and answer
(a) \(np = 36\), \(np(1-p) = 21.6\) M1
\(1-p = 0.6\) so \(p = 0.4\) A1
\(n = 90\) A1
(b) \(P(39.5\lt Y\lt 40.5)\) M1
\(= 0.0592\) A1
Answer: (a) n = 90, p = 0.4; (b) 0.0592 (exact 0.0585)
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 7medium3 marksYear 13
5% of parcels handled by a courier arrive damaged. In a week the courier handles 600 parcels.
(a) Use a normal approximation to estimate the probability that more than 40 parcels arrive damaged. (3)
Mark scheme and answer
(a) \(\mathrm{N}(30,\ 28.5)\) B1
\(P(Y\gt 40.5)\) M1
\(= 0.0246\) A1
Answer: 0.0246 (exact 0.0288)
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 8medium4 marksYear 13
In a multiple-choice quiz of 60 true/false questions, a student guesses every answer. \(X\) is the number of correct answers.
(a) Use a normal approximation to find \(P(28 \le X \le 35)\). (3)
(b) Using the binomial distribution, the exact value is \(0.6630\). Explain why the approximation is so good here. (1)
Mark scheme and answer
(a) \(\mathrm{N}(30,\ 15)\) B1
\(P(27.5\lt Y\lt 35.5)\) M1
\(= 0.6629\) A1
(b) \(p = 0.5\) so the binomial distribution is symmetric B1
Answer: (a) 0.6629; (b) p = 0.5, symmetric
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 9hard5 marksYear 13
An airline knows that 92% of passengers with tickets turn up for a flight, independently of each other. For a flight with 235 seats, the airline sells 250 tickets.
(a) Using a normal approximation, find the probability that more passengers turn up than there are seats. (4)
(b) Comment on the independence assumption in this context. (1)
Mark scheme and answer
(a) \(X\sim\mathrm{B}(250,\ 0.92)\approx\mathrm{N}(230,\ 18.4)\) M1A1
\(P(Y \gt 235.5)\) M1
\(= 0.0999\) A1
(b) Unlikely to be valid – passengers often travel in families/groups, so if one does not turn up others in the group may not either B1
Answer: (a) 0.0999 (exact 0.0954); (b) groups travel together
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 10hard4 marksYear 13
In a large town 45% of adults support a proposed new cycle lane. A random sample of 300 adults is taken.
(a) Using a suitable approximation, find the probability that a majority of the sample support the cycle lane. (4)
Mark scheme and answer
(a) Majority means \(X \ge 151\) B1
\(\mathrm{N}(135,\ 74.25)\) B1
\(P(Y\gt 150.5)\) M1
\(= 0.0360\) A1
Answer: 0.0360 (exact 0.0363)
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only.
Question 11hard5 marksYear 13
The random variable \(X \sim \mathrm{B}(200,\ 0.35)\).
(a) Using a normal approximation, find the largest integer \(k\) such that \(P(X \le k) \lt 0.05\). (5)
Mark scheme and answer
(a) \(\mathrm{N}(70,\ 45.5)\) B1
\(\dfrac{k+0.5-70}{\sqrt{45.5}} \lt -1.6449\) M1M1
\(k \lt 58.40\) A1
\(k = 58\) A1
Answer: k = 58
Board notes: All boards require a continuity correction; a correct approximating distribution with an incorrect or missing continuity correction usually scores the method marks only. Check: approx P(X ≤ 58) = 0.0441, approx P(X ≤ 59) = 0.0598.
Question 12challenging6 marksYear 13
Historically 30% of customers at a café buy a pastry with their drink. After a new display is installed, 57 of a random sample of 150 customers buy a pastry.
(a) Using a normal approximation, test at the 5% level of significance whether there is evidence that the proportion buying a pastry has increased. (6)
Mark scheme and answer
(a) \(H_0: p = 0.3\), \(H_1: p \gt 0.3\) B1
Under \(H_0\), \(X\sim\mathrm{B}(150,0.3)\approx\mathrm{N}(45,\ 31.5)\) M1
\(P(X\ge57)\approx P(Y\gt 56.5)\) M1
\(= 0.0202\) A1
\(0.0202 \lt 0.05\) (or \(z = 2.049 \gt 1.6449\)), reject \(H_0\) dM1
There is evidence that the proportion of customers buying a pastry has increased A1
Answer: p-value ≈ 0.0202 < 0.05: evidence of increase (exact binomial p-value 0.0219)
Board notes: Spec ruling: stretch/enrichment. The skills are on the DfE content (N2 normal approximation to the binomial; O2 test for a binomial proportion) and no board excludes combining them, but Edexcel, AQA and OCR almost always expect an exact binomial test (here exact p-value 0.0219, same conclusion). Keep as a stretch question; an exact-binomial method scores full marks.
Original questions in the style of Edexcel 9MA0, independently checked before publishing. Not past-paper questions. Covers: Approximating with continuity correction, When the approximation is not valid, Parameters of the approximation, Approximation in context, Inverse problems with the approximation, Hypothesis test using the approximation.
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