S6 Mathematics Extended Part Module 1 Practice Paper: Binomial, Poisson, Normal Distribution, Confidence Intervals And Trapezoidal Rule (100 Marks)

S6 Mathematics Extended Part Module 1 Practice Paper: Binomial, Poisson, Normal Distribution, Confidence Intervals and Trapezoidal Rule (100 marks)

2026-11-12

M1 – Distributions and Statistics

S6 M1 paper: 6 questions, 16 parts, 100 marks, 60 minutes. Binomial and Poisson probabilities, normal distribution, confidence intervals, trapezoidal rule and expectation.

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Section A – Binomial Distribution

Answer all three parts. Total 16 marks. Let X follow a binomial distribution with n = 20 and p = 0.3.

#	Question	Answer and marking notes
1	Find the expectation, variance and standard deviation of X.	Answer: Expectation = np = 20×0.3 = 6; variance = np(1-p) = 6×0.7 = 4.2; standard deviation is the square root of 4.2, about 2.0494. (Marks: expectation 2, variance 2, standard deviation 2)
2	Find P(X = 5), correct to four decimal places.	Answer: P(X = 5) = C(20,5)(0.3)^5(0.7)^15 = 15504×0.00243×0.0047476, about 0.1789. (Marks: binomial formula 3, value 2)
3	Find P(X is at most 2), correct to four decimal places.	Answer: P(X up to 2) is the sum of the probabilities for 0, 1 and 2, which are 0.0008, 0.0068 and 0.0278, giving about 0.0355. (Marks: individual terms 3, sum and value 2)

Section B – Poisson Distribution

Answer all three parts. Total 16 marks. A hotline receives 4 enquiries per hour on average, and the number of enquiries follows a Poisson distribution.

#	Question	Answer and marking notes
4	Find the probability of receiving no enquiry in one hour.	Answer: Let X be the number of enquiries in one hour, Poisson with mean 4. Then P(X = 0) = e^(-4), about 0.0183. (Marks: setting up 2, formula 1, value 2)
5	Find the probability of receiving at most two enquiries in one hour.	Answer: P(X up to 2) = e^(-4)(1 + 4 + 8) = 13e^(-4), about 0.2381. (Marks: three terms 3, value 2)
6	Find the probability of receiving at least one enquiry in two hours.	Answer: The count over two hours, Y, is Poisson with mean 8. So P(Y at least 1) = 1 – e^(-8), about 0.99966. (Marks: mean 8 for 2, complement 2, value 2)

Section C – Normal Distribution

Answer all three parts. Total 18 marks. Let X be normally distributed with mean 50 and standard deviation 10.

#	Question	Answer and marking notes
7	Find P(X is greater than 65), correct to four decimal places.	Answer: Standardising gives z = (65 – 50)/10 = 1.5. The upper tail probability is 1 – 0.9332 = 0.0668. (Marks: standardising 2, table value 2, complement 2)
8	Find P(40 is at most X and X is at most 60), correct to four decimal places.	Answer: The z values are -1 and 1. The probability between them is 0.8413 – 0.1587 = 0.6827. (Marks: two z values 2, table values 2, subtraction 2)
9	Find k such that P(X is less than k) = 0.9, correct to one decimal place.	Answer: The standard normal value with lower tail 0.9 is 1.2816, so k = 50 + 1.2816×10 = 62.8. (Marks: z value 3, reversing standardisation 2, value 1)

Section D – Confidence Intervals

Answer all three parts. Total 18 marks. A sample of 100 observations has mean 12.4 and standard deviation 3.2.

#	Question	Answer and marking notes
10	Find the 95% confidence interval for the population mean.	Answer: The standard error is 3.2/10 = 0.32. The 95% interval is 12.4 plus or minus 1.96×0.32, that is (11.7728, 13.0272). (Marks: standard error 2, multiplier 2, limits 2)
11	Find the 99% confidence interval for the same data.	Answer: The 99% multiplier is 2.5758, so the interval is 12.4 plus or minus 2.5758×0.32 = (11.576, 13.224). (Marks: multiplier 2, computation 2, interval 2)
12	Find the least sample size needed for a 95% interval whose half width is at most 0.25.	Answer: Requiring 1.96×3.2 divided by the square root of n to be at most 0.25 gives the square root of n at least 25.088, hence n at least 629.4, so the least sample size is 630. (Marks: setting up 2, squaring 2, rounding up 2)

Section E – Trapezoidal Rule

Answer both parts. Total 16 marks. Use the trapezoidal rule with 4 sub-intervals to estimate the integral of 1/x from 1 to 3.

#	Question	Answer and marking notes
13	Tabulate the function values at the ordinates and apply the trapezoidal rule.	Answer: The ordinates are 1, 1.5, 2, 2.5 and 3, with function values 1, 0.6667, 0.5, 0.4 and 0.3333. The estimate is 0.25x[1 + 2(0.6667+0.5+0.4) + 0.3333] = 0.25×4.4667, about 1.1167. (Marks: table 4, formula 3, middle sum 2, result 1)
14	Given that the exact value is ln 3, find the error and the percentage error.	Answer: The exact value is about 1.0986, so the error is 1.1167 – 1.0986 = 0.0181 and the percentage error is about 1.64%. The rule overestimates because the function is convex on the interval. (Marks: exact value 2, error 2, percentage 2)

Section F – Expectation and Variance

Answer both parts. Total 16 marks. A discrete random variable X takes the value 0 with probability 0.2, 1 with 0.3, 2 with 0.4 and 3 with 0.1.

#	Question	Answer and marking notes
15	Find E(X), E of X squared and Var(X).	Answer: E(X) = 0x0.2 + 1×0.3 + 2×0.4 + 3×0.1 = 1.4; E of X squared = 0 + 0.3 + 1.6 + 0.9 = 2.8; Var(X) = 2.8 – 1.96 = 0.84. (Marks: E(X) 3, E of X squared 3, variance 3)
16	Let Y = 3X – 2. Find E(Y) and Var(Y).	Answer: E(Y) = 3E(X) – 2 = 3×1.4 – 2 = 2.2; Var(Y) = 9×0.84 = 7.56, since the constant shift does not change the variance. (Marks: E(Y) 3, variance scaling 3, value 1)

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