S4 Mathematics Practice Paper: Statistics And Probability (12 Questions, 100 Marks, 75 Minutes)

S4 Mathematics Practice Paper: Statistics and Probability (12 questions, 100 marks, 75 minutes)

2026-11-09

HKDSE Mathematics: Statistics and Probability

S4 Mathematics: statistics, standard deviation, probability rules and expected value — 12 questions, 100 marks, 75 minutes, full answers.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show all working. Answers without working earn no method marks. Write down a formula before substituting numbers.

Give numerical answers to three significant figures unless stated otherwise, and state the correct units.

Spend about 30 minutes on Section A, then work through dispersion and standard deviation in Section B, and finish with probability and expected value in Section C.

Section A　Organising Data and Measures of Central Tendency

Section A: Show all working, and state the data positions used when finding a median.

#	Question	Answer and marking notes
1	The data set is: 12, 15, 15, 18, 20, 22, 22, 22, 27, 27. (a) Find the mean. (b) Find the median. (c) Find the mode. (d) State which of the three measures of central tendency is least affected by extreme values, and explain briefly.	(a) Mean = (12 + 15 + 15 + 18 + 20 + 22 + 22 + 22 + 27 + 27) ÷ 10 = 200 ÷ 10 = 20 (3 marks). (b) With 10 values the median is the mean of the 5th and 6th values = (20 + 22) ÷ 2 = 21 (3 marks). (c) The mode is 22, which occurs 3 times, more often than any other value (2 marks). (d) The median depends only on the position of the values, so an extreme value however far away does not change its position; the median is therefore least affected (1 mark).
2	The test marks of 40 students are grouped as follows: 0–10 marks, 4 students; 10–20 marks, 10 students; 20–30 marks, 16 students; 30–40 marks, 8 students; 40–50 marks, 2 students. (a) Find the total frequency. (b) Using class mid-points, estimate the mean mark. (c) Write down the modal class. (d) State one reason why the estimated mean may differ from the true mean.	(a) Total frequency = 4 + 10 + 16 + 8 + 2 = 40 (2 marks). (b) The mid-points are 5, 15, 25, 35 and 45, so the total is 4×5 + 10×15 + 16×25 + 8×35 + 2×45 = 20 + 150 + 400 + 280 + 90 = 940; estimated mean = 940 ÷ 40 = 23.5 marks (4 marks). (c) The modal class is 20–30 marks, with frequency 16, the largest (2 marks). (d) Using the mid-point for every value in a class ignores the spread of marks within the class, so the estimate differs from the true mean (1 mark).
3	Using the frequency table in the previous question (0–10: 4, 10–20: 10, 20–30: 16, 30–40: 8, 40–50: 2, total 40 students): (a) write down the class containing the median; (b) estimate the median by interpolation; (c) state the class width and the mid-point of that class; (d) find the difference between the estimated median and the estimated mean of 23.5 marks.	(a) The cumulative frequencies are 4, 14, 30, 38 and 40, so the median is the mean of the 20th and 21st values, both of which lie between 15 and 30; the median class is 20–30 marks (2 marks). (b) Median = 20 + (20 − 14) ÷ 16 × 10 = 20 + 6 ÷ 16 × 10 = 23.75 marks (3 marks). (c) Class width = 10 marks and mid-point = (20 + 30) ÷ 2 = 25 marks (2 marks). (d) Estimated median − estimated mean = 23.75 − 23.5 = 0.25 marks (1 mark).
4	A set of 11 data values is: 3, 5, 5, 7, 8, 9, 12, 14, 15, 18, 21. (a) Write down the five-number summary. (b) Find the interquartile range. (c) Using the 1.5 times interquartile range rule, determine whether 21 is an outlier, showing your calculation. (d) In a box-and-whisker diagram, what does the width of the box represent?	(a) Minimum 3, lower quartile 5, median 9, upper quartile 15, maximum 21 (3 marks). (b) Interquartile range = 15 − 5 = 10 (2 marks). (c) Upper limit = 15 + 1.5 × 10 = 30; since 21 is less than 30, 21 is not an outlier (3 marks). (d) The width of the box represents the interquartile range, the spread of the middle 50% of the data (1 mark).

Section B　Dispersion, Standard Deviation and Comparing Data

Section B: Give standard deviations to three significant figures and write down the formula used.

#	Question	Answer and marking notes
5	A set of data is: 4, 6, 8, 10, 12. (a) Find the mean. (b) Find the variance. (c) Find the standard deviation, correct to three significant figures. (d) If 5 is added to every value, how does the standard deviation change?	(a) Mean = (4 + 6 + 8 + 10 + 12) ÷ 5 = 40 ÷ 5 = 8 (2 marks). (b) Variance = [(4−8)² + (6−8)² + (8−8)² + (10−8)² + (12−8)²] ÷ 5 = (16 + 4 + 0 + 4 + 16) ÷ 5 = 8 (3 marks). (c) Standard deviation = √8 ≈ 2.83 (2 marks). (d) Adding 5 to every value does not change the distance of each value from the mean, so the standard deviation is unchanged (1 mark).
6	In Class A the mean mark in a mathematics test is 60 with a standard deviation of 8; in Class B the mean is 65 with a standard deviation of 4. (a) Explain what the standard deviation measures. (b) A student in Class A scores 76; find the standard score. (c) A student in Class B scores 73; find the standard score. (d) Compare the relative performance of the two students within their own classes. (e) What does a standard deviation of 0 in Class B mean?	(a) The standard deviation measures how far the values are spread from the mean on average; a larger value means the data are more spread out (2 marks). (b) Standard score = (76 − 60) ÷ 8 = 2.0 (2 marks). (c) Standard score = (73 − 65) ÷ 4 = 2.0 (2 marks). (d) Both students have a standard score of 2.0, so each is the same distance above the mean of his or her own class; their relative performance is the same (2 marks). (e) A standard deviation of 0 means every student in Class B has exactly the same mark, so there is no spread (1 mark).
7	The mean of 7 data values is 12. A new value, 28, is added. (a) Find the new mean. (b) Does the median necessarily change as well? Explain briefly. (c) Is the standard deviation of the new set larger or smaller than before? Explain briefly.	(a) New total = 7 × 12 + 28 = 84 + 28 = 112, so the new mean = 112 ÷ 8 = 14 (3 marks). (b) Not necessarily; since 28 is above the old mean the mean must rise, but the median depends only on position, so adding one large value may leave the median unchanged or raise it only slightly (2 marks). (c) The standard deviation becomes larger; 28 is 14 away from the new mean of 14, far more than the deviations of the other values, so the average deviation increases (3 marks).
8	A school wants to survey the daily study time of S4 students. (a) State two factors that affect how representative a sample is. (b) State one advantage of random sampling. (c) Explain why collecting data only through an online questionnaire may be biased. (d) State one common way of dealing with an outlier.	(a) The size of the sample and whether the sampling method is biased, for example surveying only one class or only students in the library at a certain time (2 marks). (b) Random sampling gives every student an equal chance of being chosen, which reduces selection bias so that the sample better represents the population (2 marks). (c) An online questionnaire reaches only students who have internet access and are willing to respond, and their study time may differ from that of other students, producing self-selection bias (2 marks). (d) Check whether the value is a recording or data-entry error and correct it; if it is a genuine anomaly it may be excluded with the reason stated, or results with and without it may both be reported (2 marks).

Section C　Probability and Expected Value

Section C: Probabilities may be given as fractions or decimals to three significant figures; state the sample space or formula used.

#	Question	Answer and marking notes
9	A bag contains 5 red, 3 blue and 2 green balls, 10 balls in total, identical apart from colour. (a) One ball is drawn at random; find the probability that it is red. (b) Find the probability that the ball drawn is not red. (c) Two balls are drawn one after another without replacement; find the probability that both are red. (d) If each ball is replaced after drawing, find the probability that both are red.	(a) P(red) = 5 ÷ 10 = 0.5 (2 marks). (b) P(not red) = 1 − 0.5 = 0.5, or (3 + 2) ÷ 10 = 0.5 (2 marks). (c) Without replacement, P(both red) = (5 ÷ 10) × (4 ÷ 9) = 0.2222 = 2/9, about 0.222 (2 marks). (d) With replacement the two draws are independent, so P(both red) = (5 ÷ 10) × (5 ÷ 10) = 0.25 (2 marks).
10	Two fair dice are thrown together. (a) Find the probability that the sum of the two numbers is 7. (b) Find the probability that the two numbers are the same. (c) Find the probability that the sum is 7 or the two numbers are the same. (d) Determine whether the events ‘the sum is 7’ and ‘the two numbers are the same’ are mutually exclusive, and explain.	(a) The sample space has 36 outcomes and the sums equal to 7 are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), six outcomes, so the probability is 6/36 = 1/6 ≈ 0.167 (2 marks). (b) The outcomes with equal numbers are (1,1) to (6,6), six outcomes, so the probability is 6/36 = 1/6 ≈ 0.167 (2 marks). (c) The two events share no outcome, since a sum of 7 requires the two numbers to differ, so they are mutually exclusive and the probability is 1/6 + 1/6 = 1/3 ≈ 0.333 (2 marks). (d) Mutually exclusive, because when the sum is 7 the two numbers must be different, so the two events cannot occur together (2 marks).
11	A class has 40 students: 24 boys and 16 girls. 18 of the boys and 10 of the girls pass a test. (a) Find the probability that a randomly chosen student passes. (b) Given that the chosen student passes, find the probability that the student is a boy. (c) Given that the chosen student is a girl, find the probability that she passes. (d) Using the results of (a) and (c), determine whether gender and passing are independent, and explain.	(a) The number who pass is 18 + 10 = 28, so P(pass) = 28 ÷ 40 = 0.7 (2 marks). (b) P(boy \| pass) = 18 ÷ 28 = 9/14 ≈ 0.643 (2 marks). (c) P(pass \| girl) = 10 ÷ 16 = 0.625 (2 marks). (d) Since P(pass) = 0.70 differs from P(pass \| girl) = 0.625, gender and passing are not independent (2 marks).
12	A game draws one ball at random from the bag described above (5 red, 3 blue, 2 green). A red ball wins 20 dollars, a blue ball wins 5 dollars and a green ball loses 10 dollars. (a) Find the expected gain of one game. (b) If each game costs 12 dollars to play, find the player’s net expected gain. (c) State whether the organiser profits in the long run. (d) What should the charge be for the game to be fair?	(a) E = 0.5 × 20 + 0.3 × 5 + 0.2 × (−10) = 10 + 1.5 − 2 = 9.5 dollars (3 marks). (b) Net expected gain = 9.5 − 12 = −2.5 dollars (2 marks). (c) The player’s net expectation is negative, so each player loses 2.5 dollars per game on average and the organiser profits in the long run (2 marks). (d) A fair charge equals the expected gain, that is 9.5 dollars (1 mark).

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