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DSE Mathematics (Compulsory Part) Practice Paper: Statistics, Probability and Sequences (12 Questions, 50 Marks, 75 Minutes)
DSE Maths: Statistics, Probability and Sequences
DSE Mathematics paper: 12 questions, 50 marks, 75 minutes on mean, median, standard deviation, sequences, z-scores, dice probability and grouped data.
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Paper details Level: S6 | Subject: Mathematics (Compulsory Part) | Time: 75 minutes | Marks: 50 | Questions: 12
Section A (Q1-Q6, 2 marks each, 12 marks)
Show the working and the value; keep fractions in fraction form and give units where they are needed.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | The scores of six students are 12, 15, 15, 18, 21 and 27. Find the mean and the median. | Sum = 108, so the mean is 108 / 6 = 18.00. In order the scores are 12, 15, 15, 18, 21 and 27, give a median of (15 + 18) / 2 = 16.50. [1 mark for the mean, 1 mark for the median] |
| 2 | A bag contains 5 red, 3 blue and 4 green balls. One ball is drawn at random. Find the probability that it is not red. | There are 12 balls and 7 of them are not red, so the probability is 7/12 = 0.583. [1 mark for the numerator, 1 mark for the denominator] |
| 3 | A committee of 3 students is chosen from 7 students. In how many ways can this be done? | By the combination formula, C(7,3) = 7 x 6 x 5 / (3 x 2 x 1) = 35 ways. [1 mark for the formula, 1 mark for the answer] |
| 4 | For the data 2, 4, 4, 4, 5, 5, 7, 9, find the mean, the variance and the standard deviation. | Mean = 40 / 8 = 5.00; the sum of squared deviations is 32, so the variance is 32 / 8 = 4.00 and the standard deviation is 2.00. |
| 5 | An arithmetic sequence has first term 7 and common difference 5. Find the 20th term and the sum of the first 20 terms. | T(20) = 7 + 19 x 5 = 102; S(20) = 20 / 2 x (7 + 102) = 1090. [1 mark each, with working shown separately] |
| 6 | A geometric sequence starts with 3, 6, 12. Find the sum of the first 10 terms. | First term a = 3 and common ratio r = 2, so S(10) = 3 x (2^10 – 1) / (2 – 1) = 3069. [1 mark for the common ratio, 1 mark for using the sum formula] |
Section B (Q7-Q9, 4 marks each, 12 marks)
Show every step; an answer without working earns at most half the marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 7 | In a test the mean is 60 and the standard deviation is 12. Find the z-scores of a mark of 78 and of a mark of 45, and interpret the z-score of 78. | z(78) = (78 – 60) / 12 = 1.50 and z(45) = (45 – 60) / 12 = -1.25. A mark of 78 is 1.50 standard deviations above the mean, which is an above-average result. [2 marks for working, 1 mark for both z-scores, 1 mark for the interpretation] |
| 8 | A and B are independent events with P(A) = 0.6 and P(B) = 0.5. Find P(A and B), P(A or B) and P(A given B). | P(A and B) = 0.6 x 0.5 = 0.30; P(A or B) = 0.6 + 0.5 – 0.30 = 0.80; P(A given B) = 0.30 / 0.5 = 0.60. [1 mark each, plus 1 mark for the conditional probability formula] |
| 9 | The table shows the marks of 30 students. Calculate the mean (2 decimal places) and state the modal class and the median class. (0 to under 10: 4; 10 to under 20: 6; 20 to under 30: 10; 30 to under 40: 8; 40 to under 50: 2) | Mid-points are 5, 15, 25, 35 and 45; the total frequency is 30 and the total of mid-point times frequency is 730, so the mean is 730 / 30 = 24.33. The modal class is 20 to under 30 (frequency 10). Cumulative frequencies are 4, 10, 20, 28 and 30, so the 15.5th value lies in the class 20 to under 30, which is also the median class. |
Section C (Q10-Q12, 26 marks)
Calculate first, then explain in full sentences with reference to the figures given in the question.
| # | Question | Answer and marking notes |
|---|---|---|
| 10 | Two fair dice are thrown. Let the random variable X be the sum of the two numbers. (a) Find P(X = 8). (b) Find P(X = 8 or X = 12). (c) Given that X is even, find P(X = 8). (d) If the pair of dice is thrown 3 times, find the probability of getting X = 8 at least once. | The sample space has 36 outcomes. (a) Five outcomes give a sum of 8, so P(X = 8) = 5/36 = 0.139. (b) Only one outcome gives 12, so the probability is 5/36 + 1/36 = 6/36 = 1/6. (c) There are 18 even outcomes, so P(X = 8 given X is even) = 5/18. (d) The probability of no 8 in three throws is (31/36)^3, so the required probability is 1 – (31/36)^3 = 0.3615. |
| 11 | A student saves 12,000 dollars in the first year and increases the amount by 8% each year after that. (a) Find the amount saved in the fifth year. (b) Find the total saved in the first ten years. (c) If the saving continues for ever, does the total converge? Explain. | (a) The fifth year amount is 12,000 x 1.08^4 = 16,325.87 dollars. (b) The ten-year total is 12,000 x (1.08^10 – 1) / 0.08 = 173,838.75 dollars. (c) It does not converge: the common ratio 1.08 is greater than 1, so the yearly amounts keep growing and the partial sums have no upper bound. |
| 12 | Class A has 40 students with mean 68 and standard deviation 8; Class B has 25 students with mean 62 and standard deviation 5. (a) Find the mean of all 65 students. (b) Find the coefficient of variation of each class. (c) Which class is more consistent? Explain. | (a) The mean of all 65 students is (68 x 40 + 62 x 25) / 65 = 65.69. (b) The coefficient of variation is 8 / 68 = 0.1176 for Class A and 5 / 62 = 0.0806 for Class B. (c) Class B has the smaller coefficient of variation, so its marks are relatively less spread out and more consistent, although its mean is lower than that of Class A. |
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