Secondary 4 Additional Mathematics Statistics Probability Quiz

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Secondary 4 Additional Mathematics Quiz - Statistics Probability

Name: _________________________
Class: _________________________
Date: _________________________
Score: ________ / 60

Duration: 1 hour 15 minutes
Total Marks: 60

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Instructions

• Answer all 20 questions in the spaces provided.
• Show all working clearly. Answers without working may not receive full marks.
• Non-programmable scientific calculators may be used.
• Give answers to 3 significant figures where appropriate unless otherwise stated.
• This quiz covers Statistics and Probability topics from the Secondary 4 Additional Mathematics syllabus.
• Questions are designed to be syllabus-aligned practice. They complement past-paper patterns but are not reproduced from actual examination papers.

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Section A: Descriptive Statistics and Data Analysis (Questions 1–5)

Answer all questions in this section.

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1. The following data set shows the number of hours 10 students spent studying for a mathematics test:

12, 8, 15, 10, 7, 14, 9, 11, 13, 6

(a) Calculate the mean number of study hours.
[2 marks]

(b) Find the median of the data set.
[1 mark]

(c) Determine the interquartile range.
[2 marks]

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2. A grouped frequency table shows the distribution of test scores for 40 students:

Score	1–10	11–20	21–30	31–40	41–50
Frequency	5	8	12	10	5

(a) Estimate the mean score.
[3 marks]

(b) Identify the modal class.
[1 mark]

(c) Estimate the median score using linear interpolation.
[3 marks]

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3. The standard deviation of a set of 8 numbers is 3.5. Each number in the set is multiplied by 2 and then increased by 5.

(a) Find the new mean, given that the original mean was 12.
[2 marks]

(b) Find the new standard deviation.
[2 marks]

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4. Two data sets, A and B, are compared:

• Data set A: mean = 25, standard deviation = 4, size = 20
• Data set B: mean = 30, standard deviation = 6, size = 30

(a) Find the combined mean of all 50 values.
[2 marks]

(b) Explain, without full calculation, whether the combined standard deviation would be closer to 4 or to 6.
[2 marks]

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5. A cumulative frequency curve is drawn for the weights (in kg) of 60 students. The following values are read from the curve:

• Lower quartile: 48 kg
• Median: 55 kg
• Upper quartile: 62 kg
• Minimum: 40 kg
• Maximum: 75 kg

(a) Draw a box-and-whisker plot to represent this data.
[3 marks]

(b) Comment on the skewness of the distribution.
[1 mark]

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Section B: Probability (Questions 6–10)

Answer all questions in this section.

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6. A bag contains 5 red balls, 4 blue balls, and 3 green balls. Two balls are drawn at random without replacement.

(a) Find the probability that both balls are red.
[2 marks]

(b) Find the probability that the two balls are of different colours.
[3 marks]

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7. Events A and B are such that P(A) = 0.6, P(B) = 0.4, and P(A ∩ B) = 0.2.

(a) Find P(A ∪ B).
[1 mark]

(b) Find P(A | B).
[2 marks]

(c) Determine whether events A and B are independent. Justify your answer.
[2 marks]

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8. A fair six-sided die is rolled twice.

(a) List all possible outcomes in a sample space diagram.
[2 marks]

(b) Find the probability that the sum of the two rolls is exactly 8.
[2 marks]

(c) Find the probability that at least one roll shows a 6.
[2 marks]

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9. In a class of 30 students, 18 study Physics and 15 study Chemistry. 10 students study both subjects.

(a) Find the probability that a randomly selected student studies Physics but not Chemistry.
[2 marks]

(b) Find the probability that a randomly selected student studies neither Physics nor Chemistry.
[2 marks]

(c) Given that a student studies Chemistry, find the probability that they also study Physics.
[2 marks]

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10. A game involves spinning two fair spinners. Spinner 1 has sectors numbered 1, 2, 3, 4. Spinner 2 has sectors numbered 1, 2, 3.

(a) Complete the following probability tree diagram showing all outcomes and their probabilities.

Spinner 1          Spinner 2          Outcome
   1        ───►    1              (1,1)
            ───►    2              (1,2)
            ───►    3              (1,3)
   2        ───►    1              (2,1)
   ... (complete)

[3 marks]

(b) Find the probability that the sum of the two numbers is 5.
[2 marks]

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Section C: Permutations, Combinations, and Advanced Probability (Questions 11–15)

Answer all questions in this section.

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11. Find the number of different arrangements of the letters in the word "MATHEMATICS".

[3 marks]

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12. A committee of 4 people is to be chosen from 6 boys and 5 girls.

(a) Find the number of ways to form the committee if there are no restrictions.
[2 marks]

(b) Find the number of ways to form the committee if it must contain exactly 2 boys and 2 girls.
[2 marks]

(c) Find the number of ways to form the committee if it must contain at least 1 girl.
[2 marks]

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13. A box contains 8 identical cards numbered 1 to 8. Two cards are drawn at random without replacement.

(a) Find the probability that both numbers are even.
[2 marks]

(b) Find the probability that the sum of the two numbers is greater than 10.
[3 marks]

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14. Three letters are selected at random from the word "PROBABILITY" without replacement.

(a) Find the probability that all three letters are consonants.
[3 marks]

(b) Find the probability that exactly one of the three letters is a vowel.
[3 marks]

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15. A biased coin has probability p of landing heads. The coin is tossed 3 times.

(a) Write down, in terms of p, the probability of getting exactly 2 heads.
[2 marks]

(b) Given that the probability of getting exactly 2 heads is 0.288, find the value of p.
[3 marks]

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Section D: Discrete Random Variables and Binomial Distribution (Questions 16–20)

Answer all questions in this section.

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16. A discrete random variable X has the following probability distribution:

x	0	1	2	3	4
P(X = x)	0.1	0.2	0.3	0.25	0.15

(a) Find E(X).
[2 marks]

(b) Find Var(X).
[3 marks]

(c) Find P(X ≥ 2).
[1 mark]

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17. The random variable X ~ B(10, 0.3).

(a) Find P(X = 4).
[2 marks]

(b) Find P(X ≥ 2).
[3 marks]

(c) Find E(X) and Var(X).
[2 marks]

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18. A factory produces light bulbs, and 5% are defective. A random sample of 20 bulbs is selected.

(a) State two assumptions needed for the number of defective bulbs to follow a binomial distribution.
[2 marks]

(b) Find the probability that exactly 2 bulbs are defective.
[2 marks]

(c) Find the probability that at least 1 bulb is defective.
[2 marks]

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19. In a multiple-choice quiz, each question has 4 options, only one of which is correct. A student guesses all answers for 15 questions.

(a) Find the probability that the student gets exactly 5 correct answers.
[2 marks]

(b) Find the probability that the student gets at least 3 correct answers.
[3 marks]

(c) What is the most likely number of correct answers?
[2 marks]

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20. A random variable X ~ B(n, 0.4). It is given that E(X) = 6.

(a) Find the value of n.
[1 mark]

(b) Find P(X = 5).
[2 marks]

(c) Find P(4 ≤ X ≤ 7).
[3 marks]

(d) Given that P(X ≤ k) ≥ 0.95, find the smallest integer value of k.
[3 marks]

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End of Quiz

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Secondary 4 Additional Mathematics Quiz - Statistics Probability

Answer Key and Marking Scheme

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Section A: Descriptive Statistics and Data Analysis

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Question 1 [5 marks]

(a) Mean:

$\bar{x} = \frac{12 + 8 + 15 + 10 + 7 + 14 + 9 + 11 + 13 + 6}{10} = \frac{105}{10} = 10.5$

Answer: 10.5 hours [2 marks — 1 for correct sum, 1 for correct division]

(b) Median:

Ordered data: 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Median = average of 5th and 6th values = (10 + 11) / 2 = 10.5

Answer: 10.5 hours [1 mark]

(c) Interquartile range:

Lower quartile Q1: median of lower half (6, 7, 8, 9, 10) = 8
Upper quartile Q3: median of upper half (11, 12, 13, 14, 15) = 13

IQR = Q3 − Q1 = 13 − 8 = 5

Answer: 5 hours [2 marks — 1 for each quartile]

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Question 2 [7 marks]

(a) Estimated mean:

Score	Midpoint (x)	Frequency (f)	fx
1–10	5.5	5	27.5
11–20	15.5	8	124
21–30	25.5	12	306
31–40	35.5	10	355
41–50	45.5	5	227.5
Total		40	1040

$\bar{x} = \frac{\sum fx}{\sum f} = \frac{1040}{40} = 26$

Answer: 26 [3 marks — 1 for midpoints, 1 for Σfx, 1 for final answer]

(b) Modal class: The class with highest frequency is 21–30.

Answer: 21–30 [1 mark]

(c) Estimated median using linear interpolation:

Median position = 40/2 = 20th value, which lies in class 21–30.

$Median = L + \left(\frac{\frac{n}{2} - F}{f_m}\right) \times c$

Where L = 20.5, n = 40, F = 13 (cumulative freq before median class), f_m = 12, c = 10

$Median = 20.5 + \left(\frac{20 - 13}{12}\right) \times 10 = 20.5 + \frac{70}{12} = 20.5 + 5.833 = 26.33$

Answer: 26.3 (3 s.f.) [3 marks — 1 for identifying median class, 1 for correct formula, 1 for correct value]

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Question 3 [4 marks]

(a) New mean:

If each value is multiplied by 2 and increased by 5:

New mean = 2 × (original mean) + 5 = 2 × 12 + 5 = 29

Answer: 29 [2 marks — 1 for multiplication, 1 for addition]

(b) New standard deviation:

Multiplying by 2 scales the standard deviation by 2. Adding 5 does not affect spread.

New standard deviation = 2 × 3.5 = 7

Answer: 7 [2 marks — 1 for correct scaling, 1 for noting addition has no effect]

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Question 4 [4 marks]

(a) Combined mean:

$\bar{x}_{combined} = \frac{n_A \bar{x}_A + n_B \bar{x}_B}{n_A + n_B} = \frac{20 \times 25 + 30 \times 30}{50} = \frac{500 + 900}{50} = \frac{1400}{50} = 28$

Answer: 28 [2 marks — 1 for correct formula, 1 for correct value]

(b) Combined standard deviation:

The combined standard deviation would be closer to 6 because data set B has a larger sample size (30 vs 20) and a larger spread. The larger sample size gives more weight to data set B's characteristics, pulling the combined standard deviation toward B's value.

Answer: Closer to 6, because data set B has larger size and greater spread. [2 marks — 1 for correct direction, 1 for valid reasoning]

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Question 5 [4 marks]

(a) Box-and-whisker plot:

    40    48      55      62    75
    |-----|       |       |-----|
    o─────[   |   ]─────────o
         Q1  Median  Q3
    Min              Max

• Minimum: 40, Q1: 48, Median: 55, Q3: 62, Maximum: 75
• Box from Q1 to Q3 with median line inside
• Whiskers from min to Q1 and from Q3 to max

[3 marks — 1 for correct scale/positions, 1 for box, 1 for whiskers]

(b) Skewness:

The median (55) is closer to Q3 (62) than to Q1 (48). The left whisker (48 − 40 = 8) is shorter than the right whisker (75 − 62 = 13). This suggests the distribution is slightly negatively skewed (skewed to the left), as the longer tail extends to the lower values.

Answer: Slightly negatively skewed / skewed to the left. [1 mark]

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Section B: Probability

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Question 6 [5 marks]

Total balls = 5 + 4 + 3 = 12

(a) P(both red):

$P(\text{both red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$

Answer: 5/33 or 0.152 (3 s.f.) [2 marks — 1 for first draw, 1 for second draw]

(b) P(different colours):

P(different) = 1 − P(same colour)

P(both red) = 5/33
P(both blue) = (4/12) × (3/11) = 12/132 = 1/11
P(both green) = (3/12) × (2/11) = 6/132 = 1/22

P(same) = 5/33 + 1/11 + 1/22 = 10/66 + 6/66 + 3/66 = 19/66

P(different) = 1 − 19/66 = 47/66

Answer: 47/66 or 0.712 (3 s.f.) [3 marks — 1 for strategy, 1 for P(same), 1 for final answer]

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Question 7 [5 marks]

(a) P(A ∪ B):

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.4 - 0.2 = 0.8$

Answer: 0.8 [1 mark]

(b) P(A | B):

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.4} = 0.5$

Answer: 0.5 [2 marks — 1 for formula, 1 for correct value]

(c) Independence check:

If A and B are independent, then P(A ∩ B) = P(A) × P(B) = 0.6 × 0.4 = 0.24.

Since P(A ∩ B) = 0.2 ≠ 0.24, events A and B are not independent.

Answer: Not independent, because P(A ∩ B) ≠ P(A) × P(B). [2 marks — 1 for test, 1 for conclusion]

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Question 8 [6 marks]

(a) Sample space:

	1	2	3	4	5	6
1	(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
2	(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
3	(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
4	(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
5	(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(5,6)
6	(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

Total outcomes = 36 [2 marks — 1 for structure, 1 for completeness]

(b) P(sum = 8):

Favourable outcomes: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes

$P(\text{sum} = 8) = \frac{5}{36}$

Answer: 5/36 [2 marks — 1 for listing outcomes, 1 for probability]

(c) P(at least one 6):

P(at least one 6) = 1 − P(no 6s) = 1 − (5/6 × 5/6) = 1 − 25/36 = 11/36

Answer: 11/36 [2 marks — 1 for complementary approach, 1 for correct value]

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Question 9 [6 marks]

Let P = Physics, C = Chemistry. n = 30, n(P) = 18, n(C) = 15, n(P ∩ C) = 10.

(a) P(Physics but not Chemistry):

n(P only) = 18 − 10 = 8

$P(\text{P only}) = \frac{8}{30} = \frac{4}{15}$

Answer: 4/15 or 0.267 (3 s.f.) [2 marks — 1 for finding count, 1 for probability]

(b) P(neither):

n(P ∪ C) = 18 + 15 − 10 = 23
n(neither) = 30 − 23 = 7

$P(\text{neither}) = \frac{7}{30}$

Answer: 7/30 or 0.233 (3 s.f.) [2 marks — 1 for union, 1 for complement]

(c) P(Physics | Chemistry):

$P(P|C) = \frac{n(P \cap C)}{n(C)} = \frac{10}{15} = \frac{2}{3}$

Answer: 2/3 or 0.667 (3 s.f.) [2 marks — 1 for formula, 1 for value]

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Question 10 [5 marks]

(a) Probability tree diagram:

Spinner 1 (each branch: prob 1/4)     Spinner 2 (each branch: prob 1/3)     Outcome (prob 1/12)

1 ─── 1  (1,1) [1/12]
 ├─── 2  (1,2) [1/12]
 └─── 3  (1,3) [1/12]

2 ─── 1  (2,1) [1/12]
 ├─── 2  (2,2) [1/12]
 └─── 3  (2,3) [1/12]

3 ─── 1  (3,1) [1/12]
 ├─── 2  (3,2) [1/12]
 └─── 3  (3,3) [1/12]

4 ─── 1  (4,1) [1/12]
 ├─── 2  (4,2) [1/12]
 └─── 3  (4,3) [1/12]

Total outcomes = 4 × 3 = 12, each with probability 1/12. [3 marks — 1 for structure, 1 for outcomes, 1 for probabilities]

(b) P(sum = 5):

Favourable outcomes: (2,3), (3,2), (4,1) → 3 outcomes

$P(\text{sum} = 5) = \frac{3}{12} = \frac{1}{4}$

Answer: 1/4 or 0.25 [2 marks — 1 for identifying outcomes, 1 for probability]

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Section C: Permutations, Combinations, and Advanced Probability

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Question 11 [3 marks]

Word: MATHEMATICS (11 letters)

Repeated letters: M appears 2 times, A appears 2 times, T appears 2 times.

$Number\ of\ arrangements = \frac{11!}{2! \times 2! \times 2!} = \frac{39916800}{8} = 4989600$

Answer: 4,989,600 [3 marks — 1 for identifying repeated letters, 1 for correct formula, 1 for correct value]

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Question 12 [6 marks]

Total people = 6 boys + 5 girls = 11

(a) No restrictions:

$\binom{11}{4} = \frac{11!}{4! \times 7!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$

Answer: 330 [2 marks — 1 for formula, 1 for value]

(b) Exactly 2 boys and 2 girls:

$\binom{6}{2} \times \binom{5}{2} = 15 \times 10 = 150$

Answer: 150 [2 marks — 1 for each combination, 1 for product]

(c) At least 1 girl:

P(at least 1 girl) = Total − P(no girls) = Total − P(all boys)

$= \binom{11}{4} - \binom{6}{4} = 330 - 15 = 315$

Answer: 315 [2 marks — 1 for complementary approach, 1 for value]

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Question 13 [5 marks]

Total cards = 8. Even numbers: {2, 4, 6, 8} (4 cards). Odd numbers: {1, 3, 5, 7} (4 cards).

(a) P(both even):

$P(\text{both even}) = \frac{4}{8} \times \frac{3}{7} = \frac{12}{56} = \frac{3}{14}$

Answer: 3/14 or 0.214 (3 s.f.) [2 marks — 1 for first draw, 1 for second draw]

(b) P(sum > 10):

Total ways to choose 2 cards from 8 = C(8,2) = 28.

Pairs with sum > 10: (3,8), (4,7), (4,8), (5,6), (5,7), (5,8), (6,7), (6,8), (7,8) → 9 pairs

$P(\text{sum} > 10) = \frac{9}{28}$

Answer: 9/28 or 0.321 (3 s.f.) [3 marks — 1 for total pairs, 1 for listing favourable pairs, 1 for probability]

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Question 14 [6 marks]

Word: PROBABILITY (11 letters)

Vowels: O, A, I, I → 4 vowels (I appears twice)
Consonants: P, R, B, B, L, T, Y → 7 consonants (B appears twice)

Total ways to choose 3 letters from 11 = C(11,3) = 165

(a) P(all three consonants):

Ways to choose 3 consonants from 7 = C(7,3) = 35

$P(\text{all consonants}) = \frac{35}{165} = \frac{7}{33}$

Answer: 7/33 or 0.212 (3 s.f.) [3 marks — 1 for identifying consonants, 1 for C(7,3), 1 for probability]

(b) P(exactly one vowel):

Ways = C(4,1) × C(7,2) = 4 × 21 = 84

$P(\text{exactly one vowel}) = \frac{84}{165} = \frac{28}{55}$

Answer: 28/55 or 0.509 (3 s.f.) [3 marks — 1 for C(4,1), 1 for C(7,2), 1 for probability]

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Question 15 [5 marks]

(a) P(exactly 2 heads in 3 tosses):

Using binomial probability: X ~ B(3, p)

$P(X = 2) = \binom{3}{2} p^2 (1-p)^1 = 3p^2(1-p)$

Answer: 3p²(1 − p) [2 marks — 1 for binomial coefficient, 1 for correct expression]

(b) Find p:

$3p^2(1-p) = 0.288$

$3p^2 - 3p^3 = 0.288$

$3p^3 - 3p^2 + 0.288 = 0$

Testing p = 0.4: 3(0.064) − 3(0.16) + 0.288 = 0.192 − 0.48 + 0.288 = 0 ✓

Answer: p = 0.4 [3 marks — 1 for setting up equation, 1 for rearranging, 1 for solving]

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Section D: Discrete Random Variables and Binomial Distribution

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Question 16 [6 marks]

(a) E(X):

$E(X) = \sum x \cdot P(X=x) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.25) + 4(0.15)$ $= 0 + 0.2 + 0.6 + 0.75 + 0.6 = 2.15$

Answer: 2.15 [2 marks — 1 for correct products, 1 for sum]

(b) Var(X):

$E(X^2) = 0^2(0.1) + 1^2(0.2) + 2^2(0.3) + 3^2(0.25) + 4^2(0.15)$ $= 0 + 0.2 + 1.2 + 2.25 + 2.4 = 6.05$

$Var(X) = E(X^2) - [E(X)]^2 = 6.05 - (2.15)^2 = 6.05 - 4.6225 = 1.4275$

Answer: 1.4275 or 1.43 (3 s.f.) [3 marks — 1 for E(X²), 1 for [E(X)]², 1 for Var(X)]

(c) P(X ≥ 2):

$P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) = 0.3 + 0.25 + 0.15 = 0.7$

Answer: 0.7 [1 mark]

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Question 17 [7 marks]

X ~ B(10, 0.3)

(a) P(X = 4):

$P(X = 4) = \binom{10}{4} (0.3)^4 (0.7)^6 = 210 \times 0.0081 \times 0.117649 = 0.2001$

Answer: 0.200 (3 s.f.) [2 marks — 1 for formula, 1 for value]

(b) P(X ≥ 2):

$P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

$P(X = 0) = (0.7)^{10} = 0.02825$

$P(X = 1) = \binom{10}{1} (0.3)^1 (0.7)^9 = 10 \times 0.3 \times 0.04035 = 0.12106$

$P(X \geq 2) = 1 - 0.02825 - 0.12106 = 0.8507$

Answer: 0.851 (3 s.f.) [3 marks — 1 for P(X=0), 1 for P(X=1), 1 for final answer]

(c) E(X) and Var(X):

$E(X) = np = 10 \times 0.3 = 3$

$Var(X) = np(1-p) = 10 \times 0.3 \times 0.7 = 2.1$

Answer: E(X) = 3, Var(X) = 2.1 [2 marks — 1 for each]

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Question 18 [6 marks]

(a) Assumptions for binomial distribution:

1. Each bulb is independent of the others (the defect status of one bulb does not affect another).
2. The probability of a bulb being defective is constant (p = 0.05) for each bulb.
3. There are only two outcomes: defective or not defective.
4. The number of trials is fixed (n = 20).

(Any two valid assumptions accepted) [2 marks — 1 for each valid assumption]

(b) P(exactly 2 defective):

X ~ B(20, 0.05)

$P(X = 2) = \binom{20}{2} (0.05)^2 (0.95)^{18} = 190 \times 0.0025 \times 0.3972 = 0.1887$

Answer: 0.189 (3 s.f.) [2 marks — 1 for formula, 1 for value]

(c) P(at least 1 defective):

$P(X \geq 1) = 1 - P(X = 0) = 1 - (0.95)^{20} = 1 - 0.3585 = 0.6415$

Answer: 0.642 (3 s.f.) [2 marks — 1 for complementary approach, 1 for value]

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Question 19 [7 marks]

X ~ B(15, 0.25) where p = 1/4 (one correct option out of four).

(a) P(exactly 5 correct):

$P(X = 5) = \binom{15}{5} (0.25)^5 (0.75)^{10} = 3003 \times 0.0009766 \times 0.05631 = 0.1651$

Answer: 0.165 (3 s.f.) [2 marks — 1 for formula, 1 for value]

(b) P(at least 3 correct):

$P(X \geq 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)$

$P(X = 0) = (0.75)^{15} = 0.01336$

$P(X = 1) = \binom{15}{1} (0.25)^1 (0.75)^{14} = 15 \times 0.25 \times 0.01782 = 0.06682$

$P(X = 2) = \binom{15}{2} (0.25)^2 (0.75)^{13} = 105 \times 0.0625 \times 0.02376 = 0.1559$

$P(X \geq 3) = 1 - 0.01336 - 0.06682 - 0.1559 = 0.7639$

Answer: 0.764 (3 s.f.) [3 marks — 1 for P(X=0), 1 for P(X=1) and P(X=2), 1 for final answer]

(c) Most likely number of correct answers:

For X ~ B(15, 0.25), the mode is at floor((n+1)p) = floor(16 × 0.25) = floor(4) = 4.

Alternatively, check: P(X=3) ≈ 0.225, P(X=4) ≈ 0.225, P(X=5) ≈ 0.165.

The most likely number is 4 (or 3 and 4 are equally likely).

Answer: 4 [2 marks — 1 for method, 1 for answer]

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Question 20 [9 marks]

X ~ B(n, 0.4), E(X) = 6.

(a) Find n:

$E(X) = np = n \times 0.4 = 6$

$n = \frac{6}{0.4} = 15$

Answer: n = 15 [1 mark]

(b) P(X = 5):

X ~ B(15, 0.4)

$P(X = 5) = \binom{15}{5} (0.4)^5 (0.6)^{10} = 3003 \times 0.01024 \times 0.006047 = 0.1859$

Answer: 0.186 (3 s.f.) [2 marks — 1 for formula, 1 for value]

(c) P(4 ≤ X ≤ 7):

$P(4 \leq X \leq 7) = P(X=4) + P(X=5) + P(X=6) + P(X=7)$

$P(X=4) = \binom{15}{4} (0.4)^4 (0.6)^{11} = 1365 \times 0.0256 \times 0.003628 = 0.1268$

$P(X=5) = 0.1859 \text{ (from part b)}$

$P(X=6) = \binom{15}{6} (0.4)^6 (0.6)^9 = 5005 \times 0.004096 \times 0.01008 = 0.2066$

$P(X=7) = \binom{15}{7} (0.4)^7 (0.6)^8 = 6435 \times 0.001638 \times 0.01680 = 0.1771$

$P(4 \leq X \leq 7) = 0.1268 + 0.1859 + 0.2066 + 0.1771 = 0.6964$

Answer: 0.696 (3 s.f.) [3 marks — 1 for each probability, 1 for sum]

(d) Smallest k such that P(X ≤ k) ≥ 0.95:

We need the cumulative probability to reach at least 0.95.

$P(X \leq 7) = P(X=0) + P(X=1) + ... + P(X=7)$

Using cumulative values:

• P(X ≤ 6) ≈ 0.4032 + 0.1859 + 0.2066 = ... (building up)
• P(X ≤ 7) ≈ 0.6964 + lower terms ≈ 0.8836
• P(X ≤ 8) ≈ 0.8836 + P(X=8) where P(X=8) = C(15,8)(0.4)^8(0.6)^7 ≈ 0.1181 → P(X ≤ 8) ≈ 0.9617

Since P(X ≤ 7) ≈ 0.884 < 0.95 and P(X ≤ 8) ≈ 0.962 ≥ 0.95:

Answer: k = 8 [3 marks — 1 for cumulative approach, 1 for checking values, 1 for correct k]

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End of Answer Key