1. [Maximum mark: 4]
The probability distribution table for a random variable \( X \) is shown below.
| \( x \) | −2 | −1 | 0.5 | 1 | 2 |
| \( P(X = x) \) | 0.12 | \( p \) | \( q \) | 0.16 | 0.3 |
Given that \( E(X) = 0.28 \), find the value of \( p \) and the value of \( q \).
▶️Answer/Explanation
Solution:
Sum of probabilities: \( 0.12 + p + q + 0.16 + 0.3 = 1 \) → \( p + q = 0.42 \).
Using \( E(X) = 0.28 \): \(-0.24 – p + 0.5q + 0.16 + 0.6 = 0.28 \) → \(-p + 0.5q = -0.24\).
Solving the system:
\( p = 0.3 \), \( q = 0.12 \).
2. [Maximum mark: 8]
The residents of Persham were surveyed about the reliability of their internet service. 12% rated the service as ‘poor’, 36% rated it as ‘satisfactory’ and 52% rated it as ‘good’.
(a) A random sample of 8 residents of Persham is chosen. Find the probability that more than 2 and fewer than 8 of them rate their internet service as poor or satisfactory.
(b) A random sample of 125 residents of Persham is now chosen. Use an approximation to find the probability that more than 72 of these residents rate their internet service as good.
▶️Answer/Explanation
(a) Solution:
\( P(\text{poor or satisfactory}) = 0.48 \).
\( P(3 \leq X \leq 7) = 1 – P(0,1,2,8) = 1 – (0.005346 + 0.039478 + 0.127544 + 0.002818) = 0.825 \).
(b) Solution:
Normal approximation with \( \mu = 125 \times 0.52 = 65 \), \( \sigma^2 = 125 \times 0.52 \times 0.48 = 31.2 \).
With continuity correction: \( P(X > 72) = P\left(Z > \frac{72.5 – 65}{\sqrt{31.2}}\right) = 1 – \Phi(1.343) = 0.0896 \).
3. [Maximum mark: 9]
The Lions and the Tigers are two basketball clubs. The heights (in cm) of their first team squads are given:
(a) Draw a back-to-back stem-and-leaf diagram for both teams (Lions on the left).
(b) Find the median and interquartile range (IQR) of the Lions’ heights.
(c) Compare the heights of the Lions and Tigers using their quartiles.
▶️Answer/Explanation
(a) Solution:
Stem-and-leaf diagram omitted for brevity.
(b) Solution:
Lions’ median = 186 cm, IQR = 190 cm (UQ) − 179 cm (LQ) = 11 cm.
(c) Solution:
1. Tigers are generally taller (higher median).
2. Tigers’ heights are less consistent (larger IQR).
4. [Maximum mark: 9]
In a large population, adults’ systolic blood pressure (SBP) is normally distributed with mean 125.4 and standard deviation 18.6.
(a) Find the probability that a randomly chosen adult has SBP less than 132.
(b) For 12-year-old children in the same population, SBP is normally distributed with mean 117. Given that 88% have SBP > 108, find the standard deviation.
(c) Three adults are chosen at random. Find the probability that all three have SBP within 1.5 standard deviations of the mean.
▶️Answer/Explanation
(a) Solution:
\( P(X < 132) = P\left(Z < \frac{132 – 125.4}{18.6}\right) = \Phi(0.3548) = 0.639 \).
(b) Solution:
\( P(X > 108) = 0.88 \) → \( z = -1.175 \).
\( \sigma = \frac{108 – 117}{-1.175} = 7.66 \).
(c) Solution:
\( P(-1.5 < Z < 1.5) = 2\Phi(1.5) – 1 = 0.8664 \).
For three adults: \( 0.8664^3 = 0.650 \).
5. [Maximum mark: 10]
A game is played with a fair 6-sided die:
- If the first throw is 2, 3, 4, or 5, that is the player’s score.
- If the first throw is 1 or 6, the player throws again and sums the two results.
(a) Draw a fully labelled tree diagram for the game.
(b) Show that \( P(A) = \frac{1}{3} \), where \( A \) is the event “score is 5, 6, 7, 8, or 9”.
(c) Determine whether events \( A \) and \( B \) (player has two throws) are independent.
(d) Find \( P(B \mid A’) \).
▶️Answer/Explanation
(a) Solution:
(b) Solution:
\( P(A) = \frac{7}{36} (5) + \frac{1}{36} (6) + \frac{2}{36} (7) + \frac{1}{36} (8) + \frac{1}{36} (9) = \frac{12}{36} = \frac{1}{3} \).
(c) Solution:
\( P(B) = \frac{1}{3} \), \( P(A \cap B) = \frac{6}{36} \). Since \( \frac{6}{36} \neq \frac{1}{9} \), \( A \) and \( B \) are not independent.
(d) Solution:
\( P(B \mid A’) = \frac{P(B \cap A’)}{P(A’)} = \frac{\frac{6}{36}}{\frac{2}{3}} = \frac{1}{4} \).
6. [Maximum mark: 10]
A Social Club has 15 members (8 men, 7 women). A committee of 5 is formed.
(a) Find the number of ways to form the committee if it must include more men than women.
(b) In how many ways can the 15 members be divided into groups of 3, 5, and 7 for a photograph?
(c) For a photo, 7 members stand in the back row. If Abel and Betty must stand next to each other, and Freya and Gino must not, find the number of valid arrangements.
▶️Answer/Explanation
(a) Solution:
Cases: 4M1W (\( \binom{8}{4}\binom{7}{1} = 490 \)) + 3M2W (\( \binom{8}{3}\binom{7}{2} = 1176 \)) = 1722 ways.
(b) Solution:
\( \binom{15}{3}\binom{12}{5}\binom{7}{7} = 360360 \).
(c) Solution:
Total arrangements with AB together: \( 6! \times 2 = 1440 \).
Subtract arrangements with AB and FG together: \( 5! \times 2 \times 2 = 480 \).
Valid arrangements: \( 1440 – 480 = 960 \).