1. [Maximum mark: 3]
The time taken, \( t \) minutes, to complete a puzzle was recorded for each of 150 students. These times are summarised in the table.
| Time taken (\( t \) minutes) | \( t \leq 25 \) | \( t \leq 50 \) | \( t \leq 75 \) | \( t \leq 100 \) | \( t \leq 150 \) | \( t \leq 200 \) |
|---|---|---|---|---|---|---|
| Cumulative frequency | 16 | 44 | 86 | 104 | 132 | 150 |
(a) Draw a cumulative frequency graph to illustrate the data.
(b) Use your graph to estimate the 20th percentile of the data.
▶️Answer/Explanation
(a) Plot points at (25,16), (50,44), (75,86), (100,104), (150,132), (200,150). Join with a smooth curve.
cumulative frequency
(b) The 20th percentile corresponds to cumulative frequency = 30. From the graph, estimate \( t \) between 37.5 and 42 minutes.
2. [Maximum mark: 3]
Twenty children were asked to estimate the height of a particular tree. Their estimates, in metres, were as follows:
4.1, 4.2, 4.4, 4.5, 4.6, 4.8, 5.0, 5.2, 5.3, 5.4, 5.5, 5.8, 6.0, 6.2, 6.3, 6.4, 6.6, 6.8, 6.9, 19.4
(a) Find the mean of the estimated heights.
(b) Find the median of the estimated heights.
(c) Give a reason why the median is likely to be more suitable than the mean as a measure of central tendency for this information.
▶️Answer/Explanation
(a) Mean = \( \frac{123.4}{20} = 6.17 \) metres.
(b) Median = \( \frac{5.4 + 5.5}{2} = 5.45 \) metres.
(c) The mean is unduly influenced by the extreme value (19.4), whereas the median is robust to outliers.
3. [Maximum mark: 6]
The random variable \( X \) takes the values \(-2, 1, 2, 3\). It is given that \( P(X = x) = kx^2 \), where \( k \) is a constant.
(a) Draw up the probability distribution table for \( X \), giving the probabilities as numerical fractions.
(b) Find \( E(X) \) and \( Var(X) \).
▶️Answer/Explanation
(a)
| \( x \) | \(-2\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \( P(X = x) \) | \( \frac{4}{18} \) | \( \frac{1}{18} \) | \( \frac{4}{18} \) | \( \frac{9}{18} \) |
(b) \( E(X) = \frac{14}{9} \), \( Var(X) = \frac{317}{81} \).
3.91
4. [Maximum mark: 7]
Ramesh throws an ordinary fair 6-sided die.
(a) Find the probability that he obtains a 4 for the first time on his 8th throw.
(b) Find the probability that it takes no more than 5 throws for Ramesh to obtain a 4.
(c) For 10 randomly chosen throws of two dice, find the probability that Ramesh obtains a total of less than 4 on at least three throws.
▶️Answer/Explanation
(a) \( \left( \frac{5}{6} \right)^7 \times \frac{1}{6} = 0.0465 \).
(b) \( 1 – \left( \frac{5}{6} \right)^5 = \frac{4651}{7776} \).
(c) Probability of total < 4 = \( \frac{3}{36} = \frac{1}{12} \). Using binomial: \( 1 – P(0,1,2) = 0.0445 \).
5. [Maximum mark: 10]
Farmer Jones grows apples. The weights, in grams, of the apples are normally distributed with mean 170 and standard deviation 25. Apples weighing between 142g and 205g are sold to a supermarket.
(a) Find the probability that a randomly chosen apple is sold to the supermarket.
(b) Calculate an estimate for Farmer Jones’s total income from 20,000 apples.
(c) Farmer Tan’s apples follow \( N(182, 20^2) \). 72% weigh more than \( w \) grams. Find \( w \).
▶️Answer/Explanation
(a) \( P(142 < X < 205) = P(-1.12 < Z < 1.4) = 0.788 \).
(b) Income = \( (0.788 \times 0.24 + 0.0808 \times 0.30) \times 20000 = \$4266.24 \).
(c) \( P(Z > \frac{w – 182}{20}) = 0.72 \) → \( w = 170 \).
6. [Maximum mark: 10]
Sajid practises long jump. A jump >6m is a success. The probability of success is 0.2 (first jump) or 0.3/0.1 (subsequent jumps, depending on the previous jump).
(a) Draw a tree diagram for three jumps.
(b) Find the probability of exactly one success given at least one success.
(c) For six jumps, find the probability that only the first three or last three are successes.
▶️Answer/Explanation
(a) Tree diagram with branches for success/failure and conditional probabilities.
(b) \( P(\text{1 success} | \geq 1) = \frac{0.254}{0.352} = 0.722 \).
(c) \( 0.2 \times 0.3^2 \times 0.7 \times 0.9^2 + 0.8 \times 0.9^2 \times 0.1 \times 0.3^2 = 0.016 \).
7. [Maximum mark: 11]
A group of 15 friends (4 families) visit an adventure park. They travel in three cars (6, 5, 4 people).
(a) In how many ways can the 12 non-drivers be divided between the cars?
(b) In how many orders can the 15 friends enter if Mr Lizo goes first and families stay together?
(c) How many ways can a team of 4 adults and 3 children be chosen if the children are from different families?
(d) How many ways can the team include at least one of Mr Kenny or Mr Lizo?
▶️Answer/Explanation
(a) \( \binom{12}{3} \times \binom{9}{4} = 27720 \).
(b) \( 6! \times 2! \times 2! \times 3! = 414720 \).
(c) \( \binom{8}{4} \times \binom{4}{1} \times \binom{3}{1} \times \binom{1}{1} = 420 \).
(d) Total – neither = \( \binom{8}{4} \times \binom{7}{3} – \binom{6}{4} \times \binom{4}{3} = 1680 \).