1. [Maximum mark: 4]
The times taken by 120 children to complete a particular puzzle are represented in the cumulative frequency graph.
(a) Use the graph to estimate the interquartile range of the data. [2]
(b) 35% of the children took longer than T seconds to complete the puzzle. Use the graph to estimate the value of T. [2]
▶️Answer/Explanation
(a) The interquartile range (IQR) is calculated as the difference between the upper quartile (UQ) and lower quartile (LQ). From the graph:
UQ ≈ 31 seconds, LQ ≈ 23.7 seconds.
IQR = 31 – 23.7 = 7.3 seconds.
(b) 35% of 120 children = 42 children took longer than T seconds. From the graph, the cumulative frequency for T is 120 – 42 = 78.
Reading from the graph, T ≈ 28.5 seconds.
2. [Maximum mark: 7]
Hazeem repeatedly throws two ordinary fair 6-sided dice at the same time. On each occasion, the score is the sum of the two numbers obtained.
(a) Find the probability that it takes exactly 5 throws of the two dice for Hazeem to obtain a score of 8 or more. [2]
(b) Find the probability that it takes no more than 4 throws of the two dice for Hazeem to obtain a score of 8 or more. [2]
(c) For 8 randomly chosen throws of the two dice, find the probability that Hazeem obtains a score of 8 or more on fewer than 3 occasions. [3]
▶️Answer/Explanation
(a) The probability of obtaining a score of 8 or more in one throw is \( \frac{15}{36} = \frac{5}{12} \). The probability of not obtaining a score of 8 or more in one throw is \( \frac{21}{36} = \frac{7}{12} \).
For exactly 5 throws: \( \left(\frac{7}{12}\right)^4 \times \frac{5}{12} = \frac{12005}{248832} \approx 0.0482.
(b) The probability of obtaining a score of 8 or more in no more than 4 throws is the sum of probabilities for 1 to 4 throws:
\( 1 – \left(\frac{7}{12}\right)^4 = \frac{18335}{20736} \approx 0.884.
(c) Using the binomial distribution for 8 throws, the probability of fewer than 3 successes (scores of 8 or more) is:
\( P(0) + P(1) + P(2) = \left(\frac{7}{12}\right)^8 + 8 \times \left(\frac{5}{12}\right) \left(\frac{7}{12}\right)^7 + 28 \times \left(\frac{5}{12}\right)^2 \left(\frac{7}{12}\right)^6 \approx 0.282.
3. [Maximum mark: 11]
A farmer sells eggs. The weights, in grams, of the eggs can be modelled by a normal distribution with mean 80.5 and standard deviation 6.6. Eggs are classified as small, medium, or large according to their weight. A small egg weighs less than 76 grams, and 40% of the eggs are classified as medium.
(a) Find the percentage of eggs that are classified as small. [3]
(b) Find the least possible weight of an egg classified as large. [3]
(c) Use an approximation to find the probability that more than 68 out of 150 eggs weighed last week were classified as medium. [5]
▶️Answer/Explanation
(a) Calculate \( P(X < 76) \):
\( z = \frac{76 – 80.5}{6.6} \approx -0.6818 \).
\( P(Z < -0.6818) = 1 – 0.7524 = 0.2476 \).
The percentage of small eggs is 24.8%.
(b) The least weight for a large egg corresponds to the 60th percentile (100% – 40% medium).
\( P(Z > z) = 0.4 \Rightarrow z \approx -0.2533 \).
\( x = 80.5 + (-0.2533 \times 6.6) \approx 78.83 \) grams.
The least weight for a large egg is 83.0 grams (rounded).
(c) Using the normal approximation to the binomial distribution:
Mean \( \mu = 150 \times 0.4 = 60 \), variance \( \sigma^2 = 150 \times 0.4 \times 0.6 = 36 \).
\( P(X > 68) \approx P\left(Z > \frac{68.5 – 60}{6}\right) = P(Z > 1.417) \approx 1 – 0.9217 = 0.0783.
4. [Maximum mark: 9]
The times, to the nearest minute, of 150 athletes taking part in a charity run are recorded. The results are summarised in the table.
| Time in minutes | 101–120 | 121–130 | 131–135 | 136–145 | 146–160 |
|---|---|---|---|---|---|
| Frequency | 18 | 48 | 34 | 32 | 18 |
(a) Draw a histogram to represent this information. [4]
(b) Calculate estimates for the mean and standard deviation of the times taken by the athletes. [5]
▶️Answer/Explanation
(a) The histogram should have:
– Class widths: 20, 10, 5, 10, 15.
– Frequency densities: 0.9, 4.8, 6.8, 3.2, 1.2.
– Correct bar heights and labeled axes (frequency density vs. time in minutes).
(b) Using midpoints (110.5, 125.5, 133, 140.5, 153):
Mean \( = \frac{18 \times 110.5 + 48 \times 125.5 + 34 \times 133 + 32 \times 140.5 + 18 \times 153}{150} \approx 131.9 minutes.
Variance \( = \frac{18 \times 110.5^2 + 48 \times 125.5^2 + 34 \times 133^2 + 32 \times 140.5^2 + 18 \times 153^2}{150} – 131.9^2 \approx 137.54 \).
Standard deviation \( \approx \sqrt{137.54} \approx 11.7 minutes.
5. [Maximum mark: 9]
A red spinner has four sides labeled 1, 2, 3, 4. The random variable X denotes the score obtained when the spinner is spun. The probability distribution for X is given below:
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X=x) | 0.28 | p | 2p | 3p |
(a) Show that p = 0.12. [1]
(b) A fair blue spinner and a fair green spinner, each with sides labeled 1, 2, 3, 4, are spun alongside the red spinner. Find the probability that the sum of the three scores is 4 or less. [3]
(c) Find the probability that the product of the three scores is 4 or less, given that X is odd. [5]
▶️Answer/Explanation
(a) Sum of probabilities: \( 0.28 + p + 2p + 3p = 1 \).
Solving gives \( 6p = 0.72 \), so \( p = 0.12 \).
(b) Possible combinations for sum ≤ 4:
(1,1,1), (1,1,2), (1,2,1), (2,1,1).
Probability \( = 0.28 \times (0.25)^2 \times 4 + 0.12 \times (0.25)^2 \times 3 = 0.06.
(c) Given X is odd (1 or 3), the probability is:
\( P(\text{product} \leq 4 \cap X \text{ odd}) = 0.28 \times (0.25)^2 \times 8 + 0.24 \times (0.25)^2 \).
\( P(X \text{ odd}) = 0.28 + 0.24 = 0.52 \).
Conditional probability \( = \frac{0.14 + 0.015}{0.52} \approx 0.298.
6. [Maximum mark: 10]
In a restaurant, rectangular tables seat four people each (two on each longer side). Eight friends, including Rajid, Sue, and Tan, have booked two tables, X and Y.
(a) Find the number of ways to divide the eight friends into two groups of 4 if Rajid and Sue must sit at the same table, and Tan must sit at the other table. [3]
(b) Find the number of seating arrangements if Rajid and Sue sit at table X on the same side, and Tan sits at any table. [3]
(c) Find the number of arrangements for a photograph line if Rajid and Sue stand next to each other but neither is at an end of the line. [4]
▶️Answer/Explanation
(a) Rajid and Sue are fixed at one table, Tan at the other. The remaining 5 friends can be split as \( \binom{5}{1} \) (for Tan’s table) and \( \binom{4}{3} \) (for Rajid and Sue’s table).
Total ways \( = \binom{5}{1} \times \binom{4}{3} = 20.
(b) Rajid and Sue are fixed on one side of table X (2 choices). The remaining 6 friends can be seated in \( 6! \) ways.
Total arrangements \( = 2 \times 6! = 1440.
(c) Treat Rajid and Sue as a single entity (2 arrangements). This entity can occupy any of the 6 middle positions in the line. The remaining 6 friends can be arranged in \( 6! \) ways.
Total arrangements \( = 2 \times 6 \times 6! = 8640.