9709 P51 - Nov 2023 - Q4
2433
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 |
Calculate estimates for the mean and standard deviation of the times taken by the athletes.
Solution
First, calculate the midpoints of each class interval:
- 101 – 120: midpoint = 110.5
- 121 – 130: midpoint = 125.5
- 131 – 135: midpoint = 133
- 136 – 145: midpoint = 140.5
- 146 – 160: midpoint = 153
Calculate the mean using the formula:
\(\bar{x} = \frac{\sum (f \times x)}{\sum f} = \frac{18 \times 110.5 + 48 \times 125.5 + 34 \times 133 + 32 \times 140.5 + 18 \times 153}{150}\)
\(= \frac{1989 + 6024 + 4522 + 4496 + 2754}{150} = 131.9\)
\(Calculate the variance using the formula:\)
\(s^2 = \frac{\sum (f \times x^2)}{\sum f} - (\bar{x})^2\)
\(= \frac{18 \times 110.5^2 + 48 \times 125.5^2 + 34 \times 133^2 + 32 \times 140.5^2 + 18 \times 153^2}{150} - (131.9)^2\)
\(= \frac{3200 + 41400 + 194400 + 157300 + 153600}{150} - 17397.61\)
\(= 137.54\)
\(Standard deviation \(s = \sqrt{137.54} = 11.7 \]\)
