Figure 2 shows a histogram for the variable t which represents the time taken, in minutes, by a group of people to swim 500m - Edexcel - A-Level Maths: Statistics - Question 5 - 2007 - Paper 2

To estimate the number of people who took longer than 20 minutes, we sum the frequencies for intervals 25-40 minutes:

• The frequency for 25-40 is 15.

So, the estimated number of people who took longer than 20 minutes is 15.

The mid points for each interval are calculated as follows:

• 5-10: 7.5
• 10-14: 12
• 14-18: 16
• 18-25: 21.5
• 25-40: 32.5

Using the midpoints:

The total frequency (n) = 10 + 16 + 24 + 7 + 15 = 72.

The mean time is estimated by: $ext{Mean} = \frac{(7.5 \times 10) + (12 \times 16) + (16 \times 24) + (21.5 \times 7) + (32.5 \times 15)}{72} = \frac{1891}{72} \approx 18.91$

The estimated standard deviation is calculated using the formula:

$\sigma = \sqrt{\frac{\sum (f \cdot (x - \bar{x})^2)}{n}}$

Substituting the values, we get: $\sigma \approx \sqrt{\frac{41033}{72}} \approx 7.26$.

To find the median and quartiles, we utilize the cumulative frequency:

• Median (Q2) is found such that 50% of the data lies below it. With a total of 72, the median is the average of the 36th and 37th data points in the cumulative frequency distribution.
• First Quartile (Q1) is the 18th data point, and Third Quartile (Q3) is the 54th.

Using values from the cumulative frequency, we find:

• Median: Approximately 18.5,
• Q1: 13.75,
• Q3: 23.75.

The skewness can be evaluated using the formula:

$\text{Skewness} = \frac{3(\text{mean} - \text{median})}{\text{standard deviation}}$

Substituting the values:

• mean = 18.91,
• median = 18.5,
• standard deviation = 7.26, We find: $\text{Skewness} \approx 0.36$ This indicates that the data is positively skewed, meaning there are a few data points significantly higher than the rest.