In a shopping survey a random sample of 104 teenagers were asked how many hours, to the nearest hour, they spent shopping in the last month - Edexcel - A-Level Maths Statistics - Question 5 - 2009 - Paper 1

To find the median, we first find the cumulative frequencies. The median is the value at the position of the (N+1)/2, where N is the total frequency.

The cumulative frequencies up to each group are:

• 0 – 5: 20
• 6 – 7: 36
• 8 – 10: 54
• 11 – 15: 79
• 16 – 25: 94
• 26 – 50: 104

The median is at position 52 (since (104+1)/2 = 52.5). The 8 – 10 group is where the median lies. Using linear interpolation:

$Q_2 = L + \left(\frac{\frac{N}{2} - F}{f}\right) \times c \quad \text{(where L = lower boundary, F = cumulative frequency before group, f = frequency, c = interval width)}$

Calculating the median value:

• L = 8, F = 36, f = 18, c = 2

$Q_2 = 8 + \left(\frac{52 - 36}{18}\right) \times 2 = 8 + \left(\frac{16}{18}\right) \times 2 \approx 8 + 1.78 = 9.78$

For the interquartile range, we follow a similar process to find Q1 (position 26) and Q3 (position 78).

• For Q1: L = 6, F = 20, f = 16
• For Q3: L = 20, F = 54, f = 15

Using linear interpolation for both, we can derive the Q1 and Q3 and get the IQR as Q3 - Q1.

The mean can be estimated using:

$\text{Mean} = \frac{\sum f \cdot x}{N}$

Where (f) is the frequency and (x) is the midpoint of the hours. The values are:

• For 0 – 5: (20)(2.75) = 55
• For 6 – 7: (16)(6.5) = 104
• For 8 – 10: (18)(9) = 162
• For 11 – 15: (25)(13) = 325
• For 16 – 25: (15)(20.5) = 307.5
• For 26 – 50: (10)(38) = 380

Total convolution: 55 + 104 + 162 + 325 + 307.5 + 380 = 1333.5

Thus, Mean = $\frac{1333.5}{104} \approx 12.8$

For standard deviation, use:

$S_d = \sqrt{\frac{\sum f(x-\bar{x})^2}{N-1}}$

Where (\bar{x}) is the mean calculated. Substitute into the formula to find the standard deviation.

The data appears positively skewed because the mean (higher value) is greater than the median, indicating that more values are clustered at the lower end of the distribution.

The median and interquartile range would be preferable to summarise the data, as they are not affected by extreme values or outliers, providing a more accurate reflection of the data's central tendency and variation.