A teacher selects a random sample of 56 students and records, to the nearest hour, the time spent watching television in a particular week - Edexcel - A-Level Maths Statistics - Question 5 - 2010 - Paper 2

The 26–30 group has:

• Width: The total range of hours from 26 to 30 is 5 hours.
• Frequency for this group is 21, which means the height of the bar in the histogram can be found using the width of the 11–20 group as a reference. Given that its width is 4 cm and height is 6 cm, the height of the 26–30 group can be calculated similarly, resulting in a height of 10.4 cm (as ratios can be maintained).

To find the mean, we use the formula:

$\bar{x} = \frac{\sum{f \times x}}{N}$

where:

• ( f ) = frequency of each group, ( x ) = mid-point of each group, and ( N ) = total frequency.

Calculating: [ \sum{f \times x} = (6 \times 5.5) + (15 \times 15.5) + (21 \times 28) + (13 \times 35.5) + (3 \times 50) = 1316.5 ] [ N = 56 ] [ \bar{x} = \frac{1316.5}{56} \approx 23.5 ]

For standard deviation, we use: $\sigma = \sqrt{\frac{\sum{f (x - \bar{x})^2}}{N}}$ Estimation gives: ( \sigma \approx 10.8 ).

The median lies between the lower quartile (Q1) and upper quartile (Q3). For the median (Q2):

• We find it within rank 28, [ Q_2 = Q(\frac{N+1}{2}) = Q(28) ]. Using the corresponding frequency and cumulative frequency, we get ( Q_2 ) as approximately ( 28 ).

Given that the lower quartile (15.8) is less than the median (28) and the upper quartile (29.3) is more than the median, the data is negatively skewed. This indicates that there are more students watching lower hours of television, resulting in a longer tail on the left side.