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 width of the 26–30 hour group can be derived from the histogram bars. Given that the 11–20 group has a width of 4 cm and represents 10 units, we can work out that:

• Width of 5 units = 2 cm (calculating the units per cm).

For the height, using the frequency of 13 for the 26–30 group and the calculated width, we find:

• Height = (Frequency/Width) x scale = (13 units / 5 units) x 6 cm = 10.4 cm.

To compute the mean, use the mid-points and frequencies: ext{Mean} = rac{(5.5 imes 6) + (15.5 imes 15) + (23 imes 11) + (28 imes 13) + (35 imes 8) + (50 imes 3)}{56} = rac{1316.5}{56} = 23.5

Next, calculate the variance and then the standard deviation: ext{Variance} = rac{ ext{Sum of } (f imes (x - ext{Mean})^2)}{56} where f is the frequency and x is the mid-point. After calculation, the standard deviation $ext{s} ext{is approximately } 10.8.$

The cumulative frequency shows that the median falls within the 26–30 hour group. Using the values:

• Median = Q2 = 28 – (56/2 - 25) / 13 x 5 = 26.38 gives a value of approximately 26.4.

The lower quartile (15.8) and upper quartile (29.3) indicate a gap towards the lower end; hence, the data is negatively skewed since the mean (23.5) is less than the median, denoting more lower values.