5. 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 for the 26–30 hour group can be calculated as follows:

Given that the 11–20 group has a width of 4 cm for 10 hours, the width for each hour is:

Width per hour = (\frac{4}{10} = 0.4) cm.

Thus, for the 26–30 hour group, which also spans 5 hours, the width would be:

Width = 5 hours * 0.4 cm/hour = 2 cm.

Next, to determine the height: For the 26–30 hour group, where the frequency is 13, and given the total height of the 11–20 group is 6 cm (for 15 units), the height of the 26-30 group can be calculated similarly as:

Height = (\frac{Frequency}{Total Frequency} * Total Height) Height = (\frac{13}{56} * 6 cm) = 1.393 cm.

To calculate the mean ((\bar{x})), we first find (\Sigma(f \cdot x)):

(\bar{x} = \frac{\Sigma(f \cdot x)}{N})

Where:

• f = frequency
• x = mid-point
• N = total number of students = 56

Calculating the sum:

(\Sigma(f \cdot x) = (6 \cdot 5.5) + (15 \cdot 15.5) + (11 \cdot 23) + (13 \cdot 28) + (8 \cdot 35) + (3 \cdot 50) = 1316.5)

Now, finding the mean: (\bar{x} = \frac{1316.5}{56} \approx 23.5)

Next, for the standard deviation ((\sigma)): (\sigma = \sqrt{\frac{\Sigma(f \cdot (x - \bar{x})^2)}{N}}) This requires first calculating (x - \bar{x}) and then the squared values.

Finally, calculate the variance using the sum of squared deviations and the standard deviation formula.

To estimate the median, we first need to locate the median class. The median is the (\frac{N}{2} = 28), which locates us in the 26–30 hour group. For linear interpolation, we can use the following formula:

(Q_2 = L + \left(\frac{\frac{N}{2} - F}{f}\right) \cdot c)

Where:

• L = lower boundary of the median class
• N = total frequency
• F = cumulative frequency of the class before the median class
• f = frequency of the median class
• c = class width

Substituting values: we find that (c = 5, F = 20, f = 13 \rightarrow Q_2= (26+(\frac{28-20}{13}) \cdot 5) = 28.21).

The data can be considered negatively skewed because the lower quartile (15.8) is much closer to the median than the upper quartile (29.3). This suggests that there are some lower values influencing the mean more than higher values, pushing it toward the left.