Each member of a group of 27 people was timed when completing a puzzle - Edexcel - A-Level Maths Statistics - Question 3 - 2020 - Paper 1

The interquartile range (IQR) is found by subtracting the first quartile (Q1) from the third quartile (Q3). From the plot, Q1 = 14 and Q3 = 25. Thus, the IQR is:

$25 - 14 = 11$

The mean ( \mu ) is calculated as the total sum of times divided by the number of participants:

$\mu = \frac{\sum x}{n} = \frac{607.5}{27} \approx 22.5$

The standard deviation (( \sigma )) is found using the formula:

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

Substituting the values:

$\sigma = \sqrt{\frac{17623.25}{27} - \left(\frac{607.5}{27}\right)^2} \approx 12.1$.

Using the mean calculated as approximately 22.5 and the standard deviation as approximately 12.1, we define an outlier as a value more than 3 standard deviations above the mean:

$\mu + 3\sigma = 22.5 + 3 \times 12.1 = 58.8$

Since values above 58.8 are considered outliers and none exceed this threshold, there is only one outlier.

Since Adam's time (a) and Beth's time (b) need to comply with the condition where a > b and observing that the median increases while the mean remains unchanged, a suitable choice could be:

• Let a = 45 minutes
• Let b = 40 minutes

This selection satisfies both conditions discussed.

Adding Adam and Beth's times will generally reduce the range of variability. Their values, which fall within the existing range, will bring down the overall standard deviation since these values are less than three standard deviations from the mean of the original sample.