A class of students had a sudoku competition - Edexcel - A-Level Maths Statistics - Question 5 - 2011 - Paper 2

To use linear interpolation for estimating the median, first, we find the cumulative frequency:

The cumulative frequency for the grouped data up to each interval is:

• 2 - 8: 2
• 9 - 12: 2 + 7 = 9
• 13 - 15: 9 + 5 = 14
• 16 - 18: 14 + 8 = 22
• 19 - 22: 22 + 4 = 26
• 23 - 30: 26 + 4 = 30

The median lies at the rac{30}{2} = 15^{th} student. This falls in the 13 - 15 interval.

Using linear interpolation: $Q_2 = 13 + \left(\frac{15 - 14}{8}\right) \times (15 - 14) = 13 + \left(\frac{1}{8}\right) = 13.875$

To calculate the mean ($\bar{x}$) and the standard deviation ($\sigma$), we use the formulas:

Mean: $\bar{x} = \frac{\sum (f \cdot x)}{\sum f} = \frac{(2 \cdot 5) + (7 \cdot 10.5) + (5 \cdot 14) + (8 \cdot 17) + (4 \cdot 20.5) + (4 \cdot 26.5)}{30}$

Calculating:

• $2 \cdot 5 = 10$
• $7 \cdot 10.5 = 73.5$
• $5 \cdot 14 = 70$
• $8 \cdot 17 = 136$
• $4 \cdot 20.5 = 82$
• $4 \cdot 26.5 = 106$

Total: 10 + 73.5 + 70 + 136 + 82 + 106 = 477.5

Thus, the mean is: $\bar{x} = \frac{477.5}{30} = 15.9167$

Standard Deviation: $\sigma = \sqrt{\frac{\sum f \cdot (x - \bar{x})^2}{\sum f}}$ Using the calculated $x^2$ values: $\sigma \approx 5.78$

A normal distribution may be suitable because:

• The sample size is relatively large, allowing for the central limit theorem to apply.
• The distribution of completion times appears symmetrical based on the frequency and mid-point data, suggesting that it approximates a bell-shaped curve.

The skewness can be assessed by comparing the quartiles:

• $Q_1 = 8.5$
• $Q_2 = 13.0$
• $Q_3 = 21.0$

Since $Q_3 - Q_2 > Q_2 - Q_1$, it indicates a positive skew. The longer completion times significantly extend the right tail, reflecting that most students completed the sudoku within a shorter time, while a few took much longer.