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

To find the median, we first calculate the cumulative frequency:

• For the interval 2 - 8: $C_1 = 2$
• For the interval 9 - 12: $C_2 = 2 + 7 = 9$
• For the interval 13 - 15: $C_3 = 9 + 5 = 14$
• For the interval 16 - 18: $C_4 = 14 + 8 = 22$
• For the interval 19 - 22: $C_5 = 22 + 4 = 26$
• For the interval 23 - 30: $C_6 = 26 + 4 = 30$

The total frequency is 30, so the median is at position $\frac{30}{2} = 15$. This falls in the interval 16 - 18.

Using linear interpolation:

$L = 16, \quad CF = 14, \quad f = 8, \quad h = 2$

Calculating, we get:

$\text{Median} = L + \frac{\frac{n}{2} - CF}{f} \times h = 16 + \frac{15 - 14}{8} \times 2 = 16 + 0.25 = 16.25$

Thus, the estimated median time is approximately 16.25 minutes.

To calculate the mean:

$\text{Mean} = \frac{\sum xf}{N}$

Where $N = 30$ (total frequency).

Calculating:

$\sum xf = (5 \times 2) + (10.5 \times 7) + (14 \times 5) + (17 \times 8) + (20.5 \times 4) + (26.5 \times 4) = 860.5 \text{ (calculated using mid-points)}$

Thus:

$\text{Mean} = \frac{860.5}{30} = 28.6833 ext{ (approximated)}$

For Standard Deviation:

First, calculate $\sum x^{2}f$:

$\sum x^{2}f = (5^2 \times 2) + (10.5^2 \times 7) + (14^2 \times 5) + (17^2 \times 8) + (20.5^2 \times 4) + (26.5^2 \times 4)$

Using the provided sum ${\sum x^{2} = 8603.75}$:

$\sigma^2 = \frac{\sum x^{2}f}{N} - \left( \frac{\sum xf}{N} \right)^{2}$

Calculate:

$\sigma = \sqrt{28.6833} \approx 5.38$

Thus, the estimated mean is approximately 28.68 and the standard deviation is approximately 5.38.

The data can be modeled by a normal distribution because the sample size is sufficiently large (30 students), which allows for the central limit theorem to apply. Additionally, if the histogram of the time data appears symmetrical, it further supports the use of a normal distribution.

To describe skewness, we consider the quartiles calculated previously:

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

Skewness can be assessed by:

$\text{Skewness} = \frac{Q_3 - Q_2}{Q_2 - Q_1}$

From the quartiles, it can be determined that $Q_3 - Q_2 > Q_2 - Q_1$, indicating positive skewness. Hence, the distribution of times is positively skewed, suggesting that most students completed the sudoku in a shorter time, while there are a few outliers with longer times.