The number of hours of sunshine each day, y, for the month of July at Heathrow are summarised in the table below - Edexcel - A-Level Maths: Statistics - Question 1 - 2017 - Paper 2

To calculate the mean, use the formula:

[ \bar{x} = \frac{\sum{(x_i \cdot f_i)}}{N} ]

where $x_i$ are the mid-points of the hour groups and $f_i$ are the frequencies.

The mid-points are calculated as follows:

• For 0 ≤ y < 5: Mid-point = 2.5
• For 5 ≤ y < 8: Mid-point = 6.5
• For 8 ≤ y < 11: Mid-point = 9.5
• For 11 ≤ y < 12: Mid-point = 11.5
• For 12 ≤ y < 14: Mid-point = 13

Now, calculate the mean:

[ \bar{x} = \frac{(2.5 \cdot 12) + (6.5 \cdot 6) + (9.5 \cdot 8) + (11.5 \cdot 3) + (13 \cdot 2)}{31} = 8.05 \text{ hours} ]

For standard deviation, use the formula:

[ \sigma = \sqrt{\frac{\sum{(f_i \cdot (x_i - \bar{x})^2)}}{N}} ]

Upon calculation, you would find: Standard deviation ( \sigma \approx 3.69 \text{ hours} ). Both answers should be provided to 3 significant figures.

The mean for Heathrow (8.05 hours) is higher than Hurn's (5.98 hours), indicating that the southern location may have more sunshine on average. However, the standard deviation for Heathrow (3.69 hours) is also significantly higher than Hurn's (4.12 hours), suggesting that the variability in daily sunshine is greater. This inconsistency does not support Thomas' belief that being further south would result in more consistency.

First, determine the threshold for being above 1 standard deviation:

Mean = 8.05 hours, Standard Deviation = 3.69 hours.

Cut-off = Mean + Standard Deviation = 8.05 + 3.69 = 11.74 hours.

Now, check the frequency table; the groups that meet this criterion are:

• 11 ≤ y < 12: 3
• 12 ≤ y < 14: 2

Total count = 3 + 2 = 5 days.

Helen’s model states that the number of hours should exceed 10.3 hours. With a proportion of days exceeding this set threshold, calculate:

Using the percentage derived earlier, notably 31 days in July and proportion exceeding this threshold: [ 31 \times 0.15865... \approx 5 \text{ days} ]

Comparing part (d) and part (e), 5 days exceed 1 standard deviation above the mean while estimates show approximately 4 to 5 days which leads us to conclude that Helen's model seems accurate, though variability suggests it may not be consistently reliable across different conditions.