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 find the mean ($\bar{y}$), we use:

$\bar{y} = \frac{\sum f_i y_i}{\sum f_i}$

where $f_i$ is the frequency and $y_i$ is the midpoint for each group. Calculating the midpoints and their weighted sum gives:

• For 0 ≤ y < 5: Midpoint = 2.5, Contribution = 12 * 2.5
• For 5 ≤ y < 8: Midpoint = 6.5, Contribution = 6 * 6.5
• For 8 ≤ y < 11: Midpoint = 9.5, Contribution = 8 * 9.5
• For 11 ≤ y < 12: Midpoint = 11.5, Contribution = 3 * 11.5
• For 12 ≤ y < 14: Midpoint = 13, Contribution = 2 * 13

Now, sum them up:

$\sum f_i y_i = 30 + 39 + 76 + 34.5 + 26 = 205.5$

Total frequency:

$\sum f_i = 12 + 6 + 8 + 3 + 2 = 31$

Thus, the mean is:

$\bar{y} = \frac{205.5}{31} \approx 6.63 \text{ hours}$

For the standard deviation ($\sigma$):

1. Calculate $\sum f_i (y_i - \bar{y})^2$ and then

divide by the total frequency to get variance.

2. Finally, take the square root to find standard deviation.

The mean number of hours of sunshine at Heathrow is higher compared to that of Hurn (5.98 hours). Additionally, the standard deviation of hours of sunshine at Heathrow is smaller (4.12 hours for Hurn compared to that from Heathrow). This suggests that the variance in sunshine hours is lower at Heathrow, supporting Thomas’ belief that further south yields a more consistent pattern of sunshine hours.

First, calculate 1 standard deviation above the mean. If the standard deviation (from calculation) is approximately:\n $\sigma \approx 3.69 \text{ hours}$

Then:

$\text{Threshold} = \bar{y} + \sigma = 6.63 + 3.69 \approx 10.32 \text{ hours}$

Next, find the number of days exceeding this threshold, using the frequency table for estimation.

Using Helen’s model, we find:

$P(H > 10.3) \text{ or } P(Z > 1)\text{ (standard normal transformation)}$

Given the parameters, we predict:

$= 31 * 0.158655... \approx 4.9 \text{ days}$.

Thus, it can be approximated to either 4 or 5 days.

Comparing part (d) and part (e): If Helen’s model predicts around 5 days while calculations from the actual frequency produce a different number, there may be discrepancies. Furthermore, such a close average number suggests that Helen's model might not be fully suitable, as observed sunshine hours may not align perfectly with her projection.