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 (ar{x}) and standard deviation (σ) of the number of hours, we first find the midpoints of each interval and assign them an appropriate weight based on their frequency:

• For 0 ≤ y < 5, midpoint is 2.5
• For 5 ≤ y < 8, midpoint is 6.5
• For 8 ≤ y < 11, midpoint is 9.5
• For 11 ≤ y < 12, midpoint is 11.5
• For 12 ≤ y < 14, midpoint is 13

Calculating the mean: ar{x} = rac{ ext{Sum of (midpoint} imes ext{frequency)}}{ ext{Total frequency}}

ar{x} = rac{(2.5 imes 12) + (6.5 imes 6) + (9.5 imes 8) + (11.5 imes 3) + (13 imes 2)}{31}

ar{x} = rac{(30 + 39 + 76 + 34.5 + 26)}{31} = rac{205.5}{31} ext{ hours} ≈ 6.63 ext{ hours}

Next, we calculate the standard deviation using the formula:

ext{Variance} (σ^2) = rac{ ext{Sum of (frequency} imes (midpoint - ar{x})^2)}{ ext{Total frequency}}

Calculating the squared deviation:

• For 0 ≤ y < 5: 12 × (2.5 - 6.63)²
• For 5 ≤ y < 8: 6 × (6.5 - 6.63)²
• For 8 ≤ y < 11: 8 × (9.5 - 6.63)²
• For 11 ≤ y < 12: 3 × (11.5 - 6.63)²
• For 12 ≤ y < 14: 2 × (13 - 6.63)²

These contributions can be summed, followed by taking the square root for the standard deviation yielding approximately σ ≈ 3.67 hours.

Thomas believes that the further south you are, the more consistent the number of hours of daily sunshine will be. Given that the mean in Heathow is significantly higher than in Hurn (5.98 hours) while the standard deviation is smaller (3.67 compared to 4.12 hours), this implies that Heathrow has less variability in sunny hours, thus supporting his belief that being further south correlates with higher consistency in sunshine hours.

Using the mean calculated (approximately 6.63 hours), we evaluate the threshold for more than 1 standard deviation above this mean:

ext{Threshold} = ar{x} + σ = 6.63 + 3.67 = 10.3 ext{ hours}

Next, we find the relevant frequency counts. For hours > 10.3, we consider: (11 ≤ y < 12) and (12 ≤ y < 14). So, the number of days: egin{itemize}

ext{Days counted: }

• 3 days (11 ≤ y < 12)
• 2 days (12 ≤ y < 14) ext{Total} = 3 + 2 = 5 ext{ days} ext{Therefore, there are approximately 5 days.} ext{This is affirmed by } 7 ext{ days estimations yielding a similar high value.} ext{Considering the estimated frequencies and occurrences show these are roughly consistent.}
  egin{itemize} </itemize>

Helen's model estimates days with sunshine as a linear function based on previous data, where ≥10.3 hours is the threshold considered: Using: $H ext{ (number of hours) - } P(H > 10.3)$ This would lead us to consider a suitable computational outcome where: $31 imes 0.15865 ext{(given as a percentage)}$ Therefore, $ext{Predicted days} = 31 imes 0.15865 ≈ 4.9 ext{ days}$ Confirming expected predictions align with forecasts for this specific range in line with earlier calculated predictions. So this model suggests about 5 days based on our estimation.

The comparison of part (d) estimated around 5 actual observed days with exceeding sunshine hours, while part (e)'s model proposes 4.9 days. These estimations are notably close, indicating convergence despite marginal differences suggesting Helen's model is somewhat reliable, yet might require further adjustment for more accuracy in sunshine day predictions.