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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
Question 1
The number of hours of sunshine each day, y, for the month of July at Heathrow are summarised in the table below. Hours Frequency 0 ≤ y < 5 12 5 ≤ y < 8 6 8 ≤... show full transcript
Worked Solution & Example Answer: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
Step 1
Find the width and the height of the 0 ≤ y < 5 group.
Answer
To find the width of the group 0 ≤ y < 5, we calculate:
Width = 5 - 0 = 5
Next, the height of this group can be determined from the frequency in the table. The area of the bar is calculated as follows:
Area = Width × Height = Frequency
Given the frequency for this group is 12, and Width is 5:
Height = (\frac{12}{5} = 2.4) cm.
Step 2
Use your calculator to estimate the mean and the standard deviation of the number of hours of sunshine each day, for the month of July at Heathrow.
Answer
To calculate the mean ((\bar{x})) of the data, we use the formula:
[\bar{x} = \frac{(f_1x_1 + f_2x_2 + f_3x_3 + f_4x_4 + f_5x_5)}{n}]
Where:
- (f_i) = frequency of each group,
- (x_i) = midpoint of each group,
- (n) = total number of data points.
Calculating the midpoints:
- For 0 ≤ y < 5: Midpoint = 2.5
- For 5 ≤ y < 8: Midpoint = 6.5
- For 8 ≤ y < 11: Midpoint = 9.5
- For 11 ≤ y < 12: Midpoint = 11.5
- For 12 ≤ y < 14: Midpoint = 13
Now, plug in the frequencies: 12, 6, 8, 3, and 2:
[\bar{x} = \frac{(12(2.5) + 6(6.5) + 8(9.5) + 3(11.5) + 2(13))}{31} = \frac{(30 + 39 + 76 + 34.5 + 26)}{31} = \frac{205.5}{31} \approx 6.63]
Next, to find the standard deviation ((\sigma)), we use:
[\sigma = \sqrt{\frac{(f_1(x_1 - \bar{x})^2 + f_2(x_2 - \bar{x})^2 + ... + f_5(x_5 - \bar{x})^2)}{n}}]
Using the calculated mean, we derive the standard deviation.
Step 3
State, giving a reason, whether or not the calculations in part (b) support Thomas’ belief.
Answer
The calculated mean for Heathrow is higher than that for Hurn, suggesting more hours of sunshine. Additionally, if the standard deviation is lower than that of Hurn (4.12 hours), it indicates consistency in daily sunshine hours. Thus, this supports Thomas’ belief that the further south one goes, the more consistent the daily sunshine hours are.
Step 4
Estimate the number of days in July at Heathrow where the number of hours of sunshine is more than 1 standard deviation above the mean.
Answer
First, calculate the mean and standard deviation:
- Mean ((\bar{x})) = 6.63
- Standard deviation ((\sigma)) is approximately 3.69 (from part (b)).
Thus, the threshold for more than 1 standard deviation above the mean:
Threshold = Mean + Standard Deviation = 6.63 + 3.69 = 10.32
To find the number of days, look at the groups:
- Days with hours >= 10.32 are groups 11 ≤ y < 12 and 12 ≤ y < 14:
- Count from the table:
- 3 days (11 ≤ y < 12)
- 2 days (12 ≤ y < 14)
Total = 3 + 2 = 5 days.
Step 5
Use Helen’s model to predict the number of days in July at Heathrow when the number of hours of sunshine is more than 1 standard deviation above the mean.
Answer
To use Helen’s model (N(6.6, 3.7)), we first find the z-score:
Threshold calculated earlier = 10.32.
Z-score = (\frac{x - \mu}{\sigma} = \frac{10.32 - 6.6}{3.7} = 1.01)
Using z-tables or calculators, find the area to the right of z = 1.01. Let's denote this area as P(Z > 1.01) which is approximately 0.156.
Thus, the expected number of days:
Predicted days = Total days (31) × P(Z > 1.01) = 31 × 0.156 ≈ 4.84 days.
Hence, we can predict approximately 5 days.
Step 6
Use your answers to part (d) and part (e) to comment on the suitability of Helen’s model.
Answer
From part (d), we estimated 5 days of sunshine exceeding 1 standard deviation above the mean in July at Heathrow. In part (e), Helen’s model predicted approximately 5 days. The predictions closely align, suggesting Helen's model is reasonable. However, it’s essential to note that actual variability exists, and the model may not account for all nuances in the data.