Magali is studying the mean total cloud cover, in oktas, for Leuchars in 1987 using data from the large data set - Edexcel - A-Level Maths Statistics - Question 4 - 2019 - Paper 1

To find $P(X > 6)$, we need to calculate:

$P(X > 6) = 1 - P(X \leq 6)$

We already calculated a part of this:

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

The respective frequencies sum to:

0 + 1 + 4 + 7 + 10 + 52 + 0 = 74.

So,

$P(X > 6) = 1 - \frac{74}{184} \approx 0.597826... \approx 0.598$

Using Magali's model $X \sim B(8, 0.76)$, we find the expected value:

$E(X) = n \cdot p$

Where:

• n = 184
• p = P(X = 7) calculated from the binomial distribution.

The calculation yields an expected number of days with a daily mean total cloud cover of 7:

$\text{Expected days} = 184 \cdot P(X = 7) \approx 51.7$

So the expected number is approximately 51.7 days.

Part (a) and part (b)(i) are similar, and the expected number of 7s (51.7) matches with the number of 7s found in the data set (52). Therefore, Magali's model is supported.

To find the proportion of days when the daily mean total cloud cover was 6 or greater, we refer to the data. There were 28 days with a daily mean total cloud cover of 8 and a total of:

• Days with 6 or greater = 2 + 2 + 28 = 32.

So, the proportion is:

$\frac{32}{28} = 1.142857... \approx 1.143$

This indicates that the proportion is greater than 1, which implies that on average, the previous day's cloud cover is high.

Given that the proportion of high cloud cover days (1.143) is greater than what Magali’s model predicts (given independence), this indicates that the binomial model may not be suitable as it assumes independence between days.