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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

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Magali is studying the mean total cloud cover, in oktas, for Leuchars in 1987 using data from the large data set. The daily mean total cloud cover for all 184 days f... show full transcript

Worked Solution & Example Answer: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

Step 1

Find the probability that it has a daily mean total cloud cover of 6 or greater.

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Answer

To find the probability of selecting a day with a daily mean total cloud cover of 6 or greater, we first need to calculate the total number of days with a cover of 0 to 5:

  • Total days = 184
  • Days with cloud cover below 6 = 1 (0 oktas) + 4 (1 oktas) + 7 (2 oktas) + 10 (3 oktas) + 52 (4 oktas) + 28 (5 oktas) = 102

Thus, the number of days with a daily mean total cloud cover of 6 or greater is:

184102=82184 - 102 = 82

Therefore, the probability is:

P(X6)=82184=0.446.P(X \geq 6) = \frac{82}{184} = 0.446.

Step 2

find $P(X > 6)$

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Answer

To calculate P(X>6)P(X > 6) using the binomial model XB(8,0.76)X \sim B(8, 0.76):

We need to find the probabilities for P(X=7)P(X = 7) and P(X=8)P(X = 8) and sum them:

P(X>6)=P(X=7)+P(X=8).P(X > 6) = P(X = 7) + P(X = 8).

Using the binomial probability formula:

P(X=k)=(nk)pk(1p)nk,P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k},

where n=8n = 8 and p=0.76p = 0.76,

  1. For P(X=7)P(X = 7): P(X=7)=(87)(0.76)7(0.24)10.27.P(X = 7) = \binom{8}{7} (0.76)^7 (0.24)^1 \approx 0.27.

  2. For P(X=8)P(X = 8): P(X=8)=(88)(0.76)8(0.24)00.20.P(X = 8) = \binom{8}{8} (0.76)^8 (0.24)^0 \approx 0.20.

Thus,

P(X>6)0.27+0.20=0.47.P(X > 6) \approx 0.27 + 0.20 = 0.47.

Step 3

find, to 1 decimal place, the expected number of days in a sample of 184 days with a daily mean total cloud cover of 7.

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Answer

To determine the expected number of days with a daily mean total cloud cover of 7, we implement the formula for expectation in a binomial distribution:

Expected days = np=184P(X=7)n \cdot p = 184 \cdot P(X = 7) where P(X=7)P(X = 7) is calculated from previous steps.

Using the value calculated above (P(X=7)0.27P(X = 7) \approx 0.27):

Expected=1840.2749.6850(to1decimalplace).Expected = 184 \cdot 0.27 \approx 49.68 \approx 50 (to 1 decimal place).

Step 4

Explain whether or not your answers to part (b) support the use of Magali’s model.

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Answer

The results from part (b) suggest that the expected number of days with a cloud cover of 7 (approximately 50) closely aligns with the data obtained (52 days). This similarity indicates that the model fits reasonably well but must be analyzed further, as factors affecting cloud cover may lead to discrepancies. Therefore, while there is initial support for using the binomial model, additional scrutiny is required to evaluate its overall suitability.

Step 5

Comment the proportion of these days when the daily mean total cloud cover was 6 or greater.

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Answer

To find the proportion of days when the daily mean total cloud cover was 6 or greater, we can assess the counts of days: There are 28 days noted, and to find how many of these are 6 or greater, we see that:

  • Days with 6 or greater = 6 (1) + 7 (2) + 8 (3) = 6.

Therefore, the required proportion is:

6280.214.\frac{6}{28} \approx 0.214.

This indicates around 21.4% of the days had a cloud cover of 6 or greater.

Step 6

Comment on Magali’s model in light of your answer to part (d).

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Answer

Considering the proportion of days with a cloud cover of 6 or greater (approximately 21.4%), it raises questions about the validity of Magali’s binomial model. The model suggests a greater probability of high cloud cover, while the actual data presents a lower occurrence of high cloud cover days, therefore indicating that Magali’s model may not adequately represent the observed daily cloud cover data.

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