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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 frequency of days with a cloud cover of 6 or more. From the table, we can see:

  • Frequency for 6: 0
  • Frequency for 7: 6
  • Frequency for 8: 28

Thus, the total frequency for days with cloud cover of 6 or more is:

i.e. Total = 6 + 28 = 34 days.

The total number of days is 184. Therefore, the probability is calculated as:

P(X6)=34184=0.1739P(X \geq 6) = \frac{34}{184} = 0.1739

Step 2

find $P(X > 6)$

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Answer

P(X>6)P(X > 6) can be found using the complementary rule:

P(X>6)=1P(X6)=1P(X=0)P(X=1)P(X=2)P(X=3)P(X=4)P(X=5)P(X=6)P(X > 6) = 1 - P(X \leq 6) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) - P(X = 4) - P(X = 5) - P(X = 6)

Using the frequencies:

P(X6)=0+1+4+4+7+10+52184=78184P(X \leq 6) = \frac{0 + 1 + 4 + 4 + 7 + 10 + 52}{184} = \frac{78}{184}

Thus,

P(X>6)=178184=106184=0.5761P(X > 6) = 1 - \frac{78}{184} = \frac{106}{184} = 0.5761

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 find the expected number of days with a daily mean total cloud cover of 7, we can use the binomial model with parameters (n = 184, p = 0.76).

The expected value (mean) of the binomial distribution is given by:

E(X)=np=184P(X=7)E(X) = n \cdot p = 184 \cdot P(X = 7)

From the probability previously computed, we can find the estimated number:

E(X)=1840.281151.735 daysE(X) = 184 \cdot 0.2811 \approx 51.735\text{ days}

Rounded to 1 decimal place, the expected number of days is 51.7.

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 a comparison between actual outcomes and model predictions. The computed probabilities should ideally align with what is observed in the data. Since the probability found in part (b)(i) indicates a notable difference between predicted and observed distribution of cloud cover days, it raises concerns about the suitability of Magali's binomial model.

If the actual distribution demonstrates much higher occurrences of lower cloud cover days or deviates significantly from the model's expectation of higher means, this indicates that the model cannot adequately capture the observed phenomena.

Step 5

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

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Answer

The proportion of days when the daily mean cloud cover was 6 or greater among the 28 days can be calculated:

From the subsequent table, we have:

  • Days with cloud cover of 6: 6 days

Thus, the proportion is:

P(X628)=628=0.2143P(X \geq 6 | 28) = \frac{6}{28} = 0.2143

Step 6

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

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Answer

The observed proportion of days with high daily mean cloud cover (greater than or equal to 6) derived from the dependent condition suggests there is a greater likelihood of observing high cloud cover, contradicting Magali's initial binomial distribution model.

This indicates that the binomial model may not be suitable as it does not accurately reflect the cloud cover trends or dependencies observed in the data.

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