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June 2011 p62 q2
3329
In Scotland, in November, on average 80% of days are cloudy. Assume that the weather on any one day is independent of the weather on other days.
(i) Use a normal approximation to find the probability of there being fewer than 25 cloudy days in Scotland in November (30 days).
(ii) Give a reason why the use of a normal approximation is justified.
Solution
(i) Let \(X\) be the number of cloudy days in November. \(X\) follows a binomial distribution with parameters \(n = 30\) and \(p = 0.8\).
We approximate \(X\) using a normal distribution with mean \(\mu = np = 30 \times 0.8 = 24\) and variance \(\sigma^2 = npq = 30 \times 0.8 \times 0.2 = 4.8\).
Standard deviation \(\sigma = \sqrt{4.8}\).
To find the probability of fewer than 25 cloudy days, we use a continuity correction: \(P(X < 25) \approx P(Y < 24.5)\) where \(Y\) is normally distributed.
Calculate the z-score: \(z = \frac{24.5 - 24}{\sqrt{4.8}} = 0.228\).
Using standard normal distribution tables, \(P(Z < 0.228) = 0.590\).
(ii) The normal approximation is justified because both \(np = 24\) and \(nq = 6\) are greater than 5.
