The Democrat and Chronicle reported that 25% of the flights arriving at the San

Question

The Democrat and Chronicle reported that 25% of the flights arriving at the San Diego airport during the first five months of 2001 were late (Democrat and Chronicle, July 23, 2001). Assume the population proportion is p = .25.

Calculate () (sigma p-bar) with a sample size of 1,000 flights (to 4 decimals).

What is the probability that the sample proportion lie between 0.22 and 0.28 if a sample of size 1,000 is selected (to 4 decimals)?

What is the probability that the sample proportion will lie between 0.22 and 0.28 if a sample of size 500 is selected (to 4 decimals)?

Explanation / Answer

Solution:-

Let p' be the sample proportion such that p' = X/n where X is the number of late flights and n is the sample size.

X follows a binomial distribution with sample size n and p = 0.25, i.e. X ~ B(n,p).

If n is large such that np > 5 and np(1-p) > 5, one can apply the normal approximation such that X ~ N(np,np(1-p))

E(p') = E(X/n) = E(X)/n = np/n = p
Var(p') = Var(X/n) = Var(X)/n² = np(1-p)/n² = p(1-p)/n

(A) Given that n = 1000, we computed np = 250 > 5 and np(1-p) = 187.5 > 5.

Normal approximation can be applied.

E(p') = 0.25
Var(p') = 0.25 * 0.75 / 1000 = 0.0001875
Standard deviation of p' = 0.0137

The sample proportion follows a normal distribution with mean 0.25 and variance 0.0001875, i.e. p' ~ N(0.25, 0.0001875).

(B) P(0.22 < p' < 0.28)
= P((0.22 - 0.25)/0.0137 < Z < (0.28 - 0.25)/0.0137)
= P(-2.191 < Z < 2.191)
= P(Z < 2.191) - P(Z < -2.191)
= 0.986 - 0.014
= 0.972

(C) Given that n = 500
E(p') = 0.25 (same)
Var(p') = 0.25 * 0.75 / 500 = 0.000375
Standard deviation of p' = 0.0194

P(0.22 < p' < 0.28)
= P((0.22 - 0.25)/0.0194 < Z < (0.28 - 0.25)/0.0194)
= P(-1.549 < Z < 1.549)
= P(Z < 1.549) - P(Z < -1.549)
= 0.939 - 0.061
= 0.879

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