Question

The opera theater manager believes that 11% of the opera tickets for tonight's show have been sold.

If the manager is right, what is the probability that the proportion of tickets sold in a sample of 792 tickets would be greater than 14%? Round your answer to four decimal places.

Answer 1

Solution:

Population proportion of ticket sold (P) = 11% = 0.11

Sample size (n) = 792

If nP ≥ 10 and n(1 - P) ≥ 10, then sampling distribution of sample proportion is follows approximately normal distribution with mean P and variance PQ/n.

i.e.

Where, P is population proportion, Q = 1 - P, n is sample size and p is sample proportion.

We have, n = 792 and P = 0.11

nP = 792×0.11 = 87.12 which is greater than 10.

n(1 - P) = 792×(1 - 0.11) = 704.88 which is greater than 10.

Hence, we can use normal distribution as sampling distribution of sample proportion.

We have to obtain Pr(p > 0.14).

We have, P = 0.11, n = 792 and Q = 1 - 0.11 = 0.99

We know that, if p ~ N(P, PQ/n) then

Using "pnorm" function of R we get, P(Z > 2.6983) = 0.0035

Hence, the probability that the proportion of tickets sold in a sample of 792 tickets would be greater than 14% is 0.0035.

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