Question

The opera theater manager calculates that 16%16% of the opera tickets for tonight's show have been sold. If the manager is correct, what is the probability that the proportion of tickets sold in a sample of 592592 tickets would be less than 13%13%? Round your answer to four decimal places.

Answer 1

0.0228 = 2.28% probability that the proportion of tickets sold in a sample of 592 tickets would be less than 13%

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For the sampling distribution of a sample proportion of size n, we have that
`\mu = p, \sigma = \sqrt{(p(1-p))/(n)}`

In this problem, we have that:

`p = 0.16, n = 592`

So

`\mu = 0.16, \sigma = \sqrt{(0.16*0.84)/(592)} = 0.015`

What is the probability that the proportion of tickets sold in a sample of 592 tickets would be less than 13%?

This is the pvalue of Z when X = 0.13. So

`Z = (X - \mu)/(\sigma)`

`Z = (0.13 - 0.16)/(0.015)`

`Z = -2`

`Z = -2` has a pvalue of 0.0228

0.0228 = 2.28% probability that the proportion of tickets sold in a sample of 592 tickets would be less than 13%