Question

The opera theater manager believes that 27'% 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 741741 tickets would differ from the population proportion by less than 4%4%

Answer 1

0.9858 = 98.58% probability that the proportion of tickets sold in a sample of 741 tickets would differ from the population proportion by less than 4%.

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

In this problem, we have that:

`p = 0.27, n = 741`

So

`\mu = 0.27`

`\sigma = \sqrt{(0.27*0.73)/(741)} = 0.0163`

What is the probability that the proportion of tickets sold in a sample of 741 tickets would differ from the population proportion by less than 4%.

This is the pvalue of Z when X = 0.27 + 0.04 = 0.31 subtracted by the pvalue of Z when X = 0.27 - 0.04 = 0.23. So

X = 0.31

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

`Z = (0.31 - 0.27)/(0.0163)`

`Z = 2.45`

`Z = 2.45` has a pvalue of 0.9929

X = 0.23

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

`Z = (0.23 - 0.27)/(0.0163)`

`Z = -2.45`

`Z = -2.45` has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability that the proportion of tickets sold in a sample of 741 tickets would differ from the population proportion by less than 4%.