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

Answer:

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%.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.27, n = 741$

So

$\mu = 0.27$

$\sigma = \sqrt{\frac{0.27*0.73}{741}} = 0.0163$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.31 - 0.27}{0.0163}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 0.23

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.23 - 0.27}{0.0163}$

$Z = -2.45$

$Z = -2.45$ 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%.