Question

Working on summer vacation. According to a Harris Interactive (July 2013) poll of U.S. adults, about 60% work during their summer vacation. (See Exercise 3.13, p. 169.) Assume that the true proportion of all U.S. adults who work during summer vacation is p = .6. Now consider a random sample of 500 U.S. adults.

a. What is the probability that between 55% and 65% of the sampled adults work during summer vacation?

b. What is the probability that over 75% of the sampled adults work during summer vacation?

Short answer

a. The probability that between 55% and 65% of sampled adults work during summer vacation is 0.9774

b. The probability that over 75% of the sampled adults work during summer vacation is 0.00

Step-by-step solution

Given information

The proportion of all U.S. adults who work during summer vacation is $p=0.6$.

A random sample of size $n=500$U.S. adults is selected.

Let $\hat{p}$represents the sample proportion of U.S. adults who work during summer vacation.

Finding the mean and standard deviation of the sample proportion

The mean of the sampling distribution of $\hat{p}$is:

$\begin{array}{rcl}E\left(\hat{p}\right)& =& p\\ & =& 0.6\end{array}$.

Ep^=pSince$E\left(\hat{p}\right)=p$ .

The standard deviation of the sampling distribution of $\hat{p}$is obtained as:

$\begin{array}{rcl}{\sigma }_{\hat{p}}& =& \sqrt{\frac{p\left(1-p\right)}{n}}\\ & =& \sqrt{\frac{0.6\times 0.4}{500}}\\ & =& \sqrt{0.00048}\\ & =& 0.0219\end{array}$.

Therefore, ${\sigma }_{\hat{p}}=0.0219$.

Computing the required probability

a.

The probability that between 55% and 65% of sampled adults work during summer vacation is obtained as:

$\begin{array}{rcl}P\left(0.55<\hat{p}<0.65\right)& =& P\left(\frac{0.55-p}{{\sigma }_{\hat{p}}}<\frac{\hat{p}-p}{{\sigma }_{\hat{p}}}<\frac{0.65-p}{{\sigma }_{\hat{p}}}\right)\\ & =& P\left(\frac{0.55-0.6}{0.0219}<Z<\frac{0.65-0.6}{0.0219}\right)\\ & =& P\left(\frac{-0.05}{0.0219}<Z<\frac{0.05}{0.0219}\right)\\ & =& P\left(-2.28<Z<2.28\right)\\ & =& P\left(Z<2.28\right)-P\left(Z<-2.28\right)\\ & =& 0.9887-0.0113\\ & =& 0.9774\end{array}$.

The z-table can be used to find the required probability. The value at the intersection of 1.00 and 0.05 indicates the probability of a z-score less than 1.05, while the value at the intersection of -1.00 and 0.05 is the probability of a z-score less than -1.05.

Therefore, the required probability is 0.7062.

Computing the probability that sample proportion is over 0.75

b.

The probability that over 75% of the sampled adults work during summer vacation is obtained as:

$\begin{array}{rcl}P\left(\hat{p}>0.75\right)& =& P\left(\frac{\hat{p}-p}{{\sigma }_{\hat{p}}}>\frac{0.75-p}{{\sigma }_{\hat{p}}}\right)\\ & =& P\left(Z>\frac{0.75-0.60}{0.0219}\right)\\ & =& P\left(Z>\frac{0.15}{0.0219}\right)\\ & =& P\left(Z>6.84\right)\\ & \approx & 0.00\end{array}$.

Thus the required probability is 0.00