Chapter 7, Problem 12PT

a.

To determine

To explain: The reason of not computing the probability that amount of the internet exceeds $29.

Answer to Problem 12PT

Because distribution is not known.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Population mean }\left(\mu \right)=28\\ \text{Population standard deviation}\left(\sigma \right)=10\\ \text{Sample size}\left(n\right)=500\end{array}$

Here, it has been provided that the population is not approximately normally distributed. Since, the population is not normal the required probability cannot be computed.

b.

To determine

To find: The mean and standard deviation of the sampling distribution

Answer to Problem 12PT

The required mean and standard deviation are 28 and 0.4772 respectively.

Explanation of Solution

The mean and standard deviation can be calculated as:

  $\begin{array}{c}{\mu }_{\overline{x}}=\mu =28\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{10}{\sqrt{500}}\\ =0.4772\end{array}$

The required mean and standard deviation are 28 and 0.4772 respectively

c.

To determine

To explain: The shape of the sampling distribution.

Explanation of Solution

The central limit theorem states that sampling distribution of sample mean follows the normal distribution if the sample size is at least 30, irrespective of the nature of the distribution.

The sample size is higher than 30. Thus, the sample mean is approximately normally distributed.

d.

To determine

To find: The probability that average fee exceeds $29.

Answer to Problem 12PT

The probability is 0.0125.

Explanation of Solution

The probability that average fee exceeds $29.can be calculated as:

  $\begin{array}{c}P\left(\overline{x}>29\right)=P\left(Z>\frac{29-28}{\frac{10}{\sqrt{500}}}\right)\\ =P\left(Z>2.24\right)\\ =0.0125\end{array}$

Thus, the probability is 0.0125