Chapter 7, Problem T7.12SPT

(a)

To determine

To Explain: the probability that the amount a randomly selected household pays for access to the internet exceeds $55.

Answer to Problem T7.12SPT

Don’t know the exact population distribution.

Explanation of Solution

Knowing that the population distribution is not Normal, but do not know the precise distribution of the population.

It is then not possible to calculate the probability for a single randomly selected house hold, because it requires the knowledge of the exact population distribution to get this probability.

(b)

To determine

To Calculate: the mean and standard deviation of the sampling distribution of $\overline{x}$ .

Answer to Problem T7.12SPT

  $\begin{array}{c}\mu =50\\ {\sigma }_{\overline{x}}=2.83\end{array}$

Explanation of Solution

Given:

  $\begin{array}{c}\mu =\text{50}\\ \sigma \text{=20}\\ n\text{=5}0\end{array}$

Formula used:

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Calculation:

  ${\mu }_{\overline{x}}=\mu =50$

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{20}{\sqrt{50}}\\ =2.83\end{array}$

(c)

To determine

To find: the shape of the sampling distribution of $\overline{x}$ .

Answer to Problem T7.12SPT

Approximate normal

Explanation of Solution

Centre limit theorem: If the sample size is greater than 30 then the sampling distribution of the sample mean $\overline{x}$ is about normal.

It is possible to use the central limit theorem and therefore the sampling distribution of the sample mean is about normal.

(d)

To determine

To Calculate: the probability that the average fee paid by the sample of households exceeds $55.

Answer to Problem T7.12SPT

3.84%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =\text{50}\\ \sigma \text{=}\text{20}\\ n\text{=5}0\end{array}$

Formula used:

  $z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}$

Calculation:

The z-score is

  $z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{55-50}{20/\sqrt{50}}=1.77$

The standard probability

  $\begin{array}{c}P\left(\overline{X}\ge 55\right)=P\left(Z>1.77\right)\\ =1-P\left(z<1.77\right)\\ =1-0.9616\\ =0.0384\\ =3.84\%\end{array}$