The amount that households pay service providers for access to the Internet varies quite a bit, but the mean monthly fee is 50\( and the standard deviation is 20\). The distribution is not Normal: many households pay a low rate as part of a bundle with phone or television service, but some pay much more for Internet only or for faster connections. 11 A sample survey asks an SRS of 50 households with Internet access how much they pay. Let $x^{-}\overline{x}$be the mean amount paid.

a. Explain why you can't determine the probability that the amount a randomly selected household pays for access to the Internet exceeds 55 .\(

b. What are the mean and standard deviation of the sampling distribution of ${\mathrm{x}}^{-\overline{x}}$?

c. What is the shape of the sampling distribution of ${\mathrm{x}}^{-\overline{x}}$? Justify your answer.

d. Find the probability that the average fee paid by the sample of households exceeds 55 .\)

The amount that households pay for Internet access varies significantly, but the average monthly rate is $50 and the standard deviation is $20.

Although you are aware that the population distribution is not normal, you are unaware of the actual population distribution.

It is consequently impossible to compute the probability for a single randomly picked household because doing so necessitates knowing the exact population distribution.

Given,

Formula used:

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

Let calculate

The amount that households pay for Internet access varies significantly, but the average monthly rate is $50 and the standard deviation is $20.

The centre limit theorem states that if the sample size is bigger than 30, the sample mean$\overline{x}$sampling distribution is approximately normal.

The central limit theorem can be used, and hence the sampling distribution of the sample mean is approximately normal.

Given,

$$\begin{gathered} \mu =50 \\ \sigma =20 \\ n=50 \end{gathered}$$

The value of 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 is

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