Question

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 .\)

Short answer

(a)Don't know the exact population distribution.

(b)$$\begin{gathered} \mu =50 \\ {\sigma }_{\overline{x}}=2.83 \end{gathered}$$

(c)Approximate normal

(d)$3.84\%$

Step-by-step solution

Part (a) Step 1: Given Information

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

Part (a) Step 2: Simplification

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.

Part (b) Step 1: Given Information

Given,

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

Formula used:

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

Part (b) Step 2: Simplification

Let calculate

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

Part (c) Step 1: Given information

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

Part (c) Step 2:  Explanation

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.

Part (d) Step 1: Given information

Given,

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

Part (d) Step 2:  Calculation

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

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