The amount that households pay service providers for access to the Internet varies quite a bit, but the mean monthly fee is $\backslash (28$ and the standard deviation is $\backslash )10$. The distribution is not Normal: many households pay about $\backslash (10$ for limited dial-up access or about $\backslash )30$ for unlimited dial-up access, but some pay much more for faster connections. A sample survey asks an SRS of 500 households with Internet access how much they pay. Let $\overline{x}$ 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 $\backslash (29$.

(b) What are the mean and standard deviation of the sampling distribution of $\overline{x}$ ?

(c) What is the shape of the sampling distribution of $\overline{x}$ ? Justify your answer.

(d) Find the probability that the average fee paid by the sample of households exceeds $\backslash )29$. Show your work.

Population mean $\left(\mu \right)=28$.

Population standard deviation $\left(\sigma \right)=10$.

Sample size $\left(n\right)=500$.

The population is not approximately normally distributed, according to the data. The required probability can't be estimated since the population isn't normal.

Population mean $\left(\mu \right)=28$,

Population standard deviation $\left(\sigma \right)=10$,

Sample size $\left(n\right)=500$.

The mean and standard deviation can be calculated as:

${\mu }_{\overline{x}}=\mu$

$\mu =28$

$=\frac{10}{\sqrt{500}}$

$=0.4772$

The required mean and standard deviations are $28$and $0.4772$respectively.

Population mean $\left(\mu \right)=28$.

Population standard deviation $\left(\sigma \right)=10$.

Sample size $\left(n\right)=500$.

The central limit theorem states that if the sample size is at least $30$, the sampling distribution of the sample mean would follow the normal distribution.

Population mean $\left(\mu \right)=28$,

Population standard deviation $\left(\sigma \right)=10$,

Sample size $\left(n\right)=500$.

The probability that the average fee will exceed $\$29$can be estimated as follows:

$P(\overline{x}>29)=P\left(Z>\frac{29-28}{\frac{10}{\sqrt{500}}}\right)$

$=P(Z>2.24)$

$=0.0125$