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The amount that households pay service providers for access to the Internet varies quite a bit, but the mean monthly fee is \(28 and the standard deviation is \)10. The distribution is not Normal: many households pay about \(10 for limited dial-up access or about \)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 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 \(29.

(b) What are the mean and standard deviation of the sampling distribution of x ?

(c) What is the shape of the sampling distribution of x ? Justify your answer.

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

Short Answer

Expert verified

a). The required probability can't be estimated since the population isn't normal.

b). The required mean and standard deviations are 28and 0.4772.

c). The sampling distribution of the sample mean would follow the normal distribution.

d). The required probability is0.0125.

Step by step solution

01

Part (a) Step 1: Given Information

Population mean ()=28.

Population standard deviation ()=10.

Sample size (n)=500.

02

Part (a) Step 2: Explanation

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

03

Part (b) Step 1: Given Information

Population mean ()=28,

Population standard deviation ()=10,

Sample size (n)=500.

04

Part (b) Step 2: Explanation

The mean and standard deviation can be calculated as:

x=

=28

=10500

=0.4772

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

05

Part (c) Step 1: Given Information

Population mean ()=28.

Population standard deviation ()=10.

Sample size (n)=500.

06

Part (c) Step 2: Explanation

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.

07

Part (d) Step 1: Given Information

Population mean ()=28,

Population standard deviation ()=10,

Sample size (n)=500.

08

Part (d) Step 2: Explanation

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

P(x>29)=PZ>29-2810500

=P(Z>2.24)

=0.0125

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