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 house- holds pay a low rate as part of a bundle with phone or television service, but some pay much more for Inter- net only or for faster connections.
A sample survey asks an SRS of 50 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 $55.
(b) What are the mean and standard deviation of the sampling distribution of ?
(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 $55. television service, but some pay much more for Inter- net only or for faster connections.

Answer:

(a) See Explanation

(b) [tex]\bar x =50[/tex] and [tex]\sigma_x = 0.8944[/tex]

(c) The shape is normal

(d) [tex]P(\bar x > 55) = 0[/tex]

Step-by-step explanation:

Given

[tex]\mu = \$50[/tex] -- mean

[tex]\sigma = \$20[/tex] --- standard deviation

[tex]n=500[/tex] --- sample size

Solving (a): The reason we can't determine the probability that an amount to access internet by a household will exceed $55.

The question says that a lot of households pay low rates, but the percentage of the households in this category is not given. This means that it is impossible to determine the shape of the distribution.

Hence, the probability cannot be calculated.

Solving (b): Sample mean and Sample standard deviation

The sample mean estimates the population mean

So:

[tex]\bar x =\mu[/tex]

[tex]\bar x =50[/tex]

The sample standard deviation is calculated as thus:

[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]

[tex]\sigma_x = \frac{20}{\sqrt{500}}[/tex]

[tex]\sigma_x = \frac{20}{22.36}[/tex]

[tex]\sigma_x = 0.8944[/tex]

Solving (c): The shape of the distribution.

We have:

[tex]n = 500[/tex] --- The sample size

According to Central limit theorem, When the sample size is greater than 30, then the shape of the distribution is normal.

Hence, the shape is normal

Solving (d): [tex]P(\bar x \ge 55)[/tex]

Calculate the test statistic (t)

[tex]t = \frac{\bar x - \mu}{\sigma}[/tex]

[tex]t = \frac{55 - 50}{0.8945}[/tex]

[tex]t = \frac{5}{0.8945}[/tex]

[tex]t = 5.590[/tex]

So:

[tex]P(\bar x > 55) = P(t > 5.590)[/tex]

Referencing the z table, we have:

[tex]P(t > 5.590) = 0[/tex]

Hence:

[tex]P(\bar x > 55) = 0[/tex]

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