The amount people who pay for cell phone service varies quite a bit, but the mean

monthly fee is $55 and the standard deviation is $22. The distribution is not Normal.

Many people pay about $30 for plans with 2GB data access and about $60 for 5GB

of data access, but some pay much more for unlimited data access. A sample survey

is designed to ask a simple random sample of 1,000 cell phone users how much they

pay. Let x be the mean amount paid.

Part A: What are the mean and standard deviation of the sample distribution of X?

Show your work and justify your reasoning. (4 points)

Part B: What is the shape of the sampling distribution of x? Justify your answer. (2

points)

Part C: What is the probability that the average cell phone service paid by the

sample of cell phone users will exceed $56? Show your work. (4 points) (10 points)

INTL

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this question:

Mean of the distribution: 55

Standard deviation of the distribution: 22

Sample of 1000.

This means that:

[tex]\mu = 55, \sigma = 22, n = 1000[/tex]

Part A: What are the mean and standard deviation of the sample distribution of X?

By the Central Limit Theorem, the mean is 55 and the standard deviation is [tex]s = \frac{22}{\sqrt{1000}} = 0.6957[/tex].

Part B: What is the shape of the sampling distribution of x?

By the Central Limit Theorem, the shape of the sampling distribution of x is approximately normal.

Part C: What is the probability that the average cell phone service paid by the sample of cell phone users will exceed $56?

This is 1 subtracted by the pvalue of Z when [tex]X = 56[/tex]. So:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{56 - 55}{0.6957}[/tex]

[tex]Z = 1.44[/tex]

[tex]Z = 1.44[/tex] has a pvalue of 0.9251

1 - 0.9251 = 0.0749

0.0749 = 7.49% probability that the average cell phone service paid by the sample of cell phone users will exceed $56.