The amount people whopay for cable service varies quite a bit, but the mean monthly fee is $142 and the standard deviation is $29. The distribution is not

Normal. Many people pay about $76 for basic cable and about $160 for premium

service, but some pay much more. A sample survey is designed to ask a simple

random sample of 1,500 cable service customers 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 cable service paid by the sample of

cable service customers will exceed $143? Show your work. (4 points) (10 points)

$142 ; 0.749 ; Normal ; 0.090938

Step-by-step explanation:

Mean and standard deviation :

The mean of the distribution x = $142, because the sample size is sufficiently large, the distribution will be approximately normal (central limit theorem).

Standard deviation of distribution : σ/ sqrt(n)

= 29 / sqrt(1500)

= 0.749

B.)

Shape of sampling distribution of X.

The shape of the sampling distribution will be approximately normal due to the large sample size used. This is according to the central limit theorem whereby distribution of sample converges to normal with increasing sample size. Mean ~ N(142, 0.749²)

C.)

P(X > 143)

Using the relation :

Z = (x - mean) / standard error

Z = 143 - 142) / 0.7489

Z = 1.335

P(Z > 1.335) :

Using the Z probability calculator :

P(Z > 1.335) = 0.090938

darn right!

step-by-step explanation:

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