Answer:
a) By the Central Limit Theorem, the mean is $142 and the standard deviation is $0.7488.
b) By the Central Limit Theorem, approximately normal.
c) 0.0901 = 9.01% probability that the average cable service paid by the sample of cable service customers will exceed $143
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The mean monthly fee is $142 and the standard deviation is $29.
This means that ![\mu = 142, \sigma = 29](https://tex.z-dn.net/?f=%5Cmu%20%3D%20142%2C%20%5Csigma%20%3D%2029)
Part a: what are the mean an standard deviation of the sample distribution of x hat show your work and justify your reasoning.
Sample of 1500(larger than 30).
By the Central Limit Theorem
The mean is $142
The standard deviation is ![s = \frac{29}{\sqrt{1500}} = 0.7488](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B29%7D%7B%5Csqrt%7B1500%7D%7D%20%3D%200.7488)
Part b: what is the shape of the sampling distribution of x hat justify your answer.
By the Central Limit Theorem, approximately normal.
Part C: what is the probability that the average cable service paid by the sample of cable service customers will exceed $143?
This is 1 subtracted by the pvalue of Z when X = 143. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{143 - 142}{0.7488}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B143%20-%20142%7D%7B0.7488%7D)
![Z = 1.34](https://tex.z-dn.net/?f=Z%20%3D%201.34)
has a pvalue of 0.9099
1 - 0.9099 = 0.0901
0.0901 = 9.01% probability that the average cable service paid by the sample of cable service customers will exceed $143