The amounts of electricity bills for all households in a city have a skewed probability distribution with a mean of $139 and a standard deviation of $30. Find the probability that the mean amount of electric bills for a random sample of 75 households selected
from this city will be within $6 of the population mean.
Round your answer to four decimal places.

P (i within $6 of 4) =

P (within $6 of 4) = 0.9164

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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 .

Mean of $139 and a standard deviation of $30.

This means that

Random sample of 75 households

This means that

75 > 30, which means that the sampling distribution is approximately normal.

Find the probability that the mean amount of electric bills for a random sample of 75 households selected from this city will be within $6 of the population mean.

This is the pvalue of Z when X = 139 + 6 = 145 subtracted by the pvalue of Z when X = 139 - 6 = 133.

X = 145

By the Central Limit Theorem

has a pvalue of 0.9582

X = 133

has a pvalue of 0.0418

0.9582 - 0.0418 = 0.9164

So

P (within $6 of 4) = 0.9164

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