Question

The average price of a gallon of gasoline in Missouri was $2.77 (as of 10-25-10 according to the AAA Daily Fuel Gauge Report). Suppose it is known that σ^2=0.32. What is the probability that a sample of 72 gas stations taken that same week will have a sample mean within $0.15 of the population mean? (Hint: You may need to use as many as 4+ decimal places in your computations to come up with the correct answer!)

A. 0.9756
B. 0.0240
C. 0.0398
D. 0.4176
E. Almost 1 2128

Answer 1

A. 0.9756

Explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation(which is the square root of the variance)
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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

In this problem, we have that:

`\mu = 2.77, \sigma = √(0.32) = 0.5657, n = 72, s = (0.5657)/(√(72)) = 0.0667`

What is the probability that a sample of 72 gas stations taken that same week will have a sample mean within $0.15 of the population mean?

This is the pvalue of Z when X = 2.77 + 0.15 = 2.92 subtracted by the pvalue of Z when X = 2.77 - 0.15 = 2.62. So

X = 2.92

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (2.92 - 2.77)/(0.0667)`

`Z = 2.25`

`Z = 2.25` has a pvalue of 0.9878

X = 2.62

`Z = (X - \mu)/(s)`

`Z = (2.62 - 2.77)/(0.0667)`

`Z = -2.25`

`Z = -2.25` has a pvalue of 0.0122

0.9878 - 0.0122 = 0.9756