Question

V. Money Magazine reported that the average price of gasoline in the United States during the first quarter of 2008 was $3.46. Assume that the price reported by Money is the population mean, and the standard deviation σ is $0.15. a. What is the probability that the mean price for a sample of 30 gas stations is within $0.03 of the population mean?

Answer 1

`z=(3.43 -3.46)/((0.15)/(√(30))) = -1.095`

`z=(3.49 -3.46)/((0.15)/(√(30))) = 1.095`

And we can find this probability using the normal standard table and we got:

`P(-1.095<z<1.095) = P(z<1.095) -P(z<-1.095) =0.863 -0.137= 0.726`

Explanation:

Let X the random variable that represent the price of a population, and for this case we know the distribution for X is given by:

`X \sim N(3.46,0.15)`

Where
`\mu=3.46` and
`\sigma=0.15`

And for this case we want to find the following probability:

`P(3.43 \leq \bar X \leq 3.49)`

And we can use the z score formula given by:

`z=(\bar X -\mu)/((\sigma)/(√(n)))`

If we find the z score for the limits we got:

`z=(3.43 -3.46)/((0.15)/(√(30))) = -1.095`

`z=(3.49 -3.46)/((0.15)/(√(30))) = 1.095`

And we can find this probability using the normal standard table and we got:

`P(-1.095<z<1.095) = P(z<1.095) -P(z<-1.095) =0.863 -0.137= 0.726`