Question

In a sample of 700 gas​ stations, the mean price for regular gasoline at the pump was $2.894 per gallon and the standard deviation was ​$0.009 per gallon. A random sample of size 55 is drawn from this population. What is the probability that the mean price per gallon is less than ​$​2.892?

Answer 1

`P(x < 2.892) = 4.36\%`

Explanation:

Given

`N = 700` --- Population

`\mu = 2.894` -- Mean

`\sigma = 0.009` --- Standard deviation

`n = 55` -- Sample

Required:
`P(x < 2.892)`

This question will be solved using the finite correction factor

First, calculated the z score

`z = \frac{x - \mu}{\sqrt{(N -n)/(N -1)} * (\sigma)/(\sqrt n)}`

`z = \frac{2.892 - 2.894}{\sqrt{(700 -55)/(700 -1)} * \frac{0.009}{\sqrt {55}}}`

`z = \frac{-0.002}{\sqrt{(645)/(699)} * (0.009)/(7.42)}`

`z = (-0.002)/(√(0.92) * (0.009)/(7.42))`

`z = (-0.002)/(0.95917 * 0.0012129)`

`z = -1.71`

So:

`P(x < 2.892) = P(z < -1.71)`

Using z table

`P(x < 2.892) = 0.043633`

`P(x < 2.892) = 4.36\%`