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

Answer:

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

Step-by-step 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

$z = \frac{x - \mu}{\sqrt{\frac{N -n}{N -1}} * \frac{\sigma}{\sqrt n}}$

$z = \frac{2.892 - 2.894}{\sqrt{\frac{700 -55}{700 -1}} * \frac{0.009}{\sqrt {55}}}$

$z = \frac{-0.002}{\sqrt{\frac{645}{699}} * \frac{0.009}{7.42}}$

$z = \frac{-0.002}{\sqrt{0.92} * \frac{0.009}{7.42}}$

$z = \frac{-0.002}{0.95917 * 0.0012129}$

$z = -1.71$

So:

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

Using z table

$P(x < 2.892) = 0.043633$

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