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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?

Sagot :

MrRoyalidn

Answer:

[tex]P(x < 2.892) = 4.36\%[/tex]

Step-by-step explanation:

Given

[tex]N = 700[/tex] --- Population

[tex]\mu = 2.894[/tex] -- Mean

[tex]\sigma = 0.009[/tex] --- Standard deviation

[tex]n = 55[/tex] -- Sample

Required: [tex]P(x < 2.892)[/tex]

This question will be solved using the finite correction factor

First, calculated the z score

[tex]z = \frac{x - \mu}{\sqrt{\frac{N -n}{N -1}} * \frac{\sigma}{\sqrt n}}[/tex]

[tex]z = \frac{2.892 - 2.894}{\sqrt{\frac{700 -55}{700 -1}} * \frac{0.009}{\sqrt {55}}}[/tex]

[tex]z = \frac{-0.002}{\sqrt{\frac{645}{699}} * \frac{0.009}{7.42}}[/tex]

[tex]z = \frac{-0.002}{\sqrt{0.92} * \frac{0.009}{7.42}}[/tex]

[tex]z = \frac{-0.002}{0.95917 * 0.0012129}[/tex]

[tex]z = -1.71[/tex]

So:

[tex]P(x < 2.892) = P(z < -1.71)[/tex]

Using z table

[tex]P(x < 2.892) = 0.043633[/tex]

[tex]P(x < 2.892) = 4.36\%[/tex]

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