Question

The average price for a gallon of gasoline in Germany is $5.57 and in France it is $5.54. Assume these averages are population averages in two countries, and that the probability distributions are normally distributed with the standard deviation of $.35 in Germany and standard deviation of $.25 in France. What is the probability that a randomly selected gas station in France charges more than $5.30 per gallon?

Answer 1

0.8315 = 83.15% probability that a randomly selected gas station in France charges more than $5.30 per gallon

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The average price for a gallon of gasoline in France is $5.54. The standard deviation is $.25.

This means that
`\mu = 5.54, \sigma = 0.25`

What is the probability that a randomly selected gas station in France charges more than $5.30 per gallon?

This is 1 subtracted by the pvalue of Z when X = 5.30. So

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

`Z = (5.3 - 5.54)/(0.25)`

`Z = -0.96`

`Z = -0.96` has a pvalue of 0.1685

1 - 0.1685 = 0.8315

0.8315 = 83.15% probability that a randomly selected gas station in France charges more than $5.30 per gallon