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?

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

Step-by-step explanation:

Normal Probability Distribution:

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

In a set with mean and standard deviation , the z-score of a measure X is given by:

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

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

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

maybe showing the graph would

step-by-step explanation:

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