Question

The average price for a gallon of gasoline in the United States is $3.73 and in Russia it is $3.40. Assume these averages are the population means in the two countries and that the probability distributions are normally distributed with a standard deviation of $.25 in the United States and a standard deviation of $.20 in Russia.

What percentage of the gas station in Russia charge less than $ 3.50?

Answer 1

Answer:

69.15% of the gas station in Russia charge less than $ 3.50

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 pvalue, we get the probability that the value of the measure is greater than X.

What percentage of the gas station in Russia charge less than $ 3.50?

This is the pvalue of Z when $X = 3.50, \mu = 3.40, \sigma = 0.2$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3.5 - 3.4}{0.2}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

69.15% of the gas station in Russia charge less than $ 3.50