# The average price for a gallon

The average price for a gallon of gasoline in the United Statesis 3.75 and in Russia it is 3.37. Assume these averages are thepopulation means in the two countries and that the probabilitydistributions are normally distributed with a standard deviation of.25 in the United States and a standard deviation of .20 in Russia.a. What is the probability that a randomly selected gas station inthe United States charges less than ________ per gallon (to 4decimals)? b. What percentage of the gas stations in Russia chargeless than _____ per gallon (to 2 decimals)? c. What is theprobability that a randomly selected gas station in Russia chargedmore than the mean price in the United States (to 4 decimals).

Answer:

**Answer:**

Given data,

The data represents the average price for a gallon of gasoline inthe United States is 3.75 and in Russia it is 3.37.

Assume these averages are the population means in the two countriesand that the probability distributions are normally distributedwith a standard deviation of .25 in the United States and astandard deviation of .20 in Russia.

**(a).Theprobability that a randomly selected gas station in the UnitedStates charges less than $3.50 per gallon:**

The probability that a randomly selected gas station in the UnitedStates charges less than $3.50 per gallon is,

First, compute the z score then find probability based on standardnormal table.

x=$ 3.50

For converts to

From the standard normal distribution table, the associatedprobability for the area to the left is shown below,

P(z<-0.96)=0.1685

**0.1685 is the probability that a randomly selected gasstation in the United States charges less than $3.50 pergallon.**

**(b).To find thepercentage of the gas stations in Russia charge less than $ 3.50per gallon:**

The percentage of the gas stations in Russia charge less than $3.50per gallon is,

First, compute the z score then find probability based on standardnormal table.

x=$ 3.50

For converts to

From the standard normal distribution table, the associatedprobability for the area to the left is shown below,

P(z<0.65)=0.7422=74.22%

**74.22% is the percentage of the gas stations in Russiacharge less than $3.50 per gallon.**

**(c).To find theprobability that a randomly selected gas station in Russia chargedmore than the mean price in the United States:**

The probability that a randomly selected gas station in Russiacharged more than the mean price in the United States is,

First, compute the z score then find probability based on standardnormal table.

x=$ 3.75

For converts to

From the standard normal distribution table, the associatedprobability for the area to the left is shown below,

P(z>1.9)=1-(z<1.9)

=1-0.9713

=0.0287

**0.0287 is the probability that a randomly selected gasstation in Russia charged more than the mean price in the UnitedStates.**