The average gas mileage of a certain model car is 30 miles per gallon. If the gas mileages are normally distributed with a standard deviation of 2.1, find the probability that a car has a gas mileage of between 29.8 and 30.3 miles per gallon.

The probability that a car has a gas mileage between 29.8 and 30.3 miles per gallon is 0.0955

Step-by-step explanation:

Average gas mileage = μ = 30 miles per gallon

Standard deviation = σ = 2.1

We have to find the probability that the gas mileage of a car is between 29.8 and 30.3 miles per gallon. Since the population is normally distributed and we know the value of population standard deviation, we will use z-distribution to find the desired probability.

First we will convert both the given values to z-scores and then using the z-table we will find the probability of a value lying in between the two numbers. Formula to calculate the z-score is:

z score for x = 29.8 will be:

z score for x = 30.3 will be:

Probability of gas mileage to be in between 29.8 and 30.3 is equivalent for z z-score to be in between -0.10 and 0.14.

i.e.

P( 29.8 < X < 30.3 ) = P( -0.10 < z < 0.14 )

Using the z-table we can find this probability:

P( -0.10 < z < 0.14 ) = P( z < 0.14 ) - P( z < -0.10 )

= 0.5557 - 0.4602

= 0.0955

Since, P( 29.8 < X < 30.3 ) = P( -0.10 < z < 0.14 ), we can conclude that:

The probability that a car has a gas mileage between 29.8 and 30.3 miles per gallon is 0.0955

a=r

d=q

c=p

b=s

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