Answer :
Answer:
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:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
z score for x = 29.8 will be:
[tex]z=\frac{29.8-30}{2.1}=-0.10[/tex]
z score for x = 30.3 will be:
[tex]z=\frac{30.3-30}{2.1}=0.14[/tex]
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