Question

Assume that gas mileage for cars is normally distributed with a mean of 23.5 miles per gallon and a standard deviation of 10 miles per gallon. Show all work. Just the answer, without supporting work, will receive no credit.

(a) What is the probability that a randomly selected car gets between 15 and 30 miles per gallon? (Round the answer to 4 decimal places)

(b) Find the 75th percentile of the gas mileage distribution. (Round the answer to 2 decimal places)

Answer 1

Solution :

Given that,

mean = = 23.5

standard deviation = = 10

a ) P (15 < x < 30 )

P ( 15 - 23.5 / 10) < ( x -  / ) < ( 30 - 23.5 / 10)

P ( - 8.5 / 10 < z < 6.5 / 10 )

P (- 0.85 < z <0.65 )

P (z < 0.65 ) - p ( z < - 0.85 )

Using z table

= 0.7422 - 0.1977

= 0.5445

b ) Using standard normal table,

P(Z < z) = 75%

1 - P(Z < z) = 0.01

P(Z < z) = 0.75

P(Z < 0.6745) = 0.99

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 * 10 + 23.5

= 30.2

The 75th percentile = 30.2