Question

In determining automobile mileage​ ratings, it was found that the mpg in the city for a certain model is normally​ distributed, with a mean of 28 mpg and a standard deviation of 1.5 mpg. Suppose that the car manufacturer samples four cars from its assembly line and tests them for mileage ratings.

a. What is the distribution of the mean mpg for the​ sample?

b. What is the probability that the mean mpg of the sample will be greater than 30 mpg?

c. What is the probability that the mean mpg of the sample will be less than 27.5 mpg?

Answer 1

Solution :

a.

= / n = 1.5 / 4 = 0.75

b.

P( > 30) = 1 - P( < 30)

= 1 - P[( - ) / < (30 - 28) / 0.75]

= 1 - P(z < 2.67)

= 1 - 0.9962

= 0.0038

Probability = 0.0038

c.

P( < 27.5) = P(( - ) / < (27.5 - 28) / 0.75)

= P(z < -0.67)

= 0.2514

Probability = 0.2514