Question

Mileage ratings for cars and trucks generally come with a qualifier stating actual mileage will depend on driving conditions and habits. A car manufacturer states that its new truck will average 20 miles per gallon with combined town and country driving. Assume the mean stated by the manufacturer is the actual​ average, and the distribution has a standard deviation of 3.1 mpg. Complete parts a and b below.

a. Given the above mean and standard​ deviation, what is the probability that 100 drivers will average more than 19.5 miles per​ gallon? _________ ​(Round to four decimal places as​ needed.)

b. Suppose 1 comma 000 drivers were randomly selected. What is the probability the average obtained by these drivers will exceed 19.5 ​mpg? ___________ ​(Round to four decimal places as​ needed.)

Answer 1

Mean = = 20

Standard deviation = = 3.1

a) Sample size = n = 100

We have to find P( > 19.5)

For finding this probability we have to find z score.

That is we have to find P(Z > - 1.61)

P(Z > - 1.61) = 1 - P(Z < - 1.61) = 1 - 0.0537 = 0.9463 ( From z table)

b) Sample size = n = 1000

We have to find P( > 19.5)

For finding this probability we have to find z score.

That is we have to find P(Z > - 5.10)

P(Z > - 5.10) = 1 - P(Z < - 5.10) = 1 - 0.000 = 1