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 19 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 3.6 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 18.1 miles per gallon? _________(round to four decimal places as needed)

b) Suppose 1000 drivers were randomly selected. What is the probability the average obtained by these drivers will exceed 18.1 mpg? _________(round four decimal places as needed)

Thank you

Answer 1

Solution :

Given that ,

mean = = 19

standard deviation = = 3.6

n = 100

= 19

= / n = 3.6 / 100 = 3.6 / 10 = 0.36

(a)

P( > 18.1) = 1 - P( < 18.1)

= 1 - P[( - ) / < (18.1 - 19) / 0.36]

= 1 - P(z < -7.91)

= 1 - 0

= 1

Probability = 1

(b)

n = 1000

= / n = 3.6 / 1000

P( > 18.1) = 1 - P( < 18.1)

= 1 - P[( - ) / < (18.1 - 19) / 3.6 / 1000 ]

= 1 - P(z < -2.5)

= 1 - 0.0062

= 0.9938

Probability = 0.9938