Question

Suppose that it is reasonable to assume that a Grand Am’s MPG has a normal pdf with a mean of 24.3 and a standard deviation of 0.6. Let X denote the MPG of a Grand Am.

a. Find the probability that a randomly selected Grand Am gets less than 25 but more than 21 MPG.

b. Find the probability that a randomly selected 16 Grand Ams get less than 25 but more than 21 MPG on average

Answer 1

a) P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 24.3

Standard deviation = 0.6

P(a randomly selected Grand Am gets less than 25 but more than 21 MPG) = P(21 < X < 25)

= P(X < 25) - P(X < 21)

= P(Z < (25 - 24.3)/0.6) - P(Z < (21 - 24.3)/0.6)

= P(Z < 1.17) - P(Z < -5.5)

= 0.8790 - 0

= 0.8790

b) For sampling distribution of mean,

Sample mean = 24.3

Standard error =

=

= 0.15

P( < A) = P(Z < (A - mean)/standard error)

P(a randomly selected Grand Am gets less than 25 but more than 21 MPG) = P(21 < X < 25)

= P( < 25) - P( < 21)

= P(Z < (25 - 24.3)/0.15) - P(Z < (21 - 24.3)/0.15)

= P(Z < 4.67) - P(Z < -22)

= 1 - 0

= 1