Question

1. Ford Motor Company advertised that 2013 Ford C-Max hybrid car has an average fuel economy of 39.2 miles with a standard deviation of 3.7 mpg. Assume the fuel economy for each tankful of gas for this car follows the normal distribution. My 2013 C-Max hybrid averages 43 mpg with a standard deviation of 3.7 mpg driving from home to Plymouth State University and back.

a. What is the probability that on my next tank of fuel I will get 43 mpg or less?

b. What is the probability that I will get 39.9 mpg or more on my next tank of fuel?

c. What is the probability that I will get 50 mpg or more on my next tank of fuel?

Answer 1

µ = 43, σ = 3.7

a) P(X < 43) =

= P( (X-µ)/σ < (43-43)/3.7 )

= P(z < 0)

Using excel function:

= NORM.S.DIST(0, 1)

= 0.50

b) P(X > 39.9) =

= P( (X-µ)/σ > (39.9-43)/3.7)

= P(z > -0.8378)

= 1 - P(z < -0.8378)

Using excel function:

= 1 - NORM.S.DIST(-0.8378, 1)

= 0.7989

c) P(X > 50) =

= P( (X-µ)/σ > (50-43)/3.7)

= P(z > 1.8919)

= 1 - P(z < 1.8919)

Using excel function:

= 1 - NORM.S.DIST(1.8919, 1)

= 0.0293