Question

In determining automobile-mileage ratings, it was found that the mpg (X) for a certain model is normally distributed, with a mean of 33 mpg and a standard deviation of 1.7 mpg. Find the following: a. P(X<30) b. P(2835) d. P(X>31) e. the mileage rating that the upper 5% of cars achieve. 

Answer 1

P(X < 36) = X-U/SD 

                 = 36-34/1.4

                 = 1.43

 

The probability at z value 1.43 is 0.9236.

 

P(X<31) = 31-34/1.4

              = -2.14

 

The probability at z value -2.14 is 0.0162

 

Required probability = 0.9236-0.0162 

                                      = 90.74%

 

P(X < 37) = 37-34/1.4

                = 2.14

 

The probability at z value 2.14 is 0.9838

 

P(X > 37) = 1-P(X < 37)

                = 1-0.9838 

                = 1.62%