Suppose the mean income of firms in the industry for a year is $55 million dollars with a standard deviation of 17 million dollars.

Question

If incomes for the industry distrubuted normally, what is the probablilty that a randomly selected firm will earn between 44 and 90 million dollars?

Answer 1

If Incomes are distributed normally (mean = 55, sd = 17).  

 

P (44 < I < 90) = P ((44-55)/17 < Z < (90-55)/17))

                       = P (-.4706       < Z  < 2.0588)

 

At this point it is a table lookup exercise.  

 

.181 + .4802 = .6612 or 66.12%

 

Have a good day.