Question

I'm not sure if these are able to be answered in excel, but so I have the formula?

5. (4 pts) The age of a store manager follows a normal distribution with a mean age of 55 years and a standard deviation of 4 years. What values represent the middle 60% of ages of store managers?

10. (4 pts) PGA Tour Golfers hit the first shot off the tee an average of 280 yards with a standard deviation of 10 yards following a normal distribution. What is the probability a PGA Tour Golfer hits the ball off the tee between 270 yards and 300 yards?

Answer 1

(5) middle 60% is between 20th percentile and 80th percentile

Using excel function NORMINV(area,mean,standard deviation)

we get

Lower value = NORMINV(0.20,55,4) = 51.63

Upper value = NORMINV(0.80,55,4) =58.37

(6) Using excel function NORMDIST(x,mean,standard deviation, cumulative)

we have to find P(270<X<300) = NORMDIST(y,mean,standard deviation, cumulative)-NORMDIST(x,mean,standard deviation, cumulative)

setting x = 270, y = 300, mean = 280, standard deviation = 10, cumulative =TRUE

we get

P(270<X<300) = NORMDIST(300,280,10,TRUE) -  NORMDIST(270,280,10,TRUE)

P(270<X<300) = 0.8185