## Thursday, February 2, 2017

1.The distance that one professional golfer can drive a golf ball has a normal distribution with a mean of 258 yards and a standard deviation of 6 yards. What proportion of his drives exceed 280 yards in length? How many yards should this golfer drive a ball so that the distance is among his longest 25%

(X > 280) , we need to get the Z score, which is Z =(x -mean)/standard deviation
Z = (280 - 258)/6
Z = 3.67
1-Z(3.67) ,.... approximately 0, less then .0001
Z for upper 25th percent is .67
Mean + Z*standard deviation
258 +.67(6) = 262.02

2.The average number of pounds of meat a person consumes a year is 212.3 pounds. Assume that the standard deviation is 20 pounds. If a sample of 50 individuals is selected, find the probability that the mean of the sample will be less than 210 pounds per year.
mean = 212.3, standard deviation is 20
n = 50
P(x-bar < 210)
Z = (x-bar -mean)/(standard deviation/square root(n))
Z = (210 - 212.3)/(20/square root(50))
Z = -0.81
Z(-0.81) = .2090