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 = 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

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

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 = (x-bar -mean)/(standard deviation/square root(n))

Z = (210 - 212.3)/(20/square root(50))

Z = -0.81

Z(-0.81) = .2090

answer is .2090

3. Suppose that a presidential candidate is favored by 51% of all eligible voters. What is the probability that in a random sample of 100 registered voters, less than 49% will favor that candidate?

p = .51

1-p = .49

1-p = .49

Z = (p^ - p)/(sqrt(p(1-p)/n))

Z = (.49 - .51)/(sqrt(.51*.49)/100)

Z = -0.4

Z = -0.4

Z(-0.4) = .3446

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