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?
My answer (not sure if I even did this correctly): P(X ? 280) = P(z ? (280 - 258)/6) = P(Z ? 3.67) = >0.9998 ? 1
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?
My answer (not sure if I even did this correctly):
? p^= ?(0.51)(0.49)/100 = 0.049989999
Continuity correction: 0.5/100 ? 0.005
P(p^<0.49) ? P(x<0.49 + 0.005) = P(x< 0.495)
= P(z < ((0.495 – 0.51)/0.049989999) = P(z< -0.300) = 0.3821
My answer (not sure if I even did this correctly): P(X ? 280) = P(z ? (280 - 258)/6) = P(Z ? 3.67) = >0.9998 ? 1
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?
My answer (not sure if I even did this correctly):
? p^= ?(0.51)(0.49)/100 = 0.049989999
Continuity correction: 0.5/100 ? 0.005
P(p^<0.49) ? P(x<0.49 + 0.005) = P(x< 0.495)
= P(z < ((0.495 – 0.51)/0.049989999) = P(z< -0.300) = 0.3821