SOLUTION: An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of all Americans have brown eyes. A random sample of 140 American college students found 42 with
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Question 965005: An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of all Americans have brown eyes. A random sample of 140 American college students found 42 with brown eyes.
(a). Give the numerical value of the sample proportion \hat{p}.
\hat{p} =
(b). Assuming that 45% of the population have brown eyes, determine the mean and standard deviation of the distribution of sample proportions for samples of size n = 140.
The mean is
The standard deviation is
(c). What is the probability that in a random sample of 140 Americans, more than 47.2 % have brown eyes?
The probability is Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! a) ^p = 42 / 140 = 0.30
b) P = .45 and Q = .55, then
The mean = P = .45
Standard Deviation = sqrt(PQ/n) = sqrt((.45*.55) / 140) = 0.042045893 approx 0.04
c) z-score = (X - mean) / std dev = (0.472 - 0.45) / 0.04 = 0.55
consult the z-tables for the probability associated with z-score = 0.55
The probability (X < 0.472) = 0.7088 approx .71
now we want the probability (X > 0.472) which is equal to
1 - 0.71 = 0.29
