Question 911703: Hello,I was able to find the probability for the entire population, but I haven't figured out how to find it for just the sample population of 11. Any help would be appreciated.
Suppose the heights of 18-year-old men are approximately normally distributed, with mean 72 inches and standard deviation 5 inches. If a random sample of eleven 18-year-old men is selected, what is the probability that the mean height x bar is between 71 and 73 inches? (Round your answer to four decimal places.)
Thank you!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We are given mean(x) 72 inches and standard deviation(sigma) 5 inches and sample size 11.
Note that the x for the sample is the same as for the population and sigma for the sample size is sigma for population divided by sqrt of 11, therefore
x(sample size) = 72
sigma(sample size) = 5/sqrt(11) = 1.507556723
let P be probability, then
P ( 71 < X <73 ) = P ( X < 73 ) - P ( X < 71 )
we calculate z-values for X(71) and X(73)
X(73) = (73 - 72) / 1.507556723 = 0.663324958 = 0.66
X(71) = (71 - 72) / 1.507556723 = −0.663324958 = -0.66
now we consult z-value tables for probabilities
P(X<73) = 0.7454 and P(X<71) = 0.2546, now
P (71 < X < 73) = P ( X < 73) - P ( X <71 ) = 0.7454 - 0.2546 = 0.4908
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