Question 309463
There are 8,000 students at the university of Tennessee at Chattanooga. The average age of all the students is 24 years with a standard deviation of 9 years. A random sample of 36 students is selected.
a. determine the standard error of the mean.
Ans: s = 9/sqrt(36) = 3/2
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b. what is the probability that the sample mean will be larger than 19.5?
z(19.5) = (19.5-24)/(3/2) = -3
P(x-bar < 19.5) = P(z < -3) = normalcdf(-100,-3) = 0.0013
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c, what us the probability that the sample mean will be between 25.5 and 27 years?
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z(27) = (27-24)/(3/2) = 2
z(25.5) = (25.5-24)/(3/2) = 1
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P(15.5< x < 27) = P(1< z < 2) = normalcdf(1,2) = 0.1359
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Cheers,
Stan H.