As of this time, University of Michigan-Dearborn has registered 8,000 students; their average age is 24 with a standard deviation of 9 years. We have decided to select 36 students for an experiment involving sleep deprivation. What would be the probability that the sample mean would be between 25.5 and 27 years old (to four decimal places)?
Solution :
=
/
n = 9 /
36 = 1.5
= P[(25.5 - 24) / 1.5 < (
-
)
/
< (27 - 24) / 1.5)]
= P(1 < Z < 2)
= P(Z < 2) - P(Z < 1)
= 0.9772 - 0.8413
= 0.1359
Probability = 0.1359
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