Suppose the mean time spent by all Harper College students preparing for final exams is believed to be 3.8 hours with a standard deviation of 1.65 hours. If Jane, as part of a project for her sociology class, interviewed 40 students she came across in the cafeteria during yesterday’s lunch rush, what is the probability the mean time spent by these students preparing for final exams was less than 2.85 hours?
Answer)
As the sample size is greater than 30
So according to the central limit theorem we can assume normality
As the data is normally distributed we can use standard normal z table to estimate the answers
Z = (x-mean)/(s.d/√n)
Given mean = 3.8
S.d = 1.65
N = 40
Z = (2.85 - 3.8)/(1.65/√40) = -3.64
From z table, P(z<-3.64) = 0.00014
Suppose the mean time spent by all Harper College students preparing for final exams is believed...
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