SOLUTION: Full-time college students report spending a mean of 25 hours per week on academic activities, both inside and outside the classroom. Assume the standard deviation of time sp

Question 1203213: Full-time college students report spending a mean of 25 hours per week on academic activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 4 hours. If you select a random sample of 25 full-time college students, what is the probability that the mean time spent on academic activities is at least 24 hours per week?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
population mean is 25
population standard deviation is 4
sample size is 25
sample mean is 24
you want to know the probability that the mean time spent on academic activities is less than 24 hous per week.

z = (x - m) / s
z is the z-score
x is the smaple mean
m is the population mean
s is the standad error.

standard error = standard deviation / sqrt(sample size) = 4 / sqrt(25) = 4/5 = .8
z-score formula becomes:
z = (24 - 25) / .8 = -1/25.

area to the right of z-score of -1.25 = .89435
that's the probability that the mean time spent on academic activitie is at least 24 hours per week.

here's what it looks like on a normal distribution graph.

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SOLUTION: ​Full-time college students report spending a mean of 25 hours per week on academic​ activities, both inside and outside the classroom. Assume the standard deviation of time sp

SOLUTION: Full-time college students report spending a mean of 25 hours per week on academic activities, both inside and outside the classroom. Assume the standard deviation of time sp