In: Statistics and Probability
The 2015 American Time Use Survey contains data on how many minutes of sleep per night an average college student gets. According to this survey, the minutes that college students sleep per night are right skewed with a mean of 529.9 minutes and standard deviation 135.6 minutes.
Suppose we take a sample of size n = 50 from this same population,
and we calculate x̅ = 540 minutes. How many standard deviations,
(σ/ √n), away from μ is this sample mean?
µ = 529.9
σ = 135.6
n= 50
X = 540
Z = (X - µ )/(σ/√n) = ( 540
- 529.90 ) / (
135.600 / √ 50 ) =
0.527
answer: 0.527 std dev above the mean