SOLUTION: IF A SAMPLE OF 30 MEN IS SELECTTED WHAT IS THE 90TH PERCENTILE OF THE MEAN NUMBER OF CUPS PER DAY
MEAN IS 3.8
SD 0.7
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Question 1196667: IF A SAMPLE OF 30 MEN IS SELECTTED WHAT IS THE 90TH PERCENTILE OF THE MEAN NUMBER OF CUPS PER DAY
MEAN IS 3.8
SD 0.7
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean is 3.8
standard deviation is .7
sample size is 30
standard error is .7/sqrt(30) = .1278
t-score formula is t = (x - m) / s
t is the t-score
x is the raw score
m is the mean
s is the standard error if you are looking for the mean of multiple samples of size 30.
the 90th percentile for a t-score with 29 degrees of freedom (sample size minus 1), is equal to 1.3114
t-score formula becomes 1.3114 = (x - 3.8) / .1278.
solve for x to get:
x = 1.3114 * .1278 + 3.8 = 3.9676
90 percent of the means of samples of size 30 will be less than 3.9676.
that's your 90th percentile.
here's what it looks like on a graph.
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