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

Algebra ->  Probability-and-statistics -> 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       Log On


   



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) About Me  (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.