Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z score.
6. Facebook reports that the average number of Facebook friends worldwide is 175.5 with a standard deviation of 90.57. If you were to take a sample of 25 students, what is the probability that the mean number Facebook friends in the sample will be 190 friends or more?
2 answers
Solution
My mean is 175.5
My standard deviation is=90.57
Sample=25
So the probability is
Formula to be used: P(X>190)=P((X-mean)/s
sqrt of my sample is = 5
(190-175.5) / 90.57/5)
14.5 / 18.114
calculator says 0.8004858120790548746825659710721
I check the z-table and I see -.8 under the z of 0.00 correlates to .2119
=0.2119
My mean is 175.5
My standard deviation is=90.57
Sample=25
So the probability is
Formula to be used: P(X>190)=P((X-mean)/s
sqrt of my sample is = 5
(190-175.5) / 90.57/5)
14.5 / 18.114
calculator says 0.8004858120790548746825659710721
I check the z-table and I see -.8 under the z of 0.00 correlates to .2119
=0.2119