Z = (score - mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion above that Z score.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Use same table.
The average age of statistics students nationwide is 22. The standard deviation is 2.5 years. Assume the age is a normally distributed variable.
Find the probability that one student selected at random is older than 23.
Find the probability that the mean age of a group of 16 students selected at random is bigger than 23
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