statistics and probability

Question

Suppose that an anxiety measure has a mean of 115 and variance of 256. The researcher only wants to work with 19% of the upper participants. What is the smallest average score the researcher should consider with a sample size of 34?

Answer 1

look for z-score associated with 81st percentile (that is the top 19%). from table or technology that z-score is approximately 0.88

z = (xbar - mean)/(std/sqrt(sample size), where xbar is what you are trying to find

substituting 0.88 for z, 115 for mean, 16 for standard deviation (= square root of variance) and 34 for sample size you get:

0.88 = (xbar - 115)/(16/sqrt(34))

0.88 * (16/sqrt(34) + 115 = xbar

xbar = 117.415