Question

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 131000 dollars. Assume the standard deviation is 36000 dollars. Suppose you take a simple random sample of 73 graduates. Do not use probability tables to find the probabilities below as they may not be accurate enough. Find the probability that a single randomly selected salary is at least 126000 dollars. P(X> 126000) = Find the probability that a sample of size n = 73 is randomly selected with a mean that is at least 126000 dollars. Pē > 126000) = Enter your answers as numbers accurate to 4 decimal places.

Answer 1

solution Given that e = 131000 6= 36000 - P(x> 126000) = 1- P(x5126000). i-er X-4 126000 - 131000) 36000 21-P ( zo = -P(zs-0.14) - 1 - 0.4443 = 0.5557 P(x>, 126000 10000) = 10.5557 557

pe 73 ex= 131000 O, 36000 4213. 4813 P(x>, 126000) = 1-PCX3126000). = 1-P / X-et 126000 - 131000 o u ? 4213.4813 ) = 1-P ( 22 -5000 ) . 4213.4813) = IP (z«-1.19) 1- 0.1170 = 0.8830 Pcx> 126000) = 0.8830