Question 1134235: 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 166000 dollars. Assume the standard deviation is 37000 dollars. Suppose you take a simple random sample of 62 graduates.
Find the probability that a single randomly selected policy has a mean value between 160831.1 and 169759.2 dollars.
P(160831.1 < X < 169759.2) = ______________
(Enter your answers as numbers accurate to 4 decimal places.)
**Below is what I have tried before and the system told me that it is wrong (note: the numbers are different because it switches the numbers in every attempt):
mean= 150,000 standard deviation=38,000 n=98
P(139252 < X < 141171.3)
z=(139252-150,000)/38,000 = -.28
z= (141171.3-150,000)/38,000 = -.23
From z-table: 0.4090 - 0.3897 = 0.0193
Answer by Glaviolette(140)   (Show Source):
You can put this solution on YOUR website! Your work looks to be correct. The only thing I can think of is the a graphing calculator was used (normal distribution function) which would have given a slightly different response to what you showed of 0.0195. Otherwise, I do not know what would be wrong and I teach these types of problems in Statistics so I feel confident with what you are doing.
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