Question 1136477: I need help solving the following question with a calculator and how to type it in
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 122000 dollars. Assume the standard deviation is 31000 dollars. Suppose you take a simple random sample of 61 graduates.
Find the probability that a single randomly selected salary is at least 117000 dollars.
P(X > 117000) =
Find the probability that a sample of size
n=61 is randomly selected with a mean that is at least 117000 dollars.
P(M > 117000) =
Enter your answers as numbers accurate to 4 decimal places.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! want z value which is (x-mean)/sd for the first or (117000-122000)/31000.
Want probability less than that.
Use 2nd VARS 2 (normal cdf) and type in (-6, the z value above ), -6 is not specific so long as the number is that or more negative -1E99 works, too.
ENTER
probability is the result
the first one has z<-5000/31000=-0.16
probability is 0.4364
for the second part, the denominator is sd/sqrt(61) the sample size.
The z is (117000-122000)* sqrt(61)/sd, inverting with division and multiplying.
The probability will be a lot less with that one.
Use z since sd is given and not using the sample sd as an estimate.
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