Question

In: Statistics and Probability

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of...

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 123000 dollars. Assume the standard deviation is 37000 dollars. Suppose you take a simple random sample of 69 graduates. Find the probability that a single randomly selected salary is less than 126000 dollars. Answer = Find the probability that a sample of size n = 69 is randomly selected with a mean that is less than 126000 dollars. Answer = Enter your answers as numbers accurate to 4 decimal places.

Solutions

Expert Solution

Solution :

Given that ,

mean = = 123000

standard deviation = = 37000

P(X< 126000) = P[(X- ) / < (126000-123000) /37000 ]

= P(z <0.08 )

Using z table

PROBABILITY= 0.5319

B.

n = 69

= 123000

=  / n = 37000/ 69=4454

P( <126000 ) = P[( - ) / < (126000-123000) /4454 ]

= P(z < 0.67)

Using z table  

PROBABILITY=0.7486


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