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 151000 dollars. Assume the standard deviation is 35000 dollars. Suppose you take a simple random sample of 77 graduates. Find the probability that a single randomly selected policy has a mean value between 138236.4 and 151000 dollars. P(138236.4 < X < 151000) = (Enter your answers as numbers accurate to 4 decimal places.) Find the probability that a random sample of size n = 77 has a mean value between 138236.4 and 151000 dollars. P(138236.4
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 151000 dollars. Assume the standard deviation is 35000 dollars. Suppose you take a simple random sample of 77 graduates. Find the probability that a single randomly selected policy has a mean value between 138236.4 and 151000 dollars. P(138236.4 < X < 151000) = (Enter your answers as numbers accurate to 4 decimal places.) Find the probability that a random sample of size n = 77 has a mean value between 138236.4 and 151000 dollars. P(138236.4
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Please help me figure this out on Excel. I cannot figure out the "Z-Table"
![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 151000 dollars.
Assume the standard deviation is 35000 dollars. Suppose you take a simple random sample of 77
graduates.
Find the probability that a single randomly selected policy has a mean value between 138236.4 and
151000 dollars.
P(138236.4 < X < 151000) =
(Enter your answers as numbers accurate to 4 decimal
places.)
Find the probability that a random sample of size n = 77 has a mean value between 138236.4 and
151000 dollars.
P(138236.4 < M < 151000) =
(Enter your answers as numbers accurate to 4 decimal
places.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e196f18-5a2b-42c2-971f-b2f79385967f%2Fa815ae36-5657-45c5-988a-45618da9b0ea%2Ffx39po7_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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 151000 dollars.
Assume the standard deviation is 35000 dollars. Suppose you take a simple random sample of 77
graduates.
Find the probability that a single randomly selected policy has a mean value between 138236.4 and
151000 dollars.
P(138236.4 < X < 151000) =
(Enter your answers as numbers accurate to 4 decimal
places.)
Find the probability that a random sample of size n = 77 has a mean value between 138236.4 and
151000 dollars.
P(138236.4 < M < 151000) =
(Enter your answers as numbers accurate to 4 decimal
places.)
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