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 $160,000 dollars. Assume the standard deviation is $41,000. Suppose you take a simple random sample of 82 graduates.
1) What is the standard deviation of the sampling distribution for this situation? Round to four places. Show work and answer using proper notation.
2) Find the probability that a single randomly selected salary has a value between $143,700 and $164,981. Your write up should include all of the following:
3) Find the probability that a random sample of size n=82n=82 has a mean value between $143,700 and $164,981.
Write your answers in complete sentence form.
1) standard deviation of the sampling distribution=σ/√n = 4527.6926
2)
we need to calculate probability for ,
143700 ≤ X ≤ 164981
X1 = 143700 , X2
= 164981
Z1 = (X1 - µ )/(σ/√n) = (
143700 - 160000 ) /
( 41000 / √ 1 )
= -0.40
Z2 = (X2 - µ )/(σ/√n) = (
164981 - 160000 ) /
( 41000 / √ 1 )
= 0.12
P ( 143700 < X <
164981 ) = P (
-0.40 < Z < 0.12 )
= P ( Z < 0.12 ) - P ( Z
< -0.40 ) =
0.548 - 0.345 =
0.2029 (answer)
the probability that a single randomly selected salary has a value
between $143,700 and $164,981 is 0.2029
3)
we need to calculate probability for ,
143700 ≤ X ≤ 164981
X1 = 143700 , X2
= 164981
Z1 = (X1 - µ )/(σ/√n) = (
143700 - 160000 ) /
( 41000 / √ 82 )
= -3.60
Z2 = (X2 - µ )/(σ/√n) = (
164981 - 160000 ) /
( 41000 / √ 82 )
= 1.10
P ( 143700 < X <
164981 ) = P (
-3.60 < Z < 1.10 )
= P ( Z < 1.10 ) - P ( Z
< -3.60 ) =
0.864 - 0.000 =
0.8642
(answer)
the probability that a random sample of size n=82 has a mean value
between $143,700 and $164,981 is 0.8642
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