In: Statistics and Probability
According to a national business travel association a 2010 survey, the average salary of a travel management professional is $96,850. Assume that the standard deviation of such salaries is $30,000. Consider a random sample of 50 travel management professionals and let Xbar represent the mean salary for the sample.
A. What is u xbar.
B. What is Standard deviation of x bar.
C. Describe the shape of the distribution.
D. Find the z-score for the value of xbar= 89,500
E. Find P(Xbar greater than 89,500)
Solution :
Given that ,
n = 50
B)
=
/
n = 30000 /
50 = 4242.64
C) The shape of the distribution is approximately normal
D) Using z-score formula,
z = 89500 - 96850 / 4242.64
z = -1.73
E) P(
> 89500) = 1 - P(
< 89500)
= 1 - P[(
-
) /
< (89500 - 96850) / 4242.64]
= 1 - P(z < -1.73)
= 1 - 0.0418
= 0.9582