Question

In: Statistics and Probability

According to a national business travel association a 2010 survey, the average salary of a travel...

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)

Solutions

Expert Solution

Solution :

Given that ,

mean = = 96850

standard deviation = = 30000

n = 50

A) = = 96850

B) = / n = 30000 / 50 = 4242.64

C) The shape of the distribution is approximately normal

D) Using z-score formula,

z = - /

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

orchestra answered 2 years ago
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