Question

A human resources manager keeps a record of how many years each employee at a large company has been working in their current role. The distribution of these years of experience are strongly skewed to the right with a mean of 333years and a standard deviation of 222 years. Suppose we were to take a random sample of 444employees and calculate the sample mean for their years of experience. We can assume independence between members in the sample.

What is the probability that the mean years of experience from the sample of 444 employees xˉxˉx, with, \bar, on top is greater than 3.53.53, point, 5 years?

Answer 1

Solution :

Given that,

mean = = 3

standard deviation = = 2

n = 4

= 3

= / n = 2 / 4 = 1

P( > 3.5 ) = 1 - P( < 3.5 )

= 1 - P[( - ) / < ( 3.5 - 3 ) / 1 ]

= 1 - P(z < 0.5 )

Using z table,    

= 1 - 0.6915

= 0.3085

Probability = 0.3085