A skills test used by a recruiter for a company has a mean s

A skills test used by a recruiter for a company has a mean score of 50 and a standard deviation of 19. What would be the standard deviation of the sampling distribution of the sample mean if samples of size 83 were taken? Round your answer to three decimal places.

Number ( n ) = 83
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)

Mean ( u ) =50
Standard Deviation ( sd )= sd/Sqrt(n) = 19/ Sqrt ( 83 ) = 2.086