The distribution of weekly salaries at a large company is ri

The distribution of weekly salaries at a large company is right skewed with a mean of $1,000 and a standard deviation of $350. What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50?

Can someone please show me step by step how to calculate the answer to this?

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          

x1 = lower bound = 1000 - 50 =   950      
x2 = upper bound = 1000 + 50 =    1050      

Also,

u = mean =    1000      
n = sample size =    50      
s = standard deviation =    350      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.010152545      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.010152545      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.156211106      
P(z < z2) =    0.843788894      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.687577789   [ANSWER]