The mean starting salary at a corporation is 25000 and a sta

The mean starting salary at a corporation is $25,000 and a standard deviation of $5.000. Assume that the starting salary is normally distributed. What is the probability that a randomly selected employee will have a starting salary of at least $29.000? What percentage of employees has starting salaries of less than $18,000? What are the maximum starting salaries of the bottom 18% of the employees?

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    29000      
u = mean =    25000      
          
s = standard deviation =    5000      
          
Thus,          
          
z = (x - u) / s =    0.8      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.8   ) =    0.211855399 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    18000      
u = mean =    25000      
          
s = standard deviation =    5000      
          
Thus,          
          
z = (x - u) / s =    -1.4      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.4   ) =    0.080756659 or 8.0756659% [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.18      
          
Then, using table or technology,          
          
z =    -0.915365088      
          
As x = u + z * s,          
          
where          
          
u = mean =    25000      
z = the critical z score =    -0.915365088      
s = standard deviation =    5000      
          
Then          
          
x = critical value =    20423.17456   [ANSWER]