Answer:
0.26599
Explanation:
We will assume a normal distribution for the salaries
We have the following information
Mean μ = 64000
Standard deviation σ = 6400
And we are asked to find P(X < 60000)
First find the z score corresponding to X = 60000
The z score is given by (X - μ) / σ
z score for 60000 with μ = 64000 and σ = 6400 is
z = (60000 - 64000)/6400
z = -40000/6400 = -0.625
We can either look up the P value from the z-tables or simply use a calculator
P(X < 60000) and this works out to 0.26599