Normal distribution of salaries
d = Standard dev = $7500 =
Correction factor= 0
Di
f(x) = (1/ d•√2π )• e ^- ((s- s')^2/2d^2)
Average value = u = 57337
Now calculate f(x)
Difference of salaries ,with respect to average
S^ 2= 57337 - 50000= = 7337
Now squared = (7337^2) = 53831569
d = 7500
2•d^2= 112500000
Then
S^2/ (2•d^2) = 53831569/112500000= 0.4785
Then now calculate
f(x) = (1/ d•√2π )•e^- (0.4785)
f(x)= 0.619/ (2•d^2)
Answer is
Probability of 61.9%