Question

5. The salaries at a corporation are normally distributed with an average salary of $19,000 and a standard deviation of $4,000.

a.	What is the probability that an employee will have a salary between $12,520 and $13,480?
b.	What is the probability that an employee will have a salary more than $11,880?
c.	What is the probability that an employee will have a salary less than $28,440?
d.	What is the probability that an employee will have a salary between $19,000 and $24,500?

Answer 1

Solution :

a.

P($12520 < x < $13480) = P[(12520 - 19000)/ 4000) < (x - ) /  < (13480 - 19000) / 4000) ]

= P(-1.62 < z < -1.38)

= P(z < -1.38) - P(z < -1.62)

= 0.0838 - 0.0526

= 0.0312

Probability = 0.0312

b.

P(x > $11880) = 1 - P(x < 11880)

= 1 - P[(x - ) / < (11880 - 19000) / 4000)

= 1 - P(z < -1.78)

= 1 - 0.0375

= 0.9625

Probability = 0.9625

c.

P(x < $28440) = P[(x - ) / < (28440 - 19000) / 4000]

= P(z < 2.36)

= 0.9909

Probability = 0.9909

d.

P($19000 < x < $24500) = P[(19000 - 19000)/ 4000) < (x - ) /  < (24500 - 19000) / 4000) ]

= P(0 < z < 1.375)

= P(z < 1.375) - P(z < 0)

= 0.9154 - 0.5

= 0.4154

Probability = 0.4154