Answer to Question #123145 in Statistics and Probability for Sisilia TALEI

Question

A supervisor has determined that the average salary of the employees in his department $40,000 with a standard deviation of $15,000. Assume that the distribution of the salaries is normally distributed. An employee is selected at random, find the probability that the employee’s salary is:

(i) More than $35,000.
(ii) Between $36,000 to $42,000.

Answer 1

Let X = the random variable denoting the salary of the selected employee

We have,

X ~ N($\mu$ = 40000, $\sigma^2$ = 15000)

Then,

Z = $\frac{X-\mu}{\sigma}$ ~ N(0, 1), Z is the standard normal variate

(i) The probability that the employee’s salary is more than $35,000

= P(X > 35000)

= P($\frac{X-40000}{15000}>\frac{35000-40000}{15000}$)

= P(Z > - 0.33)

= 1 - P(Z $\leq$ - 0.33)

= 1 - $\Phi$(- 0.33)

= 1 - 0.3707 [from standard normal table]

= 0.6293

Answer: The probability that the employee’s salary is more than $35,000 is 0.6293.

(ii) The probability that the employee’s salary is between $36,000 to $42,000

P(36000 $\leq$ X $\leq$ 42000)

= P($\frac{36000-40000}{15000}\leq\frac{X-40000}{15000}\leq\frac{42000-40000}{15000}$)

= P(-0.27 $\leq$ Z $\leq$ 0.13)

= P(Z $\leq$ 0.13) - P(Z < -0.27)

= $\Phi$(0.13) - $\Phi$(-0.27)

= 0.5517 - 0.3936 [from standard normal table]

= 0.1581

Answer: The probability that the employee’s salary is between $36,000 to $42,000 is 0.1581.