Answer to Question #147524 in Statistics and Probability for Lee Chun Ho

Question

1.​The salaries of the employees in a company follow the normal distribution with mean $16000 and standard deviation $800.

(a)​What is the probability that the salary of an employee is higher than $18000?​​​​​​​​​​​​ (3 marks)
(b)​There is 80% chance that the salary of an employee is less than $t. Find the value of t.​​​​​​​​​​​
​​​​​​​​​​​​ (3 marks)
(c)​If 20 employees are randomly chosen from the company and the mean salary of the employees is calculated. The manager of the company claims that over 20% of the mean salaries of the employee are higher than $16200. Do you agree?

Answer 1

$\mu=16000$

$\sigma=800$

a) $P(X>18000)=1-P(X<18000)=1-P(Z<\frac{18000-16000}{800})=$

$=1-P(Z<2.5)=1-0.9938=0.0062$

b) $P(X<t)=80\%=0.8$ thus Z-score is 0.84

$\frac{t-16000}{800}=0.84 \implies t=16672$

c) $P(X>16200)=1-P(X<16200)=1-P(Z<\frac{16200-16000}{800})=$

$=1-P(Z<0.25)=1-0.5987=0.4013$

Thus 40% of salaries of an employee are over $16200. We agree with the manager’s claim that over 20% of the mean salaries of the employee are higher than $16200.