Question

The salaries of employees of a certain company in Metro Manila have a mean of Php

5,000 and a standard deviation of Php 1000. What is the probability that an

employee selected at random will have a salary of

a) more than Php 5000

b) between Php 5000 and Php 6500

Answer 1

Let $X$ be a random variable representing the salaries of employees in the company then,

$\mu=5000,\space \sigma=1000$. Therefore, $X\sim N(5000,1000^2).$

a)

The probability that the salary for the selected employee is more than Php 5000 is given by,

$p(X\gt5000)=P(Z\gt (5000-5000)/1000)=p(Z\gt0)$

This can also be written as,

$p(Z\gt0)=1-p(Z\lt0)=1-\phi(0)=1-0.5=0.5$

Therefore, the probability that the randomly selected employee has a salary more than 5000 is 0.5.

b)

The probability that the salary for the selected employee is  between Php 5000 and Php 6500 is given as,

$p(5000\lt X\lt6500)=p((5000-5000)/1000\lt Z\lt (6500-5000)/1000)=p(0\lt Z\lt 1.5)$

We can write this probability as,

$p(0\lt Z\lt1.5)=\phi(1.5)-\phi(0)=0.93319-0.5=0.43319$

Therefore, the probability that the selected employee has a salary between Php 5000 and Php 6500 is 0.43319.