Question

Annual salaries for a large company are approximately normally distributed with a mean of
R50,000 and a standard deviation of R20,000.
a. What salary would an employee need to get in order to be in the lowest 30%?
b. What is the probability of having an above average salary range of between R60000 to
R80000

Answer 1

If $X\sim N(\mu, \sigma^2),$ then $Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)$

$$\mu=R50,000, \sigma=R20,000$$

a. What salary would an employee need to get in order to be in the lowest 30%? 

$$P(Z<z^*)=0.3=>z^*=-0.5244$$

$$x^*=\mu+z^*\sigma$$

$$x^*=R50000+(-0.5244)R20000=R39512$$

b. What is the probability of having an above average salary range of between R60000 to R80000

$$Z=\dfrac{X-\mu}{\sigma}$$

$$z_1=\dfrac{R60000-R50000}{R20000}=0.5$$

$$z_2=\dfrac{R80000-R50000}{R20000}=1.5$$

$$P(R60000<X<R80000)=P(0.5<Z<1.5)=$$

$$=P(Z<1.5)-P(Z<0.5)=0.93319-0.69146=0.24173$$