Answer to Question #88640 in Statistics and Probability for Justine

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%? 

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

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

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

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

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

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

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

$$=P(Z<1.5)-P(<Z<0.5)=0.93319280-0.69146246=$$

$$=0.24173034$$