Question

2. The annual salaries of employees in a large company are approximately normally distributed with a mean of N50,000 and a standard deviation of N 20,000.
i. What percent of people earn less than N 40,000?
ii. What percent of people earn between N 45,000 and N 65,000?
iii. What percent of people earn more than N 70,000?

Answer 1

Given that,

Population mean $\mu$ = 50000

Standard deviation $\sigma$ = 20000

ANSWER 1)

We have to find the probability for total percentage of people 'Z' earning less than N 40000. which is given by Probability P(x<40000)

On converting the given data to a standard normal distribution

Z = $\frac {x - \mu}{\sigma} \sim N(0,1)$

Hence P(x<40000) = P($\frac {x-\mu}{\sigma} < \frac {40000 - \mu}{\sigma}$ ) = P(Z < $\frac {40000-50000}{20000}$ ) = P(Z < -0.5)

From the standard normal table we get the value of

P(Z < -0.5) = 0.3085

ANSWER 2)

We have to find the probability for total percentage of people 'Z' earning between N 45000 to N 65000. which is given by Probability P(65000 < x < 45000)

On converting the given data to a standard normal distribution

Z = $\frac {x - \mu}{\sigma} \sim N(0,1)$

Hence,

P(65000 > x > 45000) = P($\frac {65000 - \mu}{\sigma} >$ $\frac {x-\mu}{\sigma} > \frac {45000 - \mu}{\sigma}$ )

=($\frac {65000 - 50000}{20000} >$$\frac {x-\mu}{\sigma} > \frac {45000 - 50000}{20000}$ )

P(65000 > x > 45000) = P(0.75 > Z > -0.25)

From the standard normal table we get the value of

P(0.75 > Z > -0.25) = (0.7734 - 0.5) + (0.5 - 0.4013) = 0.3721

ANSWER 3)

We have to find the probability for total percentage of people 'Z' earning more than N 70000. which is given by Probability P(x>70000)

On converting the given data to a standard normal distribution

Z = $\frac {x - \mu}{\sigma} \sim N(0,1)$

Hence P(x>70000) = P($\frac {x-\mu}{\sigma} > \frac {70000 - \mu}{\sigma}$ ) = P(Z > $\frac {70000-50000}{20000}$ ) = P(Z > 1)

From the standard normal table we get the value of

P(Z > 1) = 0.5 - 0.3413 = 0.1587