Question

The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000.

a.What percent of people earn less than $40,000?

b.What percent of people earn between $45,000 and $65,000?

c.What percent of people earn more than $70,000?

Answer 1

: The annual salaries of employees in a large company are approximately normally distributed

with a mean of $50000 and a standard deviation of $20000.

a. What percent of people earn less than $40000?

Solution: Let S be the random variable of a salary of employee (in $), S ~ N(50000,20000). Then the random

variable X =−50000

20000

~N(0,1).

( < 40000) = ( <

40000 − 50000

20000 ) = ( < −0.5) = (−0.5) = 0.3085375.

Here Φ(x) denotes the cumulative distribution function of a standard normal distribution.

Answer: 31%.

b. What percent of people earn between $45000 and $65000?

Solution:

(45000 < < 65000) = (

45000 − 50000

20000 < <

65000 − 50000

20000 ) = (−0.25 < < 0.75)

= (0.75) − (−0.25) = 0.7733726 − 0.4012937 = 0.3720789.

Answer: 37%.

c. What percent of people earn more than $70000?

Solution:

( > 70000) = ( >

70000 − 50000

20000 ) = ( > 1) = 0.8413447.

Answer 2

Let X be the normal variable denoting the annual salaries of employees.  Given,

Mean $\mu = \ 50,000$ and

Standard deviation $\sigma = \ 20,000$

(a) percent of people earning lee than $\ 40,000$

$P (X < 40,000) \\= P(X < 40,000) \\= P(Z < (40000 - 50000)/20,000) \\ = P (Z < -0.5) \\ =P (0.5 < Z)$

By using the symmetry of normal distribution curve we can also write the above equation as

$= 0.5 - P(0 < Z < 0.5) \\= 0.5 - 0.1915 \\ = 0.3085$

Hence people who earn less than $\ 40,000$ is $0.3085 * 100 = 30.85\%$

(b) percent of people earning between $\ 45,000 - \ 65,000$

$P(45000 < X < 65000)\\= P((45000 - 50000)20000 < Z < (65000 - 50000)/20000 ) \\= P (-0.25 < Z < 0.75) \\= P(-0.25 < Z < 0) + P (0 < Z < 0.75) \\= P(0 < Z < 0.25) + P (0 < Z < 0.75) \\= 0.0987 + 0.2734 \\= 0.3721$

Hence percent of people who earn between $\ 45,000$ and $\ 65,000$ is $0.3721 * 100 = 37.21\%$

(c) percent of people earn more than $\ 70,000$

$P(X > 70,000) \\= P(Z > (70000 - 50000)/20000)) \\= P(Z > 1) = 0.5 - P(0 < Z < 1) \\= 0.5 - 0.3413 \\= 0.1587$

Hence percent of people who earn more than $\ 70,000$ is

$0.1587 * 100 = 15.87\%$

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