Question

1. A group of students with normally distributed salaries earn an average of $6,800 with a standard deviation of $2,500.

(a) What proportion of students earn between $6,500 and $7,300?

(b) What are the first and third quartiles of students’ salaries?

(c) What value of salary in $ exceeded the 95% probability?

Answer 1

Solution :

Given that,

mean = = 6,800

standard deviation = = 2,500

a ) P (6,500 < x < 7,300 )

P ( 6,500 - 6,800 / 2,500) < ( x -  / ) < ( 7,300 - 6,800 / 2,500)

P ( - 300 / 2,500 < z < 500 / 2,500)

P (-0.12 < z < 0.2)

P ( z < 0.2 ) - P ( z < -0.12)

Using z table

= 0.5793 - 0.4522

= 0.1271

Probability = 0.1271

b ) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.67

Using z-score formula,

x = z * +

x = -0.67 * 2,500 + 6800

x = 5125

First quartile =Q1 = 5125

Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.6745 ) = 0.75

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 * 2,500 +6800

x = 8475

Third quartile = 8475

c ) P( Z > z) = 95%

P(Z > z) = 0.95

1 - P( Z < z) = 0.95

P(Z < z) = 1 - 0.95

P(Z < z) = 0.05

z = -1.64

Using z-score formula,

x = z * +

x = -1.64 * 2,500 +6800

x = 2700