Question

Q5. The annual salary for a sample of 500 part-time unskilled worker s is normally distributed with a mean of $30,000 and a standard deviation of $3,000. a. The salary that separates the top 15% from the lower is b. The number of students who earned more than $36,000 was c. Find the proportion that the salary is less than $28,500 d. Calculate the 90th percentile

Answer 1

Solution:-

Given that,

mean = = 30000

standard deviation = = 3000

n = 500

a) Using standard normal table,

P(Z > z) = 15%

= 1 - P(Z < z) = 0.15  

= P(Z < z) = 1 - 0.15

= P(Z < z ) = 0.85

= P(Z < 1.04 ) = 0.85  

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 3000 + 30000

x = $ 33120

b) P(x > 36000) = 1 - p( x< 36000)

=1- p P[(x - ) / < (36000 - 30000) / 3000 ]

=1- P(z < 2.00)

Using z table,

= 1 - 0.9772

= 0.0228

c) P(x < 28500)

= P[(x - ) / < (28500 - 30000) / 3000]

= P(z < -0.50)

Using z table,

= 0.3085

d) Using standard normal table,

P(Z < z) = 90%

= P(Z < z ) = 0.90

= P(Z < 1.28 ) = 0.90  

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 3000 + 30000

x = $ 33840