Question

The Internal Revenue Service reports that the mean federal income tax paid in the year 2010 was $8040. Assume that the standard deviation is $4800. The IRS plans to draw a sample of 1000 tax returns to study the effect of a new tax law.

What is the probability that the sample mean tax is less than $1800?

What is the probability that the sample mean tax between $7400 and $8000?

40th percentile of the sample mean is ?

Answer 1

Solution :

Given that ,

mean = = 8040

standard deviation = = 4800

n = 1000

= = 8040

= / n = 4800 / 1000 = 151.79

a) P( < 1800) = P(( - ) / < (1800 - 8040) / 151.79)

= P(z < -41.11)

Using z table

= 0

b) P(7400 < < 8000)  

= P[(7400 - 8040) /151.79 < ( - ) / < (8000 - 8040) / 151.79)]

= P(-4.22 < Z < -0.26)

= P(Z < -0.26) - P(Z < -4.22)

Using z table,  

= 0.3974 - 0

= 0.3974

c) Using standard normal table,

P(Z < z) = 40%

= P(Z < z ) = 0.40

= P(Z < -0.25 ) = 0.40  

z = -0.25

Using z-score formula  

= z * +   

= -0.25 * 151.79 + 8040

= 8002

40 th percentile is = $ 8002