Question

In: Statistics and Probability

The Internal Revenue Service reports that the mean federal income tax paid in the year 2010...

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 ?

Solutions

Expert Solution

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


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