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 $4600. The IRS plans to draw a 1000 sample of tax returns to study the effect of new tax law.

A) What is the probability that the sample mean tax is less than $8100? Round the answer to at least four decimal places. The probability that the sample mean tax is less than $8100 is...?

B) What is the probability that the sample mean tax is between $7500 and $8100? Round the answer to at least four decimal places. The probability that the sample mean tax is between $7500 and $8100 is...?

C) Find the 70^th percentile of the sample mean. Round the answer to at least two decimal places.

D) Would it be unusual if the sample mean were less than $7800? Round answer to at least four decimal places. It would/would not be unusual because the probability of the sample mean being less than $7800 is...?

E) Do you think it would be unusual for an individual to pay a tax of less than $7800? Explain. Assume the variable is normally distributed. Round the answer to at least four decimal places. Yes/No, because the probability that an individual pays a tax less than $7800 is...?

Answer 1

Given that :

A)

The probability that the sample mean tax is less than $8100 is

Sampling distribution of is

Using the z score formula ,

Therefore the required probability is P(z<0.41)

Using Excel function,=NORMSDIST (z)

= NORMSDIST (0.41)

= 0.6591

The probability that the sample mean tax is less than $8100 is 0.6591

B)

The probability that the sample mean tax is between $7500 and $8100

Using z score formula,

For

Using Excel function = NORMSDIST (-3.71)

= 0.0001

P(z<-3.71) =0.0001

For

Using Excel function,=NORMSDIST (0.41)

=0.6591

P(z<0.41) = 0.6591

Therefore the required probability is

P(z<0.41)-P(z<-3.71)

= 0.6591-0.0001

=0.6590

The probability that the sample mean tax is between $7500 and $8100 is 0.6590

C)

The 70th percentile of the sample mean.

Percentile is the left tailed area.

First find the z score for the area 0.70

Using Excel function, =NORMSINV(area)

=NORMSINV (0.70)

=0.52

z=0.52

Now Using the z score formula to find

The 70th percentile of the sample mean is 8115.64

D)

The sample mean were less than $7800

z=-1.65

Using Excel function,=NORMSDIST (-1.65)

=0.0495

P(z<-1.65) = 0.0495

The probability that the sample mean were less than $7800 is 0.0495

If the probability is less than 0.05 it is unusual.

As 0.0495 is less than 0.05 it is unusual.

Therefore It would be unusual because the probability of the sample mean being less than $7800 is 0.0495