Question

Taxes: The Internal Revenue Service reports that the mean federal income tax paid in the year

2010

was

$8040

. Assume that the standard deviation is

$4700

. The IRS plans to draw a sample of

1000

tax returns to study the effect of a new tax law.

(a) What is the probability that the sample mean tax is less than

$8000

? Round the answer to at least four decimal places.

The probability that the sample mean tax is less than $8000 is (b) What is the probability that the sample mean tax is between $7600 and $8100 ? Round the answer to at least four decimal places. The probability that the sample mean tax is between $7600 and $8100 is (c) Find the 30 th percentile of the sample mean. Round the answer to at least two decimal places. The 30 th percentile of the sample mean is

Answer 1

Solution:-

Given That:-

Let X be a random variable which represents the tax paid in 2010.

Given that, X ~ N(8040, 4700)

i.e. μ = $8040 and σ = $4700

We know that if X ~ N(μ , σ​​​​​​2) then x̄ ~ N(μ , σ​​​​​​2/n)

And if x̄ ~ N(μ , σ​​​​​​2/n) then (Dong z

(Where, x̄ is sample mean, n is sample size).

a) We have to obtain P(x̄ < $8000).

We have, μ = $8040, σ = $4700 and n = 1000

PT < 8000) = P (8000 - A) on / ov

P(i < 8000) = P(Z < (8000 – 8040) ) 4700/1000

Plī < 8000) = P(Z <-0.2691)

Using "pnorm" function of R we get, P(Z < -0.2691) = 0.3939

:. Plī < 8000) = 0.3939

Hence, the probability that the sample mean tax is less than $8000 is 0.3939.

b) We have to obtain P($7600 < x̄ < $8100).

We have, μ = $8040, σ = $4700 and n = 1000

\large P(7600<\bar{x}<8100)=P(\bar{x}<8100)-P(\bar{x}<7600)

\large P(7600<\bar{x}<8100)=P\left(\frac{(\bar{x}-\mu)}{\sigma/\sqrt{n}}<\frac{(8100-\mu)}{\sigma/\sqrt{n}} \right)- P\left(\frac{(\bar{x}-\mu)}{\sigma/\sqrt{n}}<\frac{(7600-\mu)}{\sigma/\sqrt{n}} \right)

\large P(7600<\bar{x}<8100)=P\left(Z<\frac{(8100-8040)}{4700/\sqrt{1000}} \right)- P\left(Z<\frac{(7600-8040)}{4700/\sqrt{1000}} \right)

\large P(7600<\bar{x}<8100)=P\left(Z<0.4037\right)- P\left(Z< -2.9604\right)Using "pnorm" function of R we get,

P(Z < 0.4037) = 0.6568 and P(Z < -2.9604) = 0.0015

\large \therefore P(7600<\bar{x}<8100)=0.6568-0.0015 = 0.6553

Hence, the probability that the sample mean tax is between $7600 and $8100 is 0.6553.

c) Let the 30th percentile of sample mean is k.

Hence, P(x̄ < k) = 0.30

\large P\left(\frac{(\bar{x}-\mu)}{\sigma/\sqrt{n}}<\frac{(k-\mu)}{\sigma/\sqrt{n}} \right)=0.30

\large P\left(Z<\frac{(k-8040)}{4700/\sqrt{1000}} \right)=0.30.....................(1)

Using "qnorm" function of R we get P(Z < -0.5244) = 0.30

Comparing, P(Z < -0.5244) = 0.30 and (1) we get,

\large \frac{(k-8040)}{4700/\sqrt{1000}} =-0.5244

\large k =8040+\left ( -0.5244\times\frac{4700}{\sqrt{1000}} \right )

he7962,06

Hence, the 30th percentile for the sample mean is $7962.06.