The average income tax refund for the 2009 tax year was $3029. Assume the refund

Question

The average income tax refund for the 2009 tax year was

$3029.

Assume the refund per person follows the normal probability distribution with a standard deviation of

$997.

Complete parts a through d below.

a. What is the probability that a randomly selected tax return refund will be more than

$2300?

nothing

(Round to four decimal places as needed.)b. What is the probability that a randomly selected tax return refund will be between

$1500

and

$3000?

nothing

(Round to four decimal places as needed.)c. What is the probability that a randomly selected tax return refund will be between

$3200

and

$4300?

nothing

(Round to four decimal places as needed.)d. What refund amount represents the

35th

percentile of tax returns?

$nothing

(Round to the nearest dollar as needed.)

Explanation / Answer

Ans:

Given that

mean=3029

standard deviation=997

a)P(X>2300)

z=(2300-3029)/997=-0.731

P(z>-0.731)=0.7676

b)P(1500<X<3000)

z(3000)=(3000-3029)/997=-0.029

z(1500)=(1500-3029)/997=-1.534

P(z<=-0.029)-P(z<=-1.534)=0.4884-0.0625=0.4259

c)P(3200<X<4300)

z(3200)=(3200-3029)/997=0.1715

z(4300)=(4300-3029)/997=1.275

P(z<=1.275)-P(z<=0.1715)=0.8988-0.5681=0.3308

d)P(Z<=z)=0.35

z=-0.3853

x=3029-0.3853*997=2644.86

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