Question

The average income tax refund for the 2009 tax year was ​$2847. Assume the refund per person follows the normal probability distribution with a standard deviation of ​$906. Complete parts a through d below. a. What is the probability that a randomly selected tax return refund will be more than ​$2100​? nothing ​(Round to four decimal places as​ needed.) b. What is the probability that a randomly selected tax return refund will be between ​$1600 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 ​$3300 and ​$3900​? nothing ​(Round to four decimal places as​ needed.) d. What refund amount represents the 35th percentile of tax​ returns? ​$nothing ​(Round to the nearest dollar as​ needed.)

Answer 1

mean = 2847 , s = 906

a)

P(x> 2100)
= P(z> (2100 - 2847)/906)
= P(z > -0.82)
=1 - P(z< -0.82)
= 1 - 0.2048
= 0.7952

b)

P(1600 < x< 3000)
= P((1600- 2847)/906 < z < (3000- 2847)/906 )
= P(-1.38 < z < 0.17)
= 0.5671 - 0.0844
= 0.4827

c)

P(3300 < x< 3900)
= P((3300- 2847)/906 < z < (3900- 2847)/906 )
= P(0.50 < z< 1.16)
= 0.8774 - 0.6915
= 0.1859

d)

z value at 35% = -0.3853

z = (x - mean)/s

-0.3853 = (x - 2847)/906

x = 906 * -0.3853+ 2847
x = 2498