Question

according to the internal revenue service, income tax returns one year average &1,332 in refunds for taxpayers. one explanation of this figure is that taxpayers would rather have the government keep back too much money during the year than to owe it money at the end of the year. suppose the average amount of tax at the end of a year is a refund of $1,332, with a standard deviation of $725. assume that amounts owed or due on tax returns are normally distributed.

What proportion of tax returns show a refund greater than $1900?

What proportion of the tax returns show a refund between $100 and $600?

Answer 1

Solution :

Given that ,

mean = = 1332

standard deviation = = 725

P(x > 1900) = 1 - p( x< 1900)

=1- p [(x - ) / < (1900 -1332) / 725]

=1- P(z < 0.78 )

= 1 - 0.7823 = 0.2177

proportion = 0.2177

P( 100< x < 600 ) = P[(100-1332)/725 ) < (x - ) /  < (600-1332) /725 ) ]

= P(-1.69 < z < -1.01)

= P(z <-1.01 ) - P(z <-1.69 )

Using standard normal table

= 0.1562 - 0.0455 = 0.1107

Proportion = 0.1107