According to the IRS,2012 tax returns showed an average refund of $2995 with a s

Question

According to the IRS,2012 tax returns showed an average refund of $2995 with a standard deviation of $2607.Assume that 2012 tax retirn are normally distributed.

Find the probability a randomly selected 2012 tax return showed an amount no less than $2,842 and at most $5,802. Find the probability a randomly selected 2012 tax return showed an amount greater than $3,185. Find the probability a randomly selected 2012 tax return showed an amount less than or equal to $4,048 or greater than $5,008. Find the probability a randomly selected 2012 tax return showed an amount less than $189. The probability is 0.15 that a randomly selected 2012 tax return showed no more than what amount? (Remember the label.) The probability is 0.97 that a randomly selected 2012 tax return showed at least what amount? (Remember the label.)

Explanation / Answer

1. P(2842<x<=5801) =P(2842-2995/2607<Z<8501-2995/2607)

=P(-0.059<Z<1.076)

=P(Z<1.076)-P(Z<-0.059)

=0.8591-0.4765 ..................use excel function '=NORMSDIST()' to find probability.

=0.3826

2.P(X>3185)=1-.P(Z<3185-2995/2607)

=1-P(Z<0.0728) .....................use excel function '=NORMSDIST(0.0728)' to find probability.

=0.5290

3.P(X<189)=P(Z<189-2995/2607)

=P(Z<-1.076)

=0.1409

4.P(Z<z)=0.15

we can use excel function to find z score NORMSINV(0.15) so will get

z=-1.036

x-mean/standard deviation = -1.036

x=mean-1.036*standard devaion

=2995-1.036*2607

=293.01 approximately 293

Please post remaining part saperately on board.

Hope this will b ehelpful to you. Thanks and God Bless You ;-)

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