According to the IRS, 2012 tax returns showed an average refund of $2891 with a

Question

According to the IRS, 2012 tax returns showed an average refund of $2891 with a standard deviation of $2566. Assume that 2012 tax return amounts are normally distributed. Based on this information (and information in Q7-13), generate excel output in the given space on the Tax sheet in this workbook to answer the following questions. Each orange numerical answer cell MUST reference Excel output cells in the Tax sheet

4621.7407

-find the probability a randomly selected 2012 tax return showed an amount greater than $3,289. answer: 0.5616

-the probability is 0.24 that a randomly selected 2012 tax return showed between what two amounts equidistant from the mean? Find the lower endpoint using excel. use this lower endpoint and some math to find the upper endpoint. (remember the label) answer:____ ____,  answer:____ ___ .

-The probability is 0.23 that a randomly selected 2012 tax return showed no more than what amount? (rememeber the label)  answer: 995.1190 dollars

-The probability is 0.75 that a randomly selected 2012 tax return showed at least what amount? (remember the label)  answer: 4621.7407 dollars

can you please explain how to find the answers? If some of my answers are wrong, why? Thank you.

Normal mu 2891 sigma 2566 xi P(X<=xi) Q 7 4067 0.6766 Q 7 5234 0.8194 Q 8 13 0.1310 Q 9 2664 0.4648 Q 9 5294 0.8255 Q 10 3289 0.5616 P(X<=xi) xi Q 11 0.24 1078.6276 Q 12 0.23 995.1190 Q 13 0.75

4621.7407

Explanation / Answer

tis can be solved by normal value table

as z=(X-mean)/std deviation

1)probability a randomly selected 2012 tax return showed an amount greater than $3,289. answer

=P(X>3289)=1-P(X<3289)=1-P(Z<(3289-2891)/2566)=1-P(Z<0.1551)=1-0.5616=0.4384

2) for 0.24 middle values, it should lie b/w 38 and 62 percentile for which z =+/-0.3055

hence corresponding interval =mean +/- z*std deviation =2107.14 ; 3674.86

3)probability is 0.23 that a randomly selected 2012 tax return showed no more than what amount

P(X<x)=0.23

for 23 percentile z=-0.7388

hence corresponding value =mean +z*std deviation =995.119

4)probability is 0.75 that a randomly selected 2012 tax return showed at least what amount=P(X>x)=0.75

1-P(X<x)=0.75

P(X<x) =0.25

for 0.25 percentile ; z=-0.6745

hence corresponding score =mean +z*std deviation =1160.259

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