End of Section Problem 6.9 According to the Internal Revenue Service, income tax

Question

End of Section Problem 6.9 According to the Internal Revenue Service, income tax returns one year averaged $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.

(a) What proportion of tax returns show a refund greater than $1,700?

(b) What proportion of the tax returns show that the taxpayer owes money to the government?

(c) What proportion of the tax returns show a refund between $150 and $680?

(Round all the z values to 2 decimal places. Round your answers to 4 decimal places.)

(a) P(x > $1,700) =

(b) P(x < 0) =

(c) P($150 x $680) =

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1700      
u = mean =    1332      
          
s = standard deviation =    725      
          
Thus,          
          
z = (x - u) / s =    0.507586207      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.507586207   ) =    0.305871783 [answer]

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b)

Hence, a less than $0 refund.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0      
u = mean =    1332      
          
s = standard deviation =    725      
          
Thus,          
          
z = (x - u) / s =    -1.837241379      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.837241379   ) =    0.033087135 [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    150      
x2 = upper bound =    680      
u = mean =    1332      
          
s = standard deviation =    725      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.630344828      
z2 = upper z score = (x2 - u) / s =    -0.899310345      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.051514319      
P(z < z2) =    0.184243689      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.13272937   [ANSWER]

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