According to the IRS the mean tax refund for the year 2013 w

According to the IRS, the mean tax refund for the year 2013 was $3,000. Assume that the standard deviation is $450 and that amounts refunded follow a normal probability distribution.

A) What is the probability that a refund is more than $3,100?

B) What proportion of the refunds are more than $3,100 but less than $3,500?

c) What refund amount corresponds to that obtained by the 10% of taxpayers who receive the largest refunds?

D) In a random sample of 25 refunds, what is the probability that the mean refund would be less than $2,800?

(Please post solutions with steps on how you got to the answer)

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    3100      
u = mean =    3000      
          
s = standard deviation =    450      
          
Thus,          
          
z = (x - u) / s =    0.222222222      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.222222222   ) =    0.412070448 [answer]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    3100      
x2 = upper bound =    3500      
u = mean =    3000      
          
s = standard deviation =    450      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.222222222      
z2 = upper z score = (x2 - u) / s =    1.111111111      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.587929552      
P(z < z2) =    0.866739737      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.278810185 [ANSWER]

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C)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    3000      
z = the critical z score =    1.281551566      
s = standard deviation =    450      
          
Then          
          
x = critical value =    3576.698204      
          

Thus, it is those who obtained $3576.70 or more. [answer]

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d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    2800      
u = mean =    3000      
n = sample size =    25      
s = standard deviation =    450      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.222222222      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -2.222222222   ) =    0.013134146 [ANSWER]