According to the IRS, 2012 tax returns showed an average refund of $2439 with a
ID: 3130549 • Letter: A
Question
According to the IRS, 2012 tax returns showed an average refund of $2439 with a standard deviation of $2198. Assume that 2012 tax return amounts are normally distributed. Find the probability a randomly selected 2012 tax return showed an amount greater than $3212 Find the probability a randomly selected 2012 tax return showed an amount less than or equal to $4388 or greater than $5015 Find the probability a randomly selected 2012 tax return showed an amount less than $470 Find the probability a randomly selected 2012 tax return showed an amount no less than $2248 and at most $5408 The probability is 0.79 that a randomly selected 2012 tax return showed at least what amount? (Remember the label. The probability is 0.4 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.) The probability is 0.19 that a randomly selected 2012 tax return showed no more than what amount? (Remember the label.)Explanation / Answer
1.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 3212
u = mean = 2439
s = standard deviation = 2198
Thus,
z = (x - u) / s = 0.351683348
Thus, using a table/technology, the right tailed area of this is
P(z > 0.351683348 ) = 0.362537875 [ANSWER]
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2.
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 4388
x2 = upper bound = 5015
u = mean = 2439
s = standard deviation = 2198
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 0.886715196
z2 = upper z score = (x2 - u) / s = 1.171974522
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.812383872
P(z < z2) = 0.879396357
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.067012485
Thus, those outside this interval is the complement = 0.932987515 [ANSWER]
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3.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 470
u = mean = 2439
s = standard deviation = 2198
Thus,
z = (x - u) / s = -0.895814377
Thus, using a table/technology, the left tailed area of this is
P(z < -0.895814377 ) = 0.185175955 [ANSWER]
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4.
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 2248
x2 = upper bound = 5408
u = mean = 2439
s = standard deviation = 2198
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.086897179
z2 = upper z score = (x2 - u) / s = 1.35077343
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.465376621
P(z < z2) = 0.911615989
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.446239368 [ANSWER]
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Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!
P.S. I have a feeling that there is a typo in the given, because the mean and standard deviation are almost the same. Normally, the mean is much greater than the standard deviation. If there is a typo, please include that in your next submission. Thanks!
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