According to the Internal Revenue Service, the average income tax refund for the
ID: 3039944 • Letter: A
Question
According to the Internal Revenue Service, the average income tax refund for the 2011 tax year was $2,913. As- sume the refund distribution with a standard deviation of $950 11 per person follows the normal probability a. What is the probability that a randomly selected tax return refund from the 2011 tax year will be 1. more than $2,000? 2. between $1,600 and S2.500? 3. between $3,200 and $4,000? Continuous Probal b. Confirm the answers to part a using Excel or PHStat. e. What refund amount represents the 35th percentile of tax returns?Explanation / Answer
Answer:
a).
1.
Z value for 2000, z =(2000-2913)/950 = -0.96
P( x >2000)m = P( z > -0.96)
=0.8315
2).
Z value for 1600, z =(1600-2913)/950 = -1.38
Z value for 2500, z =(2500-2913)/950 = -0.43
P( 1600<x<2500) = P( -1.38<z<-0.43) = P( z < -0.43) –P( z < -1.38)
=0.3336-0.0838
=0.2498
3).
Z value for 3200, z =(3200-2913)/950 = 0.30
Z value for 4000, z =(4000-2913)/950 = 1.14
P( 3200<x<4000) = P( 0.30<z<1.14) = P( z < 1.14) –P( z < 0.30)
=0.8729-0.6179
=0.255
b).
PHSTAT used
Normal Probabilities
Common Data
Mean
2913
Standard Deviation
950
Probability for X >
X Value
2000
Z Value
-0.961053
P(X>2000)
0.8317
Probability for a Range
From X Value
1600
To X Value
2500
Z Value for 1600
-1.382105
Z Value for 2500
-0.434737
P(X<=1600)
0.0835
P(X<=2500)
0.3319
P(1600<=X<=2500)
0.2484
Probability for a Range
From X Value
3200
To X Value
4000
Z Value for 3200
0.3021053
Z Value for 4000
1.1442105
P(X<=3200)
0.6187
P(X<=4000)
0.8737
P(3200<=X<=4000)
0.2550
c).
z value for 35th percentile = -0.385
x = mean+z*sd = 2913-0.385*950
=2547.25
Normal Probabilities
Common Data
Mean
2913
Standard Deviation
950
Probability for X >
X Value
2000
Z Value
-0.961053
P(X>2000)
0.8317
Probability for a Range
From X Value
1600
To X Value
2500
Z Value for 1600
-1.382105
Z Value for 2500
-0.434737
P(X<=1600)
0.0835
P(X<=2500)
0.3319
P(1600<=X<=2500)
0.2484
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