Question

Taxes:The Internal Revenue Service reports that the mean federal income tax paid in the year 2010 was 8040.Assume that the standard deviation is 4900.The IRS plans to draw a sample of 1000 tax returns to study the effect of a new tax law.

What is the probability that the sample mead tax is less than 7800? Round 4 decimals

What is the probability that that sample mean tax is between 7500 and 8000? Round atleast 4 places

Find the 30th percentile of sample mean

Answer #1

**solution:**

**mean=8040**

**s.d=4900**

**n=1000**

**a)p(x<7800)=p(Xbar - µ) /
[σ/sqrt(n)<7800-8040/4900/sqrt(1000)]**

**=p(z<-240/154.95)**

**=p(z<-1.548)**

**=1-0.9394=0.0606**

**b)P(7500<Xbar<8000) = P(Xbar<8000) –
P(Xbar<7600)**

**First we have to find P(Xbar<8000)**

**Z = (8000 – 8040)/ [4900/sqrt(1000)]**

**Z = -0.2581**

**P(Xbar<8000) = P(Z<-0.2581) = 0.3974**

**Now, we have to find P(Xbar<7500)**

**Z = (7500 – 8040)/ [4900/sqrt(1000)]**

**Z = -3.484**

**P(Xbar<7500) = P(Z< -3.484) = 0.00029**

**P(7500<Xbar<8000) = P(Xbar<8000) –
P(Xbar<7500)**

**P(7500<Xbar<8000) = 0.3974 - 0.00029 =
0.3971**

**c)**

**Z value for 30 ^{th} percentile =
-0.524**

**X = µ + Z*σ = 8040 – 0.524*4900 = 5472.4**

**Required 30 ^{th} percentile =
5472.4**

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