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