The IRS estimates that 75percent of all tax returns are eligible for a refund. (

Question

The IRS estimates that 75percent of all tax returns are eligible for a refund.
(a) If a random sample of size 50 tax returns is drawn, then whatis the probability that the number
of tax returns eligible for a refund will be less than 33?
(b) Again with n = 50, what is the probability that the number oftax returns eligible for a refund
will be between 30 and 40?
(c) Now suppose the sample is of size n = 100, what is theprobability that the number of tax
returns eligible for a refund will be less than 66? what aboutbetween 60 and 80? How do these
answers relate to your answers to the first two parts?

Explanation / Answer

(a)Given X~Binomial(n=50, p=0.75)

By using Normal distribtuion approximation,
X~Normal(=n*p=50*0.75=37.5,=n*p*(1-p)=sqrt(50*0.75*0.25)=3.06)

The probability is
P(X<33)=P((X-)/ < (33-37.5)/3.06)
=P(Z<-1.47)
= 0.0708 (check normal table)


(b)By using Normal distribtuion approximation,
X~Normal(=n*p=50*0.75=37.5,=n*p*(1-p)=sqrt(50*0.75*0.25)=3.06)

The probability is
P(30<X<40)=P((30-37.5)/3.06<(X-)/ <(40-37.5)/3.06)
=P(-2.45<Z<0.82)
= 0.7867 (check normal table) (c)By using Normal distribtuion approximation,
X~Normal(=n*p=100*0.75=75,=n*p*(1-p)=sqrt(100*0.75*0.25)=4.33)


The probability that thenumber of tax returns eligible for a refund will be less than 66is
P(X<66)=P(Z<(66-75)/4.33)
=P(Z< -2.08)
=0.0187 (check normal table)


P(60<X<80)=P((60-75)/4.33<Z<(80-75)/4.33)
=P(-3.46<Z<1.15)
=0.8746 (check normal table)

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