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# Binomial Distribution

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Records indicate that 70 percent of Canadians submit their income tax returns by March 31 of each year when required to do so. In a random sample of 10 Canadians who are required to submit income tax returns,

A) What is the probability that 9 or more will submit their return by March 31?
B) What is the probability that exactly 5 will have submitted their return by March 31?

https://brainmass.com/statistics/probability/probability-calculations-binomial-distribution-15391

## SOLUTION This solution is FREE courtesy of BrainMass!

p= 0.7 or 70.00% (70% of Canadians submit their income tax returns by March 31)
q=1-p 0.3 or 30.00% (remaining 30% of Canadians do not submit their income tax returns by March 31)

This is a Binomial distribution

A) What is the probability that 9 or more will submit their return by March 31?

Probability that 9 or more will submit their return by March 31=
Probability that 9 will submit returns+ Probability that 10 will submit return)

Probability that 9 will submit returns=

p= 0.7
q=1-p= 0.3
n= 10
r= 9

P(r)= ncr p r * q 1-r

Therefore

P(r)= ncr p r * q n-r = =10*(0.7^9)*(0.3^1)
= 0.121060821 or 12.10608%

Probability that 10 will submit returns=

p= 0.7
q=1-p= 0.3
n= 10
r= 10

P(r)= ncr p r * q 1-r

Therefore

P(r)= ncr p r * q n-r = =1*(0.7^10)*(0.3^0)
= 0.0282475 or 2.82475%

Therefore Probability that 9 or more will submit their return by March 31=
=
Probability that 9 will submit returns + Probability that 10 will submit return
= 0.1493083 or 14.93084%
=0.1211+0.0282

B) What is the probability that exactly 5 will have submitted their return by March 31?

Probability that 0 will submit returns =

p= 0.7
q=1-p= 0.3
n= 10
r= 5

P(r)= ncr p r * q 1-r

Therefore

P(r)= ncr p r * q n-r = =252*(0.7^5)*(0.3^5)
= 0.102919345 or 10.29194%