Question #217586

The literacy rate for a nation measures the proportion of people age 15 and over that can read and write.The literacy rate in Afghanistan is 28.1%.Suppose you choose 15 people in Afghanistan at random.Let X=Number of people who are literate.

i)Determine the probability distribution of X hence write down the mean and standard deviation of X

ii)is it more likely that 3 people or 4 people are literate?

iii)Find the probability that more than five people in the sample are literate

iv)Suppose you sample the Afghans until you find a literate person ,what is the probability that you will sample 4 people before getting one who is literate?


1
Expert's answer
2021-07-18T10:59:44-0400

(i) Let X=X= Number of people who are literate: XBin(n,p).X\sim Bin (n, p).

Given n=15,p=0.281,q=1p=10.281=0.719n=15, p=0.281, q=1-p=1-0.281=0.719


P(X=x)=(15x)(0.281)x(0.719)15xP(X=x)=\dbinom{15}{x}(0.281)^x(0.719)^{15-x}μ=np=15(0.281)=4.215\mu=np=15(0.281)=4.215

Var(X)=σ2=npq=15(0.281)(0.719)Var(X)=\sigma^2=npq=15(0.281)(0.719)

=3.030585=3.030585

σ=σ2=3.0305851.749\sigma=\sqrt{\sigma^2}=\sqrt{3.030585}\approx1.749

(ii)


P(X=3)=(153)(0.281)3(0.719)153P(X=3)=\dbinom{15}{3}(0.281)^3(0.719)^{15-3}

=0.19269792626=0.19269792626


P(X=4)=(154)(0.281)4(0.719)154P(X=4)=\dbinom{15}{4}(0.281)^4(0.719)^{15-4}

=0.22593094831=0.22593094831

It is more likely that 4 people are literate.


(iii)


P(X>5)=1P(X=0)P(X=1)P(X>5)=1-P(X=0)-P(X=1)

P(X=2)P(X=3)P(X=4)-P(X=2)-P(X=3)-P(X=4)

=0.22464552127=0.22464552127

(iv)


(0.719)4(0.281)=0.075096877821401(0.719)^4(0.281)=0.075096877821401


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