Answer to Question #217586 in Statistics and Probability for VEVEE

Question

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?

Answer 1

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

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

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

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

$$=3.030585$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{3.030585}\approx1.749$$

(ii)

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

$$=0.19269792626$$

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

$$=0.22593094831$$

It is more likely that 4 people are literate.

(iii)

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

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

$$=0.22464552127$$

(iv)

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