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 = the number of people who are literate.

a. Sketch a graph of the probability distribution of X.

b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X.

c. Find the probability that more than five people in the sample are literate. Is it is more likely that three people or four people are literate.

Short answer

a.

b. mean is $4.215$and standard deviation is $1.749$

c. The probability of three people or four people are literate is $P(X=3or4)=0.4185$and the probability that more than five people in the sample are literate is $P(X>5)=0.2246$

Step-by-step solution

Content Introduction

The binomial distribution determines the probability of looking at a specific quantity of a hit results in a specific quantity of trials.

Explanation (part a)

The graphic presentation of X is:

Explanation (part b)

Let us first check the distribution of X:

we are given $n=15,p=0.281$

the distribution of X is $(15,0.281)$

The mean is

$$\begin{gathered} \mu =np \\ \mu =15\times 0.281 \\ \mu =4.215 \end{gathered}$$

The standard deviation is:

$$\begin{gathered} \sigma =\sqrt{np(1-p)} \\ \sigma =\sqrt{15\times 0.281(1-0.281)} \\ \sigma =1.749 \end{gathered}$$

Explanation (part c)

The probability for three people or four people are literate is:

$$\begin{gathered} P(X=3or4)=\left(\begin{array}{c}15\\ 3\end{array}\right){(0.281)}^3{(0.719)}^{12}+\left(\begin{array}{c}15\\ 4\end{array}\right){(0.281)}^4{(0.719)}^{11} \\ P(X=3or4)=0.1926+0.2259 \\ P(X=3or4)=0.4185 \end{gathered}$$

The probability that more than five people in the sample are literate is:

$$\begin{gathered} P(X>5)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4) \\ P(X>5)=0.2246 \end{gathered}$$