According to The World Bank, only 9% of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let X = the number of people who have access to electricity.

a. What is the probability distribution for X?

b. Using the formulas, calculate the mean and standard deviation of X.

c. Use your calculator to find the probability that 15 people in the sample have access to electricity.

d. Find the probability that at most ten people in the sample have access to electricity.

e. Find the probability that more than 25 people in the sample have access to electricity

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

We are given,

$9\%$of the population of Uganda had access to electricity as of 2009 and we randomly chose a sample of 150 people in Uganda. So. here $n$is the sample size and $p$is the probability.

Therefore,

The probability distribution of X is$p=0.09$

We are given:$n=150,p=0.09$

The mean is:

$\mu =np$, where

$$\begin{gathered} n=numberofsamples \\ p=probabilityofsamples \end{gathered}$$

Therefore, the mean is

role="math" localid="1649250625500" $$\begin{gathered} \mu =np \\ \mu =150\times 0.09 \\ \mu =13.5 \end{gathered}$$

Standard deviation is:

$$\begin{gathered} \sigma =\sqrt{np(1-p)} \\ \sigma =\sqrt{150\times 0.09(1-0.09)} \\ \sigma =3.5 \end{gathered}$$

Using binompdf:

the probability that $15$people in the sample have access to electricity is:

role="math" localid="1649250800761" $$\begin{gathered} binompdf(150,0.09,15)=\frac{150!}{15!(150-15)!}0.{09}^{15}{(1-0.09)}^{50-15} \\ binompdf(150,0.09,15)=\frac{150!}{15!(150-15)!}0.{09}^{15}{(1-0.09)}^{35} \\ binompdf(150,0.09,15)=0.0988 \end{gathered}$$

The probability that at most ten people in the sample have access to electricity is:

$$\begin{gathered} binompdf(150,0.09,10)=\sum _{k=0}^{10}\frac{150!}{!(k150-k)!}0.{09}^k{(1-0.09)}^{50-k} \\ binompdf(150,0.09,10)=0.1897 \end{gathered}$$

The probability that more than 25 people in the sample have access to electricity is:

$$\begin{gathered} binompdf(150,0.09,25)=\sum _{k=0}^{25}\frac{150!}{!(k150-k)!}0.{09}^k{(1-0.09)}^{50-k} \\ binompdf(150,0.09,25)=0.9991 \end{gathered}$$