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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:

$渭=np$, where

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

Therefore, the mean is

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

Standard deviation is:

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

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:

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

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

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