Find the probability of x successes in100Bernoulli trials with probability$\mathit{p}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}$ of success.

(a) if$\mathit{x}\mathbf{=}\mathbf{25}$

(b) if$\mathit{x}\mathbf{=}\mathbf{21}$

The number of people is $1095$

Frequency distribution of the number of successful outcomes that can be achieved in a given number of trials, each with an equal chance of success.

The formula for binomial distribution states that.$f\left(x\right)=C(x,n)p^x{q}^{n-x}$

$\begin{array}{c}n=100\\ p=\frac{1}{5}\\ q=\frac{4}{5}\\ x=25\end{array}$

Substitute the values mentioned above in the formula.

$\begin{array}{c}f\left(25\right)=C(100,25){\left(\frac{1}{5}\right)}^{25}{\left(\frac{4}{5}\right)}^{75}\\ =0.0439\end{array}$

The formula for corresponding normal approximation states that $f\left(x\right)=\frac{e^{-{(x-np)}^2/2npq}}{\sqrt{2\pi npq}}$.

$\begin{array}{c}n=100\\ p=\frac{1}{5}\\ q=\frac{4}{5}\\ x=25\end{array}$

Substitute the values mentioned above in the formula.

$\begin{array}{c}f\left(25\right)=\frac{1}{\sqrt{2\pi \left(100\right)(4/25)}}\cdot e^{-25/32}\\ =0.0457\end{array}$

The formula for Poisson distribution states that .$P_n=\frac{{\mu }^n{e}^{-\mu }}{n!}$

$\begin{array}{c}P=\frac{{\left(np\right)}^x{e}^{-xp}}{x!}\\ =\frac{{\left(20\right)}^{25}}{25!}{e}^{-20}\\ =0.0446\end{array}$

The formula for binomial distribution states that .$f\left(x\right)=C(x,n)p^x{q}^{n-x}$

$\begin{array}{c}n=100\\ p=\frac{1}{5}\\ q=\frac{4}{5}\\ x=21\end{array}$

Substitute the values mentioned above in the formula.

$\begin{array}{c}f\left(25\right)=C(100,21){\left(\frac{1}{5}\right)}^{21}{\left(\frac{4}{5}\right)}^{79}\\ =0.0944\end{array}$

The formula for corresponding normal approximation states that .$f\left(x\right)=\frac{e^{-{(x-np)}^2/2npq}}{\sqrt{2\pi npq}}$

$\begin{array}{c}n=100\\ p=\frac{1}{5}\\ q=\frac{4}{5}\\ x=21\end{array}$

Substitute the values mentioned above in the formula.

$\begin{array}{c}f\left(25\right)=\frac{1}{\sqrt{2\pi \left(100\right)(4/25)}}\cdot e^{-1/32}\\ =0.0967\end{array}$

The formula for Poisson distribution states that .$P_n=\frac{{\mu }^n{e}^{-\mu }}{n!}$

$\begin{array}{c}P=\frac{{\left(np\right)}^x{e}^{-np}}{x!}\\ =\frac{{\left(20\right)}^{21}}{21!}{e}^{-20}\\ =0.0846\end{array}$

The value of the function for part (a) is given below.

Binomial distribution .$0.439$

Normal distribution $0.457$.

Poison distribution .$0.446$

The value of the function for part (b) is given below.

Binomial distribution .$0.0944$

Normal distribution .$0.0946$

Poison distribution .$0.0846$