Chapter 5.5, Problem 33E

Consider a binomial experiment with n = 20 and p = .70.

1. a. Compute f(12).
2. b. Compute f(16).
3. c. Compute P(x ≥ 16).
4. d. Compute P(x ≤ 15).
5. e. Compute E(x).
6. f. Compute Var (x) and σ.

To determine

Find the value of $f\left(12\right)$.

Answer to Problem 33E

The value of $f\left(12\right)$ is 0.1144.

Explanation of Solution

Calculation:

The binomial experiment with $n=20\text{ and }p=0.70$.

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f\left(x\right)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x},x=0,1,2,\mathrm{...},n$

Where, p is the probability of success.

Substitute values $n=20\text{ and }p=0.70\text{ and }x=12$.

$\begin{array}{c}f\left(12\right)=\left(\begin{array}{l}20\\ 12\end{array}\right){\left(0.70\right)}^{12}{\left(1-0.70\right)}^{20-12}\\ =\frac{20!}{12!\left(20-12\right)!}{\left(0.70\right)}^{12}{\left(0.30\right)}^8\\ =0.1144\end{array}$

Thus, the probability of 12 successes is 0.1144.

To determine

Find the value of, $f\left(16\right)$.

Answer to Problem 33E

The value of, $f\left(16\right)$ is 0.1304.

Explanation of Solution

Calculation:

Consider $n=20\text{ and }p=0.70\text{ and }x=16$.

$\begin{array}{c}f\left(16\right)=\left(\begin{array}{l}20\\ 16\end{array}\right){\left(0.70\right)}^{16}{\left(1-0.70\right)}^{20-16}\\ =\frac{20!}{16!\left(20-16\right)!}{\left(0.70\right)}^{16}{\left(0.30\right)}^4\\ =0.1304\end{array}$

Thus, the probability of 16 successes is 0.1304.

To determine

Find the value of, $P\left(x\ge 16\right)$.

Answer to Problem 33E

The value of $P\left(x\ge 16\right)$ is 0.2374.

Explanation of Solution

Calculation:

Here, $P\left(x\ge 16\right)$ is the probability of at least 16 successes. This probability is the sum of the probabilities at the points $x=16,x=17,x=18,x=19\text{ and }x=20$. That is,

$P\left(x\ge 16\right)=f\left(16\right)+f\left(17\right)+f\left(18\right)+f\left(19\right)+f(20)$

The probability $f\left(17\right)$ is obtained by substituting the values $n=20\text{ and }p=0.70\text{ and }x=17$ in the binomial probability formula. That is,

The probability value of $f\left(17\right)$ is,

$\begin{array}{c}f\left(17\right)=\left(\begin{array}{l}20\\ 17\end{array}\right){\left(0.70\right)}^{17}{\left(1-0.70\right)}^{20-17}\\ =\frac{20!}{17!\left(20-17\right)!}{\left(0.70\right)}^{17}{\left(0.30\right)}^3\\ =0.0716\end{array}$

The probability value of $f\left(18\right)$ is,

$\begin{array}{c}f\left(18\right)=\left(\begin{array}{l}20\\ 18\end{array}\right){\left(0.70\right)}^{18}{\left(1-0.70\right)}^{20-18}\\ =\frac{20!}{18!\left(20-18\right)!}{\left(0.70\right)}^{18}{\left(0.30\right)}^2\\ =0.0278\end{array}$

The probability value of $f\left(19\right)$ is,

$\begin{array}{c}f\left(19\right)=\left(\begin{array}{l}20\\ 19\end{array}\right){\left(0.70\right)}^{19}{\left(1-0.70\right)}^{20-19}\\ =\frac{20!}{19!\left(20-19\right)!}{\left(0.70\right)}^{19}{\left(0.30\right)}^1\\ =0.0068\end{array}$

The probability value of $f\left(20\right)$ is,

$\begin{array}{c}f\left(20\right)=\left(\begin{array}{l}20\\ 20\end{array}\right){\left(0.70\right)}^{20}{\left(1-0.70\right)}^{20-20}\\ =\frac{20!}{20!\left(20-20\right)!}{\left(0.70\right)}^{20}{\left(0.30\right)}^0\\ =0.0008\end{array}$

The required probability becomes,

$\begin{array}{c}P\left(x\ge 16\right)=f\left(16\right)+f\left(17\right)+f\left(18\right)+f\left(19\right)+f(20)\\ =0.1304+0.0716+0.0278+0.0068+0.0008\\ =0.2374\end{array}$

Thus, the probability of at least 16 successes is 0.2374.

To determine

Find the value of, $P\left(x\le 15\right)$.

Answer to Problem 33E

The value of $P\left(x\le 15\right)$ is 0.7626.

Explanation of Solution

Calculation:

The required probability of at most 15 successes can be obtained as the compliment of the probability of at least 15 successes. That is,

$P\left(x\le 15\right)=1-P\left(x\ge 16\right)$

Substituting the value from part (c),

$\begin{array}{c}P\left(x\le 15\right)=1-P\left(x\ge 16\right)\\ =1-0.2374\\ =0.7626\end{array}$

Thus, the probability of at most 15 successes is 0.7626.

To determine

Compute the expected value.

Answer to Problem 33E

The expected value of the binomial random variable is 14.

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$\begin{array}{c}E\left(x\right)=\mu \\ =np\end{array}$

Substituting the values $n=20\text{ and }p=0.70$

$\begin{array}{c}E\left(x\right)=20\times 0.70\\ =14\end{array}$

The expected value of the binomial random variable is 14.

To determine

Compute the variance and standard deviation.

Answer to Problem 33E

The variance of the binomial random variable is 4.2.

The standard deviation of the binomial random variable is 2.0494.

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

${\sigma }^2=np\left(1-p\right)$

Substituting the values $n=20\text{ and }p=0.70$

$\begin{array}{c}{\sigma }^2=np\left(1-p\right)\\ =20\times 0.7\left(1-0.7\right)\\ =4.2\end{array}$

The variance of the binomial random variable is 4.2.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$\begin{array}{c}{\sigma }_x=\sqrt{np\left(1-p\right)}\\ =\sqrt{0.42}\\ =2.0494\end{array}$

The standard deviation of the binomial random variable is 2.0494.