Chapter 5.3, Problem 17E

Find the mean, variance, and standard deviation for each of the values of n and p when the conditions for the binomial distribution are met.

a. n = 100, p = 0.75

b. n = 300, p = 0.3

c. n = 20, p = 0.5

d. n = 10, p = 0.8

To determine

The mean, variance and standard deviation.

Answer to Problem 17E

The mean is 75.

The variance is 18.8.

The standard deviation is 4.3.

Explanation of Solution

Given info:

The number of trials n is 100 and probability of success is 0.75

Calculation:

Mean of Binomial distribution:

$\mu =n{p}_{}$

Where, n is the number of trials and p is the probability of success for each trial and q is the probability of failure for each trial.

Substitute n as 100 and p as 0.75.

$\begin{array}{c}\mu =\left(100\right)\left(0.75\right)\\ =75\end{array}$

Thus, the mean is 75.

Variance of Binomial distribution:

${\sigma }^2=npq$

Substitute n as 100, p as 0.75 and q as $\text{0}\text{.25 }\left(\text{=}1-0.75\right)$

$\begin{array}{c}{\sigma }^2=\left(100\right)\left(0.75\right)\left(0.25\right)\\ =18.75\end{array}$

Thus, the variance is 18.75.

Standard deviation of Binomial distribution:

$\sigma =\sqrt{npq}$

Substitute n as 100, p as 0.75 and q as $\text{0}\text{.25 }\left(\text{=}1-0.75\right)$

$\begin{array}{c}\sigma =\sqrt{\left(100\right)\left(0.75\right)\left(0.25\right)}\\ =\sqrt{18.75}\\ =4.33\end{array}$

Thus, the standard deviation is 4.33.

To determine

The mean, variance and standard deviation.

Answer to Problem 17E

The mean is 90.

The variance is 63.

The standard deviation is 7.9.

Explanation of Solution

Given info:

The number of trials n is 300 and probability of success is 0.3

Calculation:

Mean of Binomial distribution:

$\mu =n{p}_{}$

Substitute n as 300 and p as 0.3.

$\begin{array}{c}\mu =\left(300\right)\left(0.3\right)\\ =90\end{array}$

Thus, the mean is 90.

Variance of Binomial distribution:

${\sigma }^2=npq$

Substitute n as 300, p as 0.3 and q as $\text{0}\text{.7 }\left(\text{=}1-0.3\right)$

$\begin{array}{c}{\sigma }^2=\left(300\right)\left(0.3\right)\left(0.7\right)\\ =63\end{array}$

Thus, the variance is 63.

Standard deviation of Binomial distribution:

$\sigma =\sqrt{npq}$

Substitute n as 100, p as 0.3 and q as $\text{0}\text{.7 }\left(\text{=}1-0.3\right)$

$\begin{array}{c}\sigma =\sqrt{\left(300\right)\left(0.3\right)\left(0.7\right)}\\ =\sqrt{63}\\ =7.9\end{array}$

Thus, the standard deviation is 7.9.

To determine

The mean, variance and standard deviation.

Answer to Problem 17E

The mean is 10.

The variance is 5.

The standard deviation is 2.2.

Explanation of Solution

Given info:

The number of trials n is 20 and probability of success is 0.5

Calculation:

Mean of Binomial distribution:

$\mu =n{p}_{}$

Substitute n as 20 and p as 0.5.

$\begin{array}{c}\mu =\left(20\right)\left(0.5\right)\\ =10\end{array}$

Thus, the mean is 10.

Variance of Binomial distribution:

${\sigma }^2=npq$

Substitute n as 20, p as 0.5 and q as $\text{0}\text{.5 }\left(\text{=}1-0.5\right)$

$\begin{array}{c}{\sigma }^2=\left(20\right)\left(0.5\right)\left(0.5\right)\\ =5\end{array}$

Thus, the variance is 5.

Standard deviation of Binomial distribution:

$\sigma =\sqrt{npq}$

Substitute n as 20, p as 0.5 and q as $\text{0}\text{.5 }\left(\text{=}1-0.5\right)$

$\begin{array}{c}\sigma =\sqrt{\left(20\right)\left(0.5\right)\left(0.5\right)}\\ =\sqrt{5}\\ =2.2\end{array}$

Thus, the standard deviation is 2.2.

To determine

The mean, variance and standard deviation.

Answer to Problem 17E

The mean is 8.

The variance is 1.6.

The standard deviation is 1.3.

Explanation of Solution

Given info:

The number of trials n is 10 and probability of success is 0.8

Calculation:

Mean of Binomial distribution:

$\mu =n{p}_{}$

Substitute n as 10 and p as 0.8.

$\begin{array}{c}\mu =\left(10\right)\left(0.8\right)\\ =8\end{array}$

Thus, the mean is 8.

Variance of Binomial distribution:

${\sigma }^2=npq$

Substitute n as 10, p as 0.8 and q as $\text{0}\text{.2 }\left(\text{=}1-0.8\right)$

$\begin{array}{c}{\sigma }^2=\left(10\right)\left(0.8\right)\left(0.2\right)\\ =1.6\end{array}$

Thus, the variance is 1.6.

Standard deviation of Binomial distribution:

$\sigma =\sqrt{npq}$

Substitute n as 10, p as 0.8 and q as $\text{0}\text{.2 }\left(\text{=}1-0.8\right)$

$\begin{array}{c}\sigma =\sqrt{\left(10\right)\left(0.8\right)\left(0.2\right)}\\ =\sqrt{1.6}\\ =1.3\end{array}$

Thus, the standard deviation is 1.3.