Chapter 5.3, Problem 18E

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

a. n = 1000, p = 0.1

b. n = 500, p = 0.25 125;

c. n = 50, p = ⅖

d. n = 36, p = $\frac{1}{6}$

To determine

To calculate: The mean, variance and standard deviation.

Answer to Problem 18E

The mean is 100.

The variance is 90.

The standard deviation is 9.487.

Explanation of Solution

Given info:

The number of trials n is 1,000 andp is 0.01.

Calculation:

Mean of binomial distribution:

$\mu =np$

Where, n is the number of trials and p is the probability of success.

Substitute n as 10 and p as 0.1

$\begin{array}{c}\mu =\left(1,000\right)\left(0.1\right)\\ =100\end{array}$

Thus, the mean is 100.

Variance of binomial distribution:

${\sigma }^2=npq$

Where, n is the number of trials and p is the probability of success and q is $\left(1-p\right)$ which is probability of failure.

Substitute n as 10, p as 0.1 and q as $0.90\left(1-0.1\right)$

$\begin{array}{c}{\sigma }^2=\left(1,000\right)\left(0.1\right)\left(0.90\right)\\ =90\end{array}$

Thus, the variance is 90.

Standard deviation:

$\sigma =\sqrt{npq}$

Substitute n as 10, p as 0.1 and q as $0.90\left(1-0.1\right)$

$\begin{array}{c}\sigma =\sqrt{\left(1,000\right)\left(0.1\right)\left(0.90\right)}\\ =\sqrt{90}\\ =9.487\end{array}$

Thus, the standard deviation is 9.487.

To determine

To calculate: The mean, variance and standard deviation.

Answer to Problem 18E

The mean is 125.

The variance is 93.75.

The standard deviation is 9.682.

Explanation of Solution

Given info:

The number of trials n is 500, and p is 0.25.

Calculation:

Mean of binomial distribution:

$\mu =np$

Where, n is the number of trials and p is the probability of success.

Substitute n as 500 and p as 0.25

$\begin{array}{c}\mu =\left(500\right)\left(0.25\right)\\ =125\end{array}$

Thus, the mean is 125.

Variance of binomial distribution:

${\sigma }^2=npq$

Where, n is the number of trials and p is the probability of success and q is $\left(1-p\right)$ which is probability of failure.

Substitute n as 500, p as 0.25 and q as $0.75\left(1-0.25\right)$

$\begin{array}{c}{\sigma }^2=\left(500\right)\left(0.25\right)\left(0.75\right)\\ =93.75\end{array}$

Thus, the variance is 93.75.

Standard deviation:

$\sigma =\sqrt{npq}$

Substitute n as 500, p as 0.25 and q as $0.75\left(1-0.25\right)$

$\begin{array}{c}\sigma =\sqrt{\left(500\right)\left(0.25\right)\left(0.75\right)}\\ =\sqrt{93.75}\\ =9.682\end{array}$

Thus, the standard deviation is 9.682.

To determine

To calculate: The mean, variance and standard deviation.

Answer to Problem 18E

The mean is 20.

The variance is 4.8.

The standard deviation is 2.191.

Explanation of Solution

Given info:

The number of trials n is 50, and p is $\frac{2}{5}$.

Calculation:

Mean of binomial distribution:

$\mu =np$

Where, n is the number of trials and p is the probability of success.

Substitute n as 50 and p as $\frac{2}{5}$

$\begin{array}{c}\mu =\left(50\right)\left(\frac{2}{5}\right)\\ =20\end{array}$

Thus, the mean is 20.

Variance of binomial distribution:

${\sigma }^2=npq$

Where, n is the number of trials and p is the probability of success and q is $\left(1-p\right)$ which is probability of failure.

Substitute n as 50, p as $\frac{2}{5}$ and q as $\frac{3}{5}\left(=1-\frac{2}{5}\right)$

$\begin{array}{c}{\sigma }^2=\left(50\right)\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)\\ =\left(50\right)\left(0.4\right)\left(0.6\right)\\ =4.8\end{array}$

Thus, the variance is 4.8.

Standard deviation:

$\sigma =\sqrt{npq}$

Substitute n as 50, p as $\frac{2}{5}$ and q as $\frac{3}{5}\left(=1-\frac{2}{5}\right)$

$\begin{array}{c}\sigma =\sqrt{\left(50\right)\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)}\\ =\sqrt{\left(50\right)\left(0.4\right)\left(0.6\right)}\\ =\sqrt{4.8}\\ =2.191\end{array}$

Thus, the standard deviation is 2.191.

To determine

To calculate: The mean, variance and standard deviation.

Answer to Problem 18E

The mean is 6.

The variance is 5.008.

The standard deviation is 2.238.

Explanation of Solution

Given info:

The number of trials n is 36, and p is $\frac{1}{6}$.

Calculation:

The number of trials is 36 and it is fixed. Also, the probability of success $\frac{1}{6}$ remains the same for all the trials. Hence, the conditions of binomial experiment are satisfied.

Mean of binomial distribution:

$\mu =np$

Where, n is the number of trials and p is the probability of success.

Substitute n as 36 and p as $\frac{1}{6}$

$\begin{array}{c}\mu =\left(36\right)\left(\frac{1}{6}\right)\\ =6\end{array}$

Thus, the mean is 6.

Variance of binomial distribution:

${\sigma }^2=npq$

Where, n is the number of trials and p is the probability of success and q is $\left(1-p\right)$ which is probability of failure.

Substitute n as 6, p as $\frac{1}{6}$ and q as $\frac{5}{6}\left(=1-\frac{1}{6}\right)$

$\begin{array}{c}{\sigma }^2=\left(36\right)\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)\\ =\left(36\right)\left(0.167\right)\left(0.833\right)\\ =5.008\end{array}$

Thus, the variance is 5.008.

Standard deviation:

$\sigma =\sqrt{npq}$

Substitute n as 6, p as $\frac{1}{6}$ and q as $\frac{5}{6}\left(=1-\frac{1}{6}\right)$

$\begin{array}{c}\sigma =\sqrt{\left(36\right)\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)}\\ =\sqrt{\left(36\right)\left(0.167\right)\left(0.833\right)}\\ =\sqrt{5.008}\\ =2.238\end{array}$

Thus, the standard deviation is 2.238.