Chapter 3, Problem 1RE

(a)

To determine

To compute: The sample mean and median.

Answer to Problem 1RE

Solution: The sample mean is 792.51 and the median is 792.4.

Explanation of Solution

Given: The provided data represents the muzzle velocity.

Explanation: The mean of the provided data is calculated as

$\begin{array}{c}\overline{x}=\frac{1}{n}{\displaystyle \sum x_i}\\ =\frac{1}{10}\times \left(793.8+793.1+792.4+\mathrm{...}+794.4\right)\\ =\frac{7925.1}{10}\\ =792.51\end{array}$

Hence, the sample mean is 792.51.

Now, calculate the median of the provided data. First, arrange the data in an ascending order:

$789.6,\text{ 791}\text{.4, 791}\text{.7,792}\text{.3, 792}\text{.4 , 792}\text{.4 , 793}\text{.1, 793}\text{.8, 794}\text{.0, 794}\text{.4}$

In the data, the number of observations is 10, which is even.

$\begin{array}{c}\text{Median}=\frac{n}{2}\text{term + }\left(\frac{\text{n}}{2}+1\right)\text{term}\\ =\frac{10}{2}\text{term+}\left(\frac{\text{10}}{2}+1\right)\text{term}\\ =5\text{term+6}\text{term}\end{array}$

From the above solution, the median is the average of 5th and 6th observations.

$\begin{array}{c}\text{Median}=\frac{792.4+792.4}{2}\\ =792.4\end{array}$

(b)

To determine

To compute: The range, sample variance, and sample standard deviation.

Answer to Problem 1RE

Solution: The values of the range, variance, and standard deviation are 4.8, 2.03, and 1.42, respectively.

Explanation of Solution

Given: The provided data represents the muzzle velocity.

Explanation: Maximum value and the minimum value of the observations are 794.4 and 789.6.

The range of the provided data is obtained as

$\begin{array}{c}\text{Range}=\text{Maximum value}-\text{Minimum value}\\ =794.4-789.6\\ =4.8\end{array}$

The variance is calculated as

$\begin{array}{c}{s}^{{}^2}=\frac{1}{n-1}{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}\\ =\frac{1}{10-1}\left({\left(793.8-792.5\right)}^2+{\left(793.1-792.5\right)}^2+\cdots +{\left(794.4-792.5\right)}^2\right)\\ =\frac{19.83}{9}\\ =2.033\end{array}$

Hence, the variance is 2.033.

The standard deviation for the given dataset can be calculated as

$\begin{array}{c}s=\sqrt{s^2}\\ =\sqrt{2.03}\\ =1.42\end{array}$

Hence, the standard deviation is 1.42.