Descriptive Statistics Use amounts of arsenic in Exercise 1 and find the following: (a) mean; (b) median; (c) standard deviation; (d) variance; (e) range. Include the appropriate units of measurement.

The amount of arsenic present in a sample of brown rice servings is provided.

As the name suggests, numerical quantities that describe a given set of values are calleddescriptive statistics. These values are used to represent the data and help in further analysis of certain characteristics it follows. Some of the most commonly calculated descriptive statistics along with their formula are shown below:

$\overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n}$

$\text{L}={\left(\frac{n+1}{2}\right)}^{th}$observation if n is an odd value

$L=\frac{{\left(\frac{n}{2}\right)}^{th}\text{obs}+{\left(\frac{n}{2}+1\right)}^{th}\text{obs}}{2}$if n is an even value

$s=\sqrt{\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1}}$

$s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1}$

$\text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value}$

The mean is computed as follows:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{6.1+5.4+...+7.3}{12} \\ =6.09 \end{gathered}$$

Therefore, the mean value of the data is equal to 6.09 micrograms per serving.

Here, n is equal to 12. Thus, n is an even number.

$$\begin{gathered} \text{Median =}\frac{{\left(\frac{n}{2}\right)}^{th}\text{obs}+{\left(\frac{n}{2}+1\right)}^{th}\text{obs}}{2} \\ =\frac{{\left(6\right)}^{th}\text{obs}+{\left(7\right)}^{th}\text{obs}}{2} \\ =\frac{6.3+6.6}{2} \\ =6.45 \end{gathered}$$

Therefore, the median value of the data is equal to 6.45 micrograms per serving.

$$\begin{gathered} s=\sqrt{\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(6.1-6.09\right)}^2+{\left(5.4-6.09\right)}^2+...+{\left(7.3-6.09\right)}^2}{12-1}} \\ =1.75 \end{gathered}$$

Therefore, the standard deviation of the data is equal to 1.75 micrograms per serving.

The variance is square of standard deviation measure:

$$\begin{gathered} s^2={\left(1.75\right)}^2 \\ =3.06 \end{gathered}$$

Therefore, the variance of the data is equal to 3.06 micrograms per serving square.

The range is computed as follows:

$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =8.2-1.5 \\ =6.70 \end{gathered}$$

Therefore, the range of the data is equal to 6.70 micrograms per serving.