Question

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.

Short answer

The values of the descriptive statistics are obtained as follows:

(a)Mean: 6.09 micrograms per serving

(b)Median: 6.45 micrograms per serving

(c)Standard Deviation: 1.75 micrograms per serving

(d)Variance: 3.06 micrograms per serving square

(e)Range: 6.70 micrograms per serving

Step-by-step solution

Given information

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

Define the descriptive statistics

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:

• Mean: It is the average value of the data. It is computed using the given formula:

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

• Median: The central value of the sorted data is known as the median. 50% of the values lie above the median, and 50% lie below the median. It has the following formula:

$\text{L}={\left(\frac{n+1}{2}\right)}^{th}$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

• Standard Deviation: It measures the variation present in the values. Its formula is shown below:

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

• Variance: It is the square of the standard deviation. It has the following formula:

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

• Range: It gives the overall extent of the data. Its formula is as follows:

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

Calculation of Mean

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.

Calculation of Median

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.

Calculation of Standard Deviation

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

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

Calculation of Variance

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.

Calculation of Range

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.