Question

In Exercises 5–20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions.

Peas in a Pod Biologists conducted experiments to determine whether a deficiency of carbon dioxide in the soil affects the phenotypes of peas. Listed below are the phenotype codes, where 1 = smooth-yellow, 2 = smooth-green, 3 = wrinkled yellow, and 4 = wrinkled-green. Can the measures of variation be obtained for these values? Do the results make sense?

2 1 1 1 1 1 1 4 1 2 2 1 2 3 3 2 3 1 3 1 3 1 3 2 2

Short answer

The sample range is 3.0.

The sample variance is 0.9.

The sample standard deviation is 0.9.

As the given data is nominal and the values are used to only label the different types of peas, the measures of variation should not be calculated as they are meaningless and do not make any sense.

Step-by-step solution

Given information

Data consisting of phenotype codes of 25 peas are given where 1 denotes smooth-yellow peas, 2 denotes smooth-green peas, 3 denotes wrinkled-yellow peas, and 4 denotes wrinkled-green peas.

The number of values (n) is 25.

Computation of the measures of variation

Three types of measures of variation are given as follows:

The range$\left(s^2\right)$is calculated by evaluating the difference between the highest and the lowest values .

The variancefor a sample of sizen is computed by squaring the difference between each observation of the data set and the mean, adding up all the squared terms, and then dividing by $n-1$.

The standard deviation$\left(s\right)$ is simply the value obtained by taking the square root of the variance.

The range is calculated as

$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =4-1 \\ =3.0 \end{gathered}$$

Therefore, the range is obtained as 3.0.

The mathematical expression used to calculate sample variance is

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

.

Here,

x represents the sample values, and

$\overline{x}$is the sample mean.

The sample mean is calculated as

$$\begin{gathered} \overline{x}=\frac{{\sum }_{1=1}^n{x}_i}{n} \\ =\frac{2+1+1+...+2}{25} \\ =\frac{47}{25} \\ \approx 1.9 \end{gathered}$$

.

Thus, the sample mean is 1.9.

The variance of the sample is calculated as

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(2-1.9\right)}^2+{\left(1-1.9\right)}^2+...+{\left(2-1.9\right)}^2}{25-1} \\ =\frac{20.65}{24} \\ \approx 0.9 \end{gathered}$$

.

Therefore, the sample variance is obtained as 0.9.

The sample standard deviation is calculated as

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{0.9} \\ \approx 0.9 \end{gathered}$$

Therefore, the sample standard deviation is obtained as 0.9.

Interpretation

Here, the sample has nominal data where values 1,2,3, and4 are used only to code different categories of peas. As it is known that no numerical calculation performed on categorical data has any significance, the measures of variation do not make any sense.