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In Exercises 5鈥20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as 鈥渕inutes鈥) 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

Expert verified

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

01

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.

02

Computation of the measures of variation

Three types of measures of variation are given as follows:

The ranges2is 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 deviations is simply the value obtained by taking the square root of the variance.

The range is calculated as

Range=MaximumValue-MinimumValue=4-1=3.0

Therefore, the range is obtained as 3.0.

The mathematical expression used to calculate sample variance is

s2=i=1nxi-x2n-1

.

Here,

x represents the sample values, and

xis the sample mean.

The sample mean is calculated as

x=1=1nxin=2+1+1+...+225=47251.9

.

Thus, the sample mean is 1.9.

The variance of the sample is calculated as

s2=i=1nxi-x2n-1=2-1.92+1-1.92+...+2-1.9225-1=20.65240.9

.

Therefore, the sample variance is obtained as 0.9.

The sample standard deviation is calculated as

s=s2=0.90.9

Therefore, the sample standard deviation is obtained as 0.9.

03

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.

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