Find the​ range, variance, and standard deviation for the given sample​ data, if possible. If the measures of variation can be obtained for these​ values, do the results make​ sense?
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 1equals​smooth-yellow, 2equals​smooth-green, 3equalswrinkled ​yellow, and 4 equals​wrinkled-green.
1    1    4    1    3    1    3    3    2    2    2    3    1    2    2    2    2    2    2    2
Can the range of the sample data be obtained for these​ values? Choose the correct answer below​ and, if​ necessary, fill in the answer box within your choice.
A. The standard deviation of the sample data is nothing. ​(Round to one decimal place as​ needed.)
B. The standard deviation of the sample data cannot be calculated. Can the variance of the sample data be obtained for these​ values?

Choose the correct answer below​ and, if​ necessary, fill in the answer box within your choice.
A. The variance of the sample data is nothing. ​(Round to one decimal place as​ needed.)
B. The variance of the sample data cannot be calculated.

Do the results make​sense?
A. The measures of variation do not make sense because the standard deviation cannot be greater than the variance.
B. While the measures of variation can be​ found, they do not make sense because the data are​ nominal; they​ don't measure or count anything.
C. The measures of variation make sense because the data is​ numeric, so the spread between the values is meaningful.
D. It makes sense that the measures of variation cannot be calculated because there is not a large enough sample size to calculate the measures of variation.

Respuesta :

Answer:

C. The measures of variation make sense because the data is​ numeric, so the spread between the values is meaningful.

Step-by-step explanation:

Categories               X        x-x`      (x-x`)²

1                                 5         0          0

2                                10        5          25

3                                4         -1            1

4                                1           -4         16

                            ∑x= 20 ∑x-x` =0  ∑ (x-x`)²=42

Yes the range for these values can be obtained as range gives the idea of the spread of the values and is simply the difference between the largest and smallest value . The range of this data is 10-1= 9

The standard deviation is given by  √∑(x-x`)²/n

where x` is the mean and is calculated as = 20/4=5  

and n is the number of observations

S.D= √42/4 =√10.5= 3.24

And Variance is 10.5