The correct order which shows the standard deviations of the data sets A,B and C in order from greatest to least is C, B, A.
How to find the standard deviation of a data set?
Suppose that
[tex]x_i; \: \: i = 1,2, ... ,n[/tex] are n data values for the given data set.
Then we have: Variance as,
[tex]\sigma^2 = \dfrac{\sum_{\forall x_i} (x_i - \overline{x})^2}{n}[/tex]
As standard deviation is the positive root of variance, thus,
[tex]\sigma = \sqrt {\dfrac{\sum_{\forall x_i} (x_i - \overline{x})^2}{n}}[/tex]
If its sample, we divide by n-1 instead of n
Use data sets A, B, and C to answer the question that follows.
- A {25, 24, 25, 22, 21, 23, 1}
- B: {20, 23, 25, 21, 24, 22, 60}
- C. {20, 21, 20, 22, 20, 24, 99}
The mean of the set A is,
[tex]\overline{x}_A=\dfrac{25+24+ 25+ 22+ 21+23+ 1}{7}=20.143[/tex]
Similarly, the mean of set B and C is,
[tex]\overline{x}_B=\dfrac{20+ 23+ 25+ 21+ 24+ 22+ 60}{7}=27.857\\\overline{x}_C=\dfrac{20+21+ 20+ 22+ 20+ 24+ 99}{7}=32.286[/tex]
The standard deviation of the set A is,
[tex]\sigma_A = \sqrt {\dfrac{(25-20.143)^2+(24-20.143)^2......+(1-20.143)^2}{7}}\\\sigma_A = 7.936[/tex]
Similarly, the standard deviation of set B and C is,
[tex]\sigma_B = \sqrt {\dfrac{(20-27.857)^2+(23-27.857)^2......+(60-27.857)^2}{7}}= 13.217\\\sigma_C = \sqrt {\dfrac{(20-32.286)^2+(21-32.286)^2......+(99-32.286)^2}{7}}= 27.27[/tex]
The value of standard deviation of set C is the highest while set A is lowest.
Thus, the correct order which shows the standard deviations of the data sets A,B and C in order from greatest to least is C, B, A.
Learn more about the standard deviation here;
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