Answer:
C.) The standard deviation of Data Set 1 is about twice the standard deviation of Data Set 2.
Step-by-step explanation:
Data Set 1
[tex]\mu =\dfrac{1+2+3+3+6}{5}= \dfrac{15}{5}=3[/tex]
[tex]S$tandard Deviation, \sigma =\sqrt{\dfrac{\sum (x-\mu)^2}{N}}[/tex]
[tex]\sigma =\sqrt{\dfrac{(1-3)^2+(2-3)^2+(3-3)^2+(3-3)^2+(6-3)^2}{5}}\\=\sqrt{\dfrac{4+1+0+0+9}{5}}\\=\sqrt{\dfrac{14}{5}}\\\\=1.67[/tex]
Data Set 2
[tex]\mu =\dfrac{2+2+3+3+4+4}{6}= \dfrac{18}{6}=3[/tex]
[tex]\sigma =\sqrt{\dfrac{(2-3)^2+(2-3)^2+(3-3)^2+(3-3)^2+(4-3)^2+(4-3)^2}{6}}\\=\sqrt{\dfrac{1+1+0+0+1+1}{6}}\\=\sqrt{\dfrac{4}{5}}\\\\=0.89[/tex]
0.89 X 2=1.78 which is close to 1.67
Therefore, the standard deviation of Data Set 1 is about twice the standard deviation of Data Set 2.