Answer:
Second set.
Step-by-step explanation:
I put the formula of standart deviation in the image below.
x is each data of the list
μ is the data average
N is the amount of elements.
So, for the first list we have:
μ= [tex]\frac{28995+37534+31361+27087+20966+37741}{6}= \frac{183684}{6}=30614[/tex]
DE= [tex]\sqrt{\frac{|28995-30614|^{2}+|37534-30614|^{2}+|31361-30614|^{2}+|27087-30614|^{2}+|20966-30614|^{2}+|37741-30614|^{2}}{6} }[/tex]
= [tex]\sqrt{\frac{2621161+47886400+558009+12439729+93083904+50794129}{6} }[/tex]
= [tex]\sqrt{\frac{207383332}{6} }[/tex]
= [tex]\sqrt{ 34563888.6667 }[/tex]
= 5879.1061.
For the other list of elements:
μ= [tex]\frac{72964+70763+101510+62161+66625+54339}{6}= \frac{428362}{6}=71393.66[/tex]
DE=[tex]\sqrt{\frac{|72964-71393.66|^{2}+|70763-71393.66|^{2}+|101510-71393.66|^{2}+|62161-71393.66|^{2}+|66625-71393.66|^{2}+|54339-71393.66|^{2}}{6} }[/tex]
= [tex]\sqrt{\frac{2465967.71+397732.03+906993934.99+85242010.67+22740118.19+290861427.71}{6} }[/tex]
= [tex]\sqrt{\frac{1308701191.3}{6} }[/tex]
= [tex]\sqrt{218116865.21 }[/tex]
= 6029.32
Then, the second set is more variable.
