Answer:
B)1,1,1,4
Step-by-step explanation:
A.1,2,3,4
Mean=[tex]\bar x=\frac{1+2+3+4}{4}=2.5[/tex]
x [tex]x-\bar x[/tex] [tex](x-\bar x)^2[/tex]
1 -1.5 2.25
2 -0.5 0.25
3 0.5 0.25
4 1.5 2.25
[tex]\sum(x-\bar x)^2=2.25+0.25+0.25+2.25=5[/tex]
B.
Mean=[tex]\bar x=\frac{1+1+1+4}{4}=1.75[/tex]
[tex](x-\bar x)^2[/tex]
0.5625
0.5625
0.5625
5.0625
[tex]\sum(x-\bar x)^2=0.5626+0.5625+0.5625+5.0625=6.75[/tex]
C.
Mean=[tex]\bar x=\frac{1+2+2+4}{4}=2.25[/tex]
[tex](x-\bar x)^2[/tex]
1.5625
0.0625
0.0625
3.0625
[tex]\sum (x-\bar x)=1.5625+0.0625+0.0625+3.0625=4.75[/tex]
D.[tex]Mean=\bar x=\frac{4+4+4+4}{4}=4[/tex]
[tex](x-\bar x)^2[/tex]
0
0
0
0
[tex]\sum (x-\bar x)^2=0+0+0+0=0[/tex]
We know that
S.D is directly proportional to [tex]\sum (x-\bar x)^2[/tex].
When [tex]\sum (x-\bar x)^2[/tex] is highest then the S.D is also highest.
We can see that the value of [tex]\sum (x-\bar x)^2[/tex] is highest in option B.
Therefore, S.D of the date set of option B is highest.
