急! Measures of Dispersion 2條

25a)

(a+b+c+d+e)/5=7
a+b+c+d+e=35...(1)

(a+b+c+d)/4=7
a+b+c+d=28...(2)

Sub. (2) into (1),
28+e=35
e=7

25b)

σ of a,b,c,d,e = 2
√{[(7-a)^2+(7-b)^2+(7-c)^2+(7-d)^2+(7-e)^2]/5}=2
[(7-a)^2+(7-b)^2+(7-c)^2+(7-d)^2+(7-7)^2]/5=4
[(7-a)^2+(7-b)^2+(7-c)^2+(7-d)^2+0]=20
[(7-a)^2+(7-b)^2+(7-c)^2+(7-d)^2]/4=5
√{[(7-a)^2+(7-b)^2+(7-c)^2+(7-d)^2]/4}=√5
σ of a,b,c,d =√5

26)

(7+9+11+14+15+15+x+y)/8=10
x+y+71=80
x+y=9...(1)

√{[(10-7)^2+(10-9)^2+(10-11)^2+(10-14)^2+2(10-15)^2+(10-x)^2+(10-y)^2]/8}
=(√89)/2
[3^2+1^2+(-1)^2+(-4)^2+2(-5)^2+(10-x)^2+(10-y)^2]/8=89/4
(10-x)^2+(10-y)^2+77=178
(10-x)^2+(10-y)^2=101
100-20x+x^2+100-20y+y^2=101
-20(x+y)+x^2+y^2=-99
-20(9)+x^2+y^2=-99
x^2+y^2=81...(2)

From (1), x=9-y...(3)

Sub. (3) into (2),
(9-y)^2+y^2=81
81-18y+y^2+y^2=81
2y^2-18y=0
y^2-9y=0
y(y-9)=0
y=0 or y=9

When y=0,
x=9-0 = 9

When y=9
x=9-9=0 (rejected)

Therefore, x=9 and y=0

2013-08-05 23:04:01 補充：
Some steps can be omitted.
I show all the steps in order to make it understandable.

我的回答名額用盡了，請看這個：
http://upload.lsforum.net/users/public/t3957120130806tempz88.png