TO WHOMSOEVER CONCERN
In a series of 5 observations the mean
and variance are 4.4 and 8.24. r54 If three observations are 1, 2, 6 find other
two.
Data:- 1, 2, 6 Mean x' =4.4 Variance =8.24 N=5
Let the other two numbers be x and
y.
The
sum of all numbers = 1+2+6 =9
The numbers are 1, 2, 6, , x, y
There are 5 numbers N=5 Mean = 4.4
We know that x' = (sum of the
frequencies)/N
NX' =sum of the frequencies =22 22= 1+2+6+x+y=9+x+y
x+y = 22-9 =13
Variance =8.24 ϭ^2 = 8.24
Ϭ^2 = Variance =1/N(ƩXi-X’) ^2 Frequencies:
- 1, 2, 6, x, y
Xi -X' -3.4 -2.4 1.6 x-4.4 y-4.4
Ʃ(Xi-X')^2= 11.56 + 5.76 + 2.56 + (x4.4)^2+
(y-4.4)^2
Ʃ(Xi-X’) ^2 = 19.88 + (x-4.4)^2 + (y-4.4)^2
σ^2 = { 19.88
+ (x-4.4)^2 + (y-7-4.4)^2 } /5 =8.24
19.88
+ (x-4.4)^2 + (y-4.4)^2 =41.20
(x-4.4)^2 + (y-4.4)^2 = 21.32
x
+ y = 13 (i)
(x-4.4)^2 + (y-4.4)^2 = 21.32 (ii)
By substituting y = 13-x in (ii) (x-4.4)^2 + (8.6-x)^ =21.32
x^2 - 8.8x +19.36 + 73.96 - 17.2x
+ x^2 = 21.32
2x^2 - 26x + 93.32 = 21.32
2x^2 - 26x +72 = 0
x^2 -13x + 36 = 0
x^2 - 9x - 4x +36 = 0
x(x-9) -4(x-9) = 0
(x-9) (x-4) = 0
That is x=9 or 4
In a series of 5 observations the other
two numbers are 4 and 9.
Verification:
- X’ = Ʃx/N = (1+2+4+6+9/5 = 22/5 = 4.4