Examples

Chapter 13 Class 11 Statistics
Serial order wise

### Transcript

Example 14 The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations. Let the other two observations be x and y. Therefore, our observations are 1, 2, 6, x, y. Given Mean = 4.4 i.e. ððĒð ðð ððð ðððĢððĄðððð ï·ŪððĒðððð ðð ððð ðððĢððĄðððð ï·Ŋ = 4.4 1 + 2 + 6 + ðĨ + ðĶï·Ū5ï·Ŋ = 4.4 9 + x + y = 4.4 Ã 5 x + y = 22 â 9 x + y = 13 Also, Given Variance = 8.24 1ï·Ūnï·Ŋ ï·Ūï·Ū( ðĨï·Ūðï·Ŋï·Ŋâ ðĨï·Ŋ)ï·Ū2ï·Ŋ = 8.24 1ï·Ū5ï·Ŋ ï·Ūï·Ū( ðĨï·Ūðï·Ŋï·Ŋâ4.4)ï·Ū2ï·Ŋ = 8.24 1ï·Ū5ï·Ŋ [ (1 â 4.4)2 + (2 â 4.4)2 + (6 â 4.4)2 + (x â 4.4)2 + (y â 4.4)2 ] = 8.24 1ï·Ū5ï·Ŋ [ (â3.4)2 + (â2.4)2 + (1.6)2 + (x â 4.4)2 + (y â 4.4)2 ] = 8.24 1ï·Ū5ï·Ŋ [11.56 + 5.76 + 2.56 + x2 + (4.4)2 - 2(4.4)x + y2 + (4.4)2 - 2(4.4)y] = 8.24 [ 19.88 + x2 + 19.36 â 8.8x + y2 + 19.36 â 8.8y] = 8.24 Ã 5 [ 19.88 + 19.36 + 19.36 + x2 + y2 â 8.8y â 8.8x ] = 41.2 [ 58.6 + x2 + y2 â 8.8(x + y) ] = 41.2 [ 58.6 + x2 + y2 â 8.8(13) ] = 41.2 58.6 + x2 + y2 â 114.4 = 41.2 x2 + y2 = 114.4 + 41.2 â 58.6 x2 + y2 = 97 From (1) x + y = 13 Squaring both sides (x + y)2 = 132 x2 + y2 + 2xy = 169 97 + 2xy = 169 2xy = 169 â 97 2xy = 72 xy = 1ï·Ū2ï·Ŋ Ã 72 xy = 36 x = 36ï·ŪðĶï·Ŋ Putting (3) in (1) x + y = 13 36ï·ŪðĶï·Ŋ + y = 13 36 + y(y) = 13(y) 36 + y2 = 13y y2 â 13y + 36 = 0 y2 â 9y â 4y + 36 = 0 y(y â 9) â 4(y â 9) = 0 (y â 4)(y â 9) = 0 So, y = 4 & y = 9 For y = 4 x = 36ï·ŪðĶï·Ŋ = 36ï·Ū4ï·Ŋ = 9 Hence x = 9, y = 4 are the remaining two observations For y = 9 x = 36ï·ŪðĶï·Ŋ = 36ï·Ū9ï·Ŋ = 4 Hence, x = 4, y = 9 are the remaining two observations Thus, remaining observations are 4 & 9