Last updated at Sept. 6, 2021 by Teachoo

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Misc 1 The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations. Let the other two observations be x and y. Therefore, our observations are 6, 7, 10, 12, 12 , 13, x, y. Given Mean = 9 i.e. đđĸđ đđ đđđ đđđŖđđĄđđđđ īˇŽđđĸđđđđ đđ đđđ đđđŖđđĄđđđđ īˇ¯ = 9 6 + 7 + 10 + 12 + 12 + 13 + đĨ + đĻīˇŽ8īˇ¯ = 9 60 + x + y = 9 Ã 8 x + y = 72 â 60 x + y = 12 Also, Given Variance = 9.25 1īˇŽnīˇ¯ īˇŽīˇŽ( đĨīˇŽđīˇ¯īˇ¯â đĨīˇ¯)īˇŽ2īˇ¯ = 9.25 1īˇŽ8īˇ¯ īˇŽīˇŽ( đĨīˇŽđīˇ¯īˇ¯â9)īˇŽ2īˇ¯ = 9.25 1īˇŽ8īˇ¯ [(6â9)2 +(7â9)2+(10â9)2+(12â9)2+(12â9)2+(13â9)2+(xâ9)2+(yâ9)2]=9.25 1īˇŽ8īˇ¯ [ (â3)2 + (â2)2 + (1)2 + (3)2 + (3)2 + (4)2 + (x â 9)2 + (y â 9)2 ] = 9.25 1īˇŽ8īˇ¯ [9 + 4 + 1 + 9 + 9 + 16 + x2 + (9)2 - 2(9)x + y2 + (9)2 - 2(9)y] = 9.25 [ 48 + x2 + 81 â 18x + y2 + 81 â 18y] = 9.25 Ã 8 [ 210 + x2 + y2 â 18y â 18x ] = 74 [ 210 + x2 + y2 â 18(x + y) ] = 74 [ 210 + x2 + y2 â 18(12) ] = 74 210 + x2 + y2 â 216 = 74 x2 + y2 = 74 â 210 + 216 x2 + y2 = 80 From (1) x + y = 12 Squaring both sides (x + y)2 = 122 x2 + y2 + 2xy = 144 80 + 2xy = 144 2xy = 144 â 80 2xy = 64 xy = 1īˇŽ2īˇ¯ Ã 64 xy = 32 x = 32īˇŽđĻīˇ¯ Putting (3) in (1) x + y = 12 32īˇŽđĻīˇ¯ + y = 12 32 + y2 = 12y y2 â 12y + 32 = 0 y2 â 8y â 4y + 32 = 0 y(y â 8) â 4(y â 8) = 0 (y â 4)(y â 8) = 0 So, y = 4 & y = 8 For y = 4 x = 32īˇŽđĻīˇ¯ = 32īˇŽ4īˇ¯ = 8 Hence x = 8, y = 4 are the remaining two observations For y = 8 x = 32īˇŽđĻīˇ¯ = 32īˇŽ8īˇ¯ = 4 Hence, x = 4, y = 8 are the remaining two observations Thus, remaining observations are 4 & 8

Chapter 15 Class 11 Statistics (Term 1)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.