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Misc 2 - Mean, variance of 7 observations are 8, 16. If five - Miscellaneous

Misc 2 - Chapter 15 Class 11 Statistics - Part 2
Misc 2 - Chapter 15 Class 11 Statistics - Part 3 Misc 2 - Chapter 15 Class 11 Statistics - Part 4 Misc 2 - Chapter 15 Class 11 Statistics - Part 5


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Misc 2 The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations. Let the other two observations be x and y. Therefore, our observations are 2, 4, 10, 12, 14, x, y. Given Mean = 8 i.e. 𝑆𝑢𝑚 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠﷮𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠﷯ = 8 2 + 4 + 10 + 12 + 14 + 𝑥 + 𝑦﷮7﷯ = 8 42 + x + y = 7 × 8 x + y = 56 – 42 x + y = 14 Also, Given Variance = 16 1﷮n﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ = 16 1﷮7﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯−8)﷮2﷯ = 16 1﷮7﷯ [(2 – 8)2 +(4 – 8)2+(10 – 8)2+(12 – 8)2 +(14 – 8)2+(x – 8)2 +(y – 8)2] = 16 1﷮7﷯ [ (–6)2 + (–4)2 + (2)2 + (4)2 + (6)2 + (x – 8)2 + (y – 8)2 ] = 16 1﷮7﷯ [36 + 16 + 4 + 16 + 36 + x2 + (8)2 - 2(8)x + y2 + (8)2 - 2(8)y] = 16 [ 108 + x2 + 64 – 16x + y2 + 64 – 16y] = 16 × 7 [ 236 + x2 + y2 – 16y – 16x ] = 112 [ 236 + x2 + y2 – 16(x + y) ] = 112 [ 236 + x2 + y2 – 16(14) ] = 112 236 + x2 + y2 – 224 = 112 x2 + y2 = 112 – 236 + 224 x2 + y2 = 100 From (1) x + y = 14 Squaring both sides (x + y)2 = 142 x2 + y2 + 2xy = 196 100 + 2xy = 196 2xy = 196 – 100 2xy = 96 xy = 1﷮2﷯ × 96 xy = 48 x = 48﷮𝑦﷯ Putting (3) in (1) x + y = 14 48﷮𝑦﷯ + y = 14 48 + y2 = 14y y2 – 14y + 48 = 0 y2 – 6y – 8y + 48 = 0 y(y – 6) – 8(y – 6) = 0 (y – 6)(y – 8) = 0 So, y = 6 & y = 8 For y = 6 x = 48﷮𝑦﷯ = 48﷮6﷯ = 8 Hence x = 8, y = 6 are the remaining two observations For y = 8 x = 48﷮𝑦﷯ = 48﷮8﷯ = 6 Hence, x = 6, y = 8 are the remaining two observations Thus, remaining observations are 6 & 8

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.