Miscellaneous

Chapter 13 Class 11 Statistics
Serial order wise

### Transcript

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. 𝑆𝑢𝑚 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠﷮𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛𝑠﷯ = 9 6 + 7 + 10 + 12 + 12 + 13 + 𝑥 + 𝑦﷮8﷯ = 9 60 + x + y = 9 × 8 x + y = 72 – 60 x + y = 12 Also, Given Variance = 9.25 1﷮n﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯− 𝑥﷯)﷮2﷯ = 9.25 1﷮8﷯ ﷮﷮( 𝑥﷮𝑖﷯﷯−9)﷮2﷯ = 9.25 1﷮8﷯ [(6–9)2 +(7–9)2+(10–9)2+(12–9)2+(12–9)2+(13–9)2+(x–9)2+(y–9)2]=9.25 1﷮8﷯ [ (–3)2 + (–2)2 + (1)2 + (3)2 + (3)2 + (4)2 + (x – 9)2 + (y – 9)2 ] = 9.25 1﷮8﷯ [9 + 4 + 1 + 9 + 9 + 16 + x2 + (9)2 - 2(9)x + y2 + (9)2 - 2(9)y] = 9.25 [ 48 + x2 + 81 – 18x + y2 + 81 – 18y] = 9.25 × 8 [ 210 + x2 + y2 – 18y – 18x ] = 74 [ 210 + x2 + y2 – 18(x + y) ] = 74 [ 210 + x2 + y2 – 18(12) ] = 74 210 + x2 + y2 – 216 = 74 x2 + y2 = 74 – 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 – 80 2xy = 64 xy = 1﷮2﷯ × 64 xy = 32 x = 32﷮𝑦﷯ Putting (3) in (1) x + y = 12 32﷮𝑦﷯ + y = 12 32 + y2 = 12y y2 – 12y + 32 = 0 y2 – 8y – 4y + 32 = 0 y(y – 8) – 4(y – 8) = 0 (y – 4)(y – 8) = 0 So, y = 4 & y = 8 For y = 4 x = 32﷮𝑦﷯ = 32﷮4﷯ = 8 Hence x = 8, y = 4 are the remaining two observations For y = 8 x = 32﷮𝑦﷯ = 32﷮8﷯ = 4 Hence, x = 4, y = 8 are the remaining two observations Thus, remaining observations are 4 & 8

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.