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Misc 1 - Mean and variance of eight observations are 9, 9.25 - Indirect questions - finding remaining observations

Misc 1 - Chapter 15 Class 11 Statistics - Part 2
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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. 𝑆ð‘Ē𝑚 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟ð‘Ģð‘Žð‘Ąð‘–ð‘œð‘›ð‘ ï·Ū𝑁ð‘Ē𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑠𝑒𝑟ð‘Ģð‘Žð‘Ąð‘–ð‘œð‘›ð‘ ï·Ŋ = 9 6 + 7 + 10 + 12 + 12 + 13 + ð‘Ĩ + ð‘Ķï·Ū8ï·Ŋ = 9 60 + x + y = 9 × 8 x + y = 72 – 60 x + y = 12 Also, Given Variance = 9.25 1ï·Ūnï·Ŋ ï·Ūï·Ū( ð‘Ĩï·Ū𝑖ï·Ŋï·Ŋ− ð‘Ĩï·Ŋ)ï·Ū2ï·Ŋ = 9.25 1ï·Ū8ï·Ŋ ï·Ūï·Ū( ð‘Ĩï·Ū𝑖ï·Ŋï·Ŋ−9)ï·Ū2ï·Ŋ = 9.25 1ï·Ū8ï·Ŋ [(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 1ï·Ū8ï·Ŋ [ (–3)2 + (–2)2 + (1)2 + (3)2 + (3)2 + (4)2 + (x – 9)2 + (y – 9)2 ] = 9.25 1ï·Ū8ï·Ŋ [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 = 1ï·Ū2ï·Ŋ × 64 xy = 32 x = 32ï·Ūð‘Ķï·Ŋ Putting (3) in (1) x + y = 12 32ï·Ūð‘Ķï·Ŋ + 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 = 32ï·Ūð‘Ķï·Ŋ = 32ï·Ū4ï·Ŋ = 8 Hence x = 8, y = 4 are the remaining two observations For y = 8 x = 32ï·Ūð‘Ķï·Ŋ = 32ï·Ū8ï·Ŋ = 4 Hence, x = 4, y = 8 are the remaining two observations Thus, remaining observations are 4 & 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.