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Example 17 - Mean of 5 observations is 4.4, variance is 8.24 - Indirect questions - finding remaining observations

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  1. Chapter 15 Class 11 Statistics
  2. Serial order wise
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Example 17 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. ๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘œ๐‘๐‘ ๐‘’๐‘Ÿ๐‘ฃ๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘ ๏ทฎ๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘œ๐‘๐‘ ๐‘’๐‘Ÿ๐‘ฃ๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘ ๏ทฏ = 4.4 1 + 2 + 6 + ๐‘ฅ + ๐‘ฆ๏ทฎ5๏ทฏ = 4.4 9 + x + y = 4.4 ร— 5 x + y = 22 โ€“ 9 x + y = 13 Also, Given Variance = 8.24 1๏ทฎn๏ทฏ ๏ทฎ๏ทฎ( ๐‘ฅ๏ทฎ๐‘–๏ทฏ๏ทฏโˆ’ ๐‘ฅ๏ทฏ)๏ทฎ2๏ทฏ = 8.24 1๏ทฎ5๏ทฏ ๏ทฎ๏ทฎ( ๐‘ฅ๏ทฎ๐‘–๏ทฏ๏ทฏโˆ’4.4)๏ทฎ2๏ทฏ = 8.24 1๏ทฎ5๏ทฏ [ (1 โ€“ 4.4)2 + (2 โ€“ 4.4)2 + (6 โ€“ 4.4)2 + (x โ€“ 4.4)2 + (y โ€“ 4.4)2 ] = 8.24 1๏ทฎ5๏ทฏ [ (โ€“3.4)2 + (โ€“2.4)2 + (1.6)2 + (x โ€“ 4.4)2 + (y โ€“ 4.4)2 ] = 8.24 1๏ทฎ5๏ทฏ [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 โ€“ 8.8x + y2 + 19.36 โ€“ 8.8y] = 8.24 ร— 5 [ 19.88 + 19.36 + 19.36 + x2 + y2 โ€“ 8.8y โ€“ 8.8x ] = 41.2 [ 58.6 + x2 + y2 โ€“ 8.8(x + y) ] = 41.2 [ 58.6 + x2 + y2 โ€“ 8.8(13) ] = 41.2 58.6 + x2 + y2 โ€“ 114.4 = 41.2 x2 + y2 = 114.4 + 41.2 โ€“ 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 โ€“ 97 2xy = 72 xy = 1๏ทฎ2๏ทฏ ร— 72 xy = 36 x = 36๏ทฎ๐‘ฆ๏ทฏ Putting (3) in (1) x + y = 13 36๏ทฎ๐‘ฆ๏ทฏ + y = 13 36 + y(y) = 13(y) 36 + y2 = 13y y2 โ€“ 13y + 36 = 0 y2 โ€“ 9y โ€“ 4y + 36 = 0 y(y โ€“ 9) โ€“ 4(y โ€“ 9) = 0 (y โ€“ 4)(y โ€“ 9) = 0 So, y = 4 & y = 9 For y = 4 x = 36๏ทฎ๐‘ฆ๏ทฏ = 36๏ทฎ4๏ทฏ = 9 Hence x = 9, y = 4 are the remaining two observations For y = 9 x = 36๏ทฎ๐‘ฆ๏ทฏ = 36๏ทฎ9๏ทฏ = 4 Hence, x = 4, y = 9 are the remaining two observations Thus, remaining observations are 4 & 9

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