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Statistics question

2021 · 26 Aug · Shift 2 · Q44
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Statistics question

2021 · 26 Aug · Shift 2 · Q44

JEE MainMathematicsStatisticsNumerical+4 / −1
Let the mean and variance of four numbers 3, 7, x and y(x > y) be 5 and 10 respectively. Then the mean of four numbers 3 + 2x, 7 + 2y, x + y and x −-− y is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 12

  1. Let the four numbers be 3,7,x,y3, 7, x, y3,7,x,y.

Given mean =5=5=5, so 3+7+x+y4=5\frac{3+7+x+y}{4}=543+7+x+y​=5 10+x+y4=5\frac{10+x+y}{4}=5410+x+y​=5 10+x+y=2010+x+y=2010+x+y=20 x+y=10x+y=10x+y=10

  1. Given variance =10=10=10.

Using variance formula for numbers aia_iai​ with mean aˉ\bar aaˉ: Variance=1n∑(ai−aˉ)2\text{Variance}=\frac{1}{n}\sum (a_i-\bar a)^2Variance=n1​∑(ai​−aˉ)2

So, (3−5)2+(7−5)2+(x−5)2+(y−5)24=10\frac{(3-5)^2+(7-5)^2+(x-5)^2+(y-5)^2}{4}=104(3−5)2+(7−5)2+(x−5)2+(y−5)2​=10 4+4+(x−5)2+(y−5)24=10\frac{4+4+(x-5)^2+(y-5)^2}{4}=1044+4+(x−5)2+(y−5)2​=10 8+(x−5)2+(y−5)2=408+(x-5)^2+(y-5)^2=408+(x−5)2+(y−5)2=40 (x−5)2+(y−5)2=32(x-5)^2+(y-5)^2=32(x−5)2+(y−5)2=32

Since x+y=10x+y=10x+y=10, let a=x−5,b=y−5a=x-5, \quad b=y-5a=x−5,b=y−5 Then a+b=(x+y)−10=0  ⟹  b=−aa+b=(x+y)-10=0 \implies b=-aa+b=(x+y)−10=0⟹b=−a Also, a2+b2=32a^2+b^2=32a2+b2=32 a2+a2=32a^2+a^2=32a2+a2=32 2a2=322a^2=322a2=32 a2=16a^2=16a2=16 a=±4a=\pm 4a=±4

Thus (x,y)=(9,1)(x,y)=(9,1)(x,y)=(9,1) or (1,9)(1,9)(1,9). Given x>yx>yx>y, we get x=9,y=1x=9,\quad y=1x=9,y=1

  1. Now find the mean of the four numbers: 3+2x,7+2y,x+y,x−y3+2x, \quad 7+2y, \quad x+y, \quad x-y3+2x,7+2y,x+y,x−y Substitute x=9,y=1x=9, y=1x=9,y=1: 3+2x=3+18=213+2x=3+18=213+2x=3+18=21 7+2y=7+2=97+2y=7+2=97+2y=7+2=9 x+y=10x+y=10x+y=10 x−y=8x-y=8x−y=8

Their mean is 21+9+10+84=484=12\frac{21+9+10+8}{4}=\frac{48}{4}=12421+9+10+8​=448​=12

  1. Therefore, the required mean is 12\boxed{12}12​

  2. Comparison with stored answer: Stored correct answer = 121212. This matches our derived answer.

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