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Statistics question

2023 · 1 Feb · Shift 2 · Q24
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Statistics question

2023 · 1 Feb · Shift 2 · Q24

JEE MainMathematicsStatisticsMCQ+4 / −1
Let 9=x1<x2<…<x79=x_{1} \lt x_{2} \lt \ldots \lt x_{7}9=x1​<x2​<…<x7​ be in an A.P. with common difference d. If the standard deviation of x1,x2...,x7x_{1}, x_{2}..., x_{7}x1​,x2​...,x7​ is 4 and the mean is xˉ\bar{x}xˉ, then xˉ+x6\bar{x}+x_{6}xˉ+x6​ is equal to :
  1. A
    2(9+87)2\left(9+\frac{8}{\sqrt{7}}\right)2(9+7​8​)
  2. B
    25
  3. C
    18(1+13)18\left(1+\frac{1}{\sqrt{3}}\right)18(1+3​1​)
  4. D
    34
View written solutionFree

Correct answer: D

  1. Write the A.P. terms

Since x1=9x_1=9x1​=9 and there are 777 terms in A.P. with common difference ddd:

9,9+d,9+2d,9+3d,9+4d,9+5d,9+6d9, 9+d, 9+2d, 9+3d, 9+4d, 9+5d, 9+6d9,9+d,9+2d,9+3d,9+4d,9+5d,9+6d

So, x6=9+5dx_6=9+5dx6​=9+5d

  1. Find the mean

For an A.P. with odd number of terms, the mean is the middle term:

xˉ=x4=9+3d\bar{x}=x_4=9+3dxˉ=x4​=9+3d
  1. Use the standard deviation formula

Deviations from the mean are:

−3d,−2d,−d,0,d,2d,3d-3d,-2d,-d,0,d,2d,3d−3d,−2d,−d,0,d,2d,3d

Hence variance is

σ2=17[(−3d)2+(−2d)2+(−d)2+02+d2+(2d)2+(3d)2]\sigma^2=\frac{1}{7}\left[(-3d)^2+(-2d)^2+(-d)^2+0^2+d^2+(2d)^2+(3d)^2\right]σ2=71​[(−3d)2+(−2d)2+(−d)2+02+d2+(2d)2+(3d)2] =17(9d2+4d2+d2+0+d2+4d2+9d2)=\frac{1}{7}(9d^2+4d^2+d^2+0+d^2+4d^2+9d^2)=71​(9d2+4d2+d2+0+d2+4d2+9d2) =28d27=4d2=\frac{28d^2}{7}=4d^2=728d2​=4d2

Therefore standard deviation is

σ=2d\sigma=2dσ=2d

Given standard deviation is 444, so

2d=4  ⟹  d=22d=4 \implies d=22d=4⟹d=2
  1. Compute xˉ+x6\bar{x}+x_6xˉ+x6​
xˉ=9+3(2)=15\bar{x}=9+3(2)=15xˉ=9+3(2)=15 x6=9+5(2)=19x_6=9+5(2)=19x6​=9+5(2)=19

Thus,

xˉ+x6=15+19=34\bar{x}+x_6=15+19=34xˉ+x6​=15+19=34
  1. Check options
  • A: 2(9+87)2\left(9+\frac{8}{\sqrt{7}}\right)2(9+7​8​), not 343434
  • B: 252525, not correct
  • C: 18(1+13)18\left(1+\frac{1}{\sqrt{3}}\right)18(1+3​1​), not 343434
  • D: 343434, correct

Therefore, the correct answer is D.

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