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Statistics question

2019 · 11 Jan · Shift 1 · Q27
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Statistics question

2019 · 11 Jan · Shift 1 · Q27

JEE MainMathematicsStatisticsMCQ+4 / −1
The outcome of each of 30 items was observed; 10 items gave an outcome 12{1 \over 2}21​– d each, 10 items gave outcome 12{1 \over 2}21​ each and the remaining 10 items gave outcome 12{1 \over 2}21​+ d each. If the variance of this outcome data is 43{4 \over 3}34​ then |d| equals :
  1. A
    23{2 \over 3}32​
  2. B
    52{{\sqrt 5 } \over 2}25​​
  3. C
    2{\sqrt 2 }2​
  4. D
    2
View written solutionFree

Correct answer: C

  1. Write the data values and frequencies

There are 303030 observations:

  • 101010 observations are 12−d\dfrac12-d21​−d
  • 101010 observations are 12\dfrac1221​
  • 101010 observations are 12+d\dfrac12+d21​+d

So the distribution is symmetric about 12\dfrac1221​.

  1. Find the mean

The mean is

xˉ=10(12−d)+10(12)+10(12+d)30.\bar x=\frac{10\left(\frac12-d\right)+10\left(\frac12\right)+10\left(\frac12+d\right)}{30}.xˉ=3010(21​−d)+10(21​)+10(21​+d)​.

Simplifying,

xˉ=10(32)30=12.\bar x=\frac{10\left(\frac32\right)}{30}=\frac12.xˉ=3010(23​)​=21​.

So the mean is

μ=12.\mu=\frac12.μ=21​.
  1. Compute the variance

Variance is

σ2=130∑(xi−μ)2.\sigma^2=\frac1{30}\sum (x_i-\mu)^2.σ2=301​∑(xi​−μ)2.

Now the deviations from the mean 12\frac1221​ are:

  • for 12−d\frac12-d21​−d: deviation =−d=-d=−d, square =d2=d^2=d2
  • for 12\frac1221​: deviation =0=0=0, square =0=0=0
  • for 12+d\frac12+d21​+d: deviation =d=d=d, square =d2=d^2=d2

Thus,

σ2=10d2+10⋅0+10d230=20d230=2d23.\sigma^2=\frac{10d^2+10\cdot 0+10d^2}{30} =\frac{20d^2}{30} =\frac{2d^2}{3}.σ2=3010d2+10⋅0+10d2​=3020d2​=32d2​.

Given variance is 43\dfrac4334​, so

2d23=43.\frac{2d^2}{3}=\frac43.32d2​=34​.

Multiply by 333:

2d2=4.2d^2=4.2d2=4.

Hence,

d2=2  ⟹  ∣d∣=2.d^2=2 \implies |d|=\sqrt2.d2=2⟹∣d∣=2​.
  1. Match with the options
∣d∣=2|d|=\sqrt2∣d∣=2​

which is Option C.

  1. Compare with stored correct answer

Stored correct answer: C

This matches our derived answer.

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