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Statistics question

2021 · 17 Mar · Shift 2 · Q40
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Statistics question

2021 · 17 Mar · Shift 2 · Q40

JEE MainMathematicsStatisticsNumerical+4 / −1
Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 68

  1. Given data

A set has 3n3n3n numbers with variance 444.

  • Mean of first 2n2n2n numbers =6=6=6
  • Mean of remaining nnn numbers =3=3=3

We form a new set by:

  • adding 111 to each of the first 2n2n2n numbers
  • subtracting 111 from each of the remaining nnn numbers

We need the new variance kkk, then find 9k9k9k.

  1. Find the mean of the original set

Let the original overall mean be μ\muμ.

Using weighted mean,

μ=2n⋅6+n⋅33n=12n+3n3n=5\mu=\frac{2n\cdot 6+n\cdot 3}{3n} =\frac{12n+3n}{3n} =5μ=3n2n⋅6+n⋅3​=3n12n+3n​=5

So the original mean is

μ=5\mu=5μ=5
  1. Track the change made to each observation

Let each original number be transformed by adding a quantity ddd:

  • for first 2n2n2n numbers, d=+1d=+1d=+1
  • for last nnn numbers, d=−1d=-1d=−1

If original values are xix_ixi​, new values are

yi=xi+diy_i=x_i+d_iyi​=xi​+di​

We use the variance transformation formula:

Var⁡(Y)=Var⁡(X)+Var⁡(D)+2Cov⁡(X,D)\operatorname{Var}(Y)=\operatorname{Var}(X)+\operatorname{Var}(D)+2\operatorname{Cov}(X,D)Var(Y)=Var(X)+Var(D)+2Cov(X,D)

So we compute these terms one by one.

  1. Compute mean of the change variable DDD
\bar d=\frac{2n(1)+n(-1)}{3n}= rac{2n-n}{3n}=\frac13
  1. Compute Var⁡(D)\operatorname{Var}(D)Var(D)

Since DDD takes values 111 and −1-1−1,

E(D2)=1E(D^2)=1E(D2)=1

Hence

Var⁡(D)=E(D2)−[E(D)]2=1−(13)2=1−19=89\operatorname{Var}(D)=E(D^2)-[E(D)]^2=1-\left(\frac13\right)^2=1-\frac19=\frac89Var(D)=E(D2)−[E(D)]2=1−(31​)2=1−91​=98​
  1. Compute Cov⁡(X,D)\operatorname{Cov}(X,D)Cov(X,D)

We use

Cov⁡(X,D)=E(XD)−E(X)E(D)\operatorname{Cov}(X,D)=E(XD)-E(X)E(D)Cov(X,D)=E(XD)−E(X)E(D)

We already know:

E(X)=5,E(D)=13E(X)=5,\qquad E(D)=\frac13E(X)=5,E(D)=31​

Now calculate E(XD)E(XD)E(XD).

For the first 2n2n2n numbers, D=1D=1D=1, and their mean is 666. So contribution to ∑XD\sum XD∑XD is 2n⋅62n\cdot 62n⋅6.

For the remaining nnn numbers, D=−1D=-1D=−1, and their mean is 333. So contribution to ∑XD\sum XD∑XD is n⋅(3)(−1)=−3nn\cdot (3)(-1)=-3nn⋅(3)(−1)=−3n.

Thus,

E(XD)=2n⋅6+n⋅(−3)3n=12n−3n3n=3E(XD)=\frac{2n\cdot 6+n\cdot(-3)}{3n} =\frac{12n-3n}{3n} =3E(XD)=3n2n⋅6+n⋅(−3)​=3n12n−3n​=3

Therefore,

Cov⁡(X,D)=3−5⋅13=3−53=43\operatorname{Cov}(X,D)=3-5\cdot\frac13=3-\frac53=\frac43Cov(X,D)=3−5⋅31​=3−35​=34​
  1. Compute the new variance

Original variance is

Var⁡(X)=4\operatorname{Var}(X)=4Var(X)=4

So,

k=Var⁡(Y)=4+89+2⋅43k=\operatorname{Var}(Y)=4+\frac89+2\cdot\frac43k=Var(Y)=4+98​+2⋅34​ k=4+89+83k=4+\frac89+\frac83k=4+98​+38​

Take LCM 999:

k=369+89+249=689k=\frac{36}{9}+\frac89+\frac{24}{9}=\frac{68}{9}k=936​+98​+924​=968​

Hence,

9k=689k=689k=68
  1. Final answer
68\boxed{68}68​

The derived answer matches the stored correct answer.

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