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Statistics question

2023 · 11 Apr · Shift 1 · Q31
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Statistics question

2023 · 11 Apr · Shift 1 · Q31

JEE MainMathematicsStatisticsMCQ+4 / −1
Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A\mathrm{A}A and adding 2 to each element of B\mathrm{B}B. Then the sum of the mean and variance of the elements of C\mathrm{C}C is ‾\underline{\hspace{2cm}}​.
  1. A
    36
  2. B
    40
  3. C
    38
  4. D
    32
View written solutionFree

Correct answer: C

  1. Given data

For set AAA:

  • Number of elements =5=5=5
  • Mean μA=5\mu_A=5μA​=5
  • Variance σA2=12\sigma_A^2=12σA2​=12

For set BBB:

  • Number of elements =5=5=5
  • Mean μB=8\mu_B=8μB​=8
  • Variance σB2=20\sigma_B^2=20σB2​=20

A new set CCC is formed as follows:

  • From each element of AAA, subtract 333
  • To each element of BBB, add 222

So transformed subsets are:

  • A′={ai−3}A' = \{a_i-3\}A′={ai​−3}
  • B′={bi+2}B' = \{b_i+2\}B′={bi​+2}
  1. Find means after transformation

If a constant kkk is added to every element, the mean changes by kkk.

Hence,

μA′=μA−3=5−3=2\mu_{A'}=\mu_A-3=5-3=2μA′​=μA​−3=5−3=2 μB′=μB+2=8+2=10\mu_{B'}=\mu_B+2=8+2=10μB′​=μB​+2=8+2=10

Since set CCC has 10 elements, with 5 from A′A'A′ and 5 from B′B'B′, its mean is the weighted average:

μC=5⋅2+5⋅1010=10+5010=6\mu_C=\frac{5\cdot 2+5\cdot 10}{10} =\frac{10+50}{10}=6μC​=105⋅2+5⋅10​=1010+50​=6
  1. Find variances after transformation

Variance does not change when the same constant is added or subtracted from every element.

Thus,

σA′2=12,σB′2=20\sigma_{A'}^2=12, \qquad \sigma_{B'}^2=20σA′2​=12,σB′2​=20
  1. Find variance of combined set CCC

Use the combined variance formula:

σC2=n1(σ12+(μ1−μ)2)+n2(σ22+(μ2−μ)2)n1+n2\sigma_C^2=\frac{n_1\left(\sigma_1^2+(\mu_1-\mu)^2\right)+n_2\left(\sigma_2^2+(\mu_2-\mu)^2\right)}{n_1+n_2}σC2​=n1​+n2​n1​(σ12​+(μ1​−μ)2)+n2​(σ22​+(μ2​−μ)2)​

Here,

n1=5,n2=5,μ1=2,μ2=10,μ=6n_1=5,\quad n_2=5,\quad \mu_1=2,\quad \mu_2=10,\quad \mu=6n1​=5,n2​=5,μ1​=2,μ2​=10,μ=6

So,

(μ1−μ)2=(2−6)2=16(\mu_1-\mu)^2=(2-6)^2=16(μ1​−μ)2=(2−6)2=16 (μ2−μ)2=(10−6)2=16(\mu_2-\mu)^2=(10-6)^2=16(μ2​−μ)2=(10−6)2=16

Therefore,

σC2=5(12+16)+5(20+16)10\sigma_C^2=\frac{5(12+16)+5(20+16)}{10}σC2​=105(12+16)+5(20+16)​ =5⋅28+5⋅3610=140+18010=32010=32=\frac{5\cdot 28+5\cdot 36}{10} =\frac{140+180}{10} =\frac{320}{10}=32=105⋅28+5⋅36​=10140+180​=10320​=32
  1. Required sum of mean and variance
μC+σC2=6+32=38\mu_C+\sigma_C^2=6+32=38μC​+σC2​=6+32=38
  1. Check options
  • A: 363636
  • B: 404040
  • C: 383838 ✅
  • D: 323232

So the correct option is:

C (38)\boxed{\text{C }(38)}C (38)​
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