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JEE Mathematics 2023 Question with Solution

Let mm be the mean and σ\sigma be the standard deviation of the distribution:

A frequency distribution table with $$x_i$$ values $$0, 1, 2, 3, 4, 5$$ and corresponding $$f_i$$ values $$k+2, 2k, k^2-1, k^2-1, k^2+1, k-3$$.

where ∑fi=62\sum f_i = 62. If [x][x] denotes the greatest integer ≤x\leq x, then [μ2+σ2][\mu^2 + \sigma^2] is equal to:

  • A

    88

  • B

    77

  • C

    66

  • D

    99

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: ∑fi=62\sum f_i = 62 and the frequencies are k+2,2k,k2−1,k2−1,k2+1,k−3k+2, 2k, k^2-1, k^2-1, k^2+1, k-3.

Find: [μ2+σ2][\mu^2 + \sigma^2].

First use the total frequency:

(k+2)+2k+(k2−1)+(k2−1)+(k2+1)+(k−3)=62(k+2) + 2k + (k^2-1) + (k^2-1) + (k^2+1) + (k-3) = 62

So,

3k2+4k−2=623k^2 + 4k - 2 = 62 3k2+4k−64=03k^2 + 4k - 64 = 0

Solving, we get k=4k = 4, taking the positive integer solution.

Therefore the frequencies become 6,8,15,15,17,16, 8, 15, 15, 17, 1.

Now compute the mean:

μ=∑xifi∑fi=0(6)+1(8)+2(15)+3(15)+4(17)+5(1)62\mu = \frac{\sum x_i f_i}{\sum f_i} = \frac{0(6) + 1(8) + 2(15) + 3(15) + 4(17) + 5(1)}{62} μ=15662≈2.516\mu = \frac{156}{62} \approx 2.516

Next compute the variance:

σ2=∑fi(xi−μ)2∑fi≈1.733\sigma^2 = \frac{\sum f_i (x_i - \mu)^2}{\sum f_i} \approx 1.733

Hence,

μ2+σ2≈(2.516)2+1.733≈6.33+1.733≈8.063\mu^2 + \sigma^2 \approx (2.516)^2 + 1.733 \approx 6.33 + 1.733 \approx 8.063

Therefore,

[μ2+σ2]=[8.063]=8[\mu^2 + \sigma^2] = [8.063] = 8

So the correct option is A.

The solution labels option C, but the working gives the value 88, which matches option A.

Common mistakes

  • Using the option label written in the solution without checking the actual working is incorrect. The computation gives [μ2+σ2]=8[\mu^2 + \sigma^2] = 8, so you must match the derived value with the options, which gives A.

  • Adding the frequencies incorrectly while forming the equation in kk leads to a wrong value of kk. First simplify ∑fi\sum f_i carefully before solving the quadratic.

  • Confusing standard deviation σ\sigma with variance σ2\sigma^2 is incorrect. The expression asked is [μ2+σ2][\mu^2 + \sigma^2], so do not take the square root of the variance.

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