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JEE Mathematics 2024 Question with Solution

If the mean and variance of the data 6565, 6868, 5858, 4444, 4848, 4545, 6060, α\alpha, β\beta, 6060 where α>β\alpha > \beta are 5656 and 66.266.2 respectively, then α2+β2\alpha^2 + \beta^2 is equal to:

  • A

    63446344

  • B

    64506450

  • C

    63006300

  • D

    62006200

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The data are 65,68,58,44,48,45,60,α,β,6065, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60 with mean xˉ=56\bar{x} = 56 and variance σ2=66.2\sigma^2 = 66.2.

Find: α2+β2\alpha^2 + \beta^2.

Using the variance formula,

σ2=∑xi2n−(xˉ)2\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2

with n=10n = 10, σ2=66.2\sigma^2 = 66.2, and xˉ=56\bar{x} = 56.

So,

66.2=α2+β2+2567810−(56)266.2 = \frac{\alpha^2 + \beta^2 + 25678}{10} - (56)^2

Rearranging,

α2+β2+2567810=66.2+3136=3202.2\frac{\alpha^2 + \beta^2 + 25678}{10} = 66.2 + 3136 = 3202.2

Hence,

α2+β2+25678=32022\alpha^2 + \beta^2 + 25678 = 32022

Therefore,

α2+β2=6344\alpha^2 + \beta^2 = 6344

Therefore, the correct option is A.

Detailed Computation

Given: There are 1010 observations.

Find: α2+β2\alpha^2 + \beta^2.

First use the mean:

65+68+58+44+48+45+60+α+β+6010=56\frac{65 + 68 + 58 + 44 + 48 + 45 + 60 + \alpha + \beta + 60}{10} = 56

Simplifying,

(65+68+58+44+48+45+60+60)+α+β=560(65 + 68 + 58 + 44 + 48 + 45 + 60 + 60) + \alpha + \beta = 560 448+α+β=560448 + \alpha + \beta = 560 α+β=112\alpha + \beta = 112

Now use the variance formula:

66.2=∑xi210−56266.2 = \frac{\sum x_i^2}{10} - 56^2 ∑xi210=66.2+3136=3202.2\frac{\sum x_i^2}{10} = 66.2 + 3136 = 3202.2 ∑xi2=32022\sum x_i^2 = 32022

Compute the sum of squares of the known values:

652=4225,682=4624,582=3364,442=193665^2 = 4225, \quad 68^2 = 4624, \quad 58^2 = 3364, \quad 44^2 = 1936 482=2304,452=2025,602=360048^2 = 2304, \quad 45^2 = 2025, \quad 60^2 = 3600

Since 6060 occurs twice,

2×3600=72002 \times 3600 = 7200

Thus,

4225+4624+3364+1936+2304+2025+7200=256784225 + 4624 + 3364 + 1936 + 2304 + 2025 + 7200 = 25678

So,

25678+α2+β2=3202225678 + \alpha^2 + \beta^2 = 32022 α2+β2=32022−25678=6344\alpha^2 + \beta^2 = 32022 - 25678 = 6344

Therefore, α2+β2=6344\alpha^2 + \beta^2 = 6344, so the correct option is A.

Common mistakes

  • Using the variance formula incorrectly by taking σ2=∑xi2n\sigma^2 = \frac{\sum x_i^2}{n} only. This ignores subtraction of (xˉ)2(\bar{x})^2. Always use σ2=∑xi2n−(xˉ)2\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2.

  • Forgetting that 6060 appears twice in the data. This makes the sum of squares smaller than it should be. Include both terms, so the contribution is 2×6022 \times 60^2.

  • Adding the known observations incorrectly while forming the mean equation. If the sum 448448 is wrong, then the relation α+β=112\alpha + \beta = 112 becomes wrong. First total the known eight values carefully.

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