JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let μ be the mean and s be the standard deviation of the distribution.

where $\sum f_i = 62$ . If $[x]$ denotes the greatest integer $\le x$ , then $[\mu^2 + \sigma^2]$ is equal to:

Choose the correct answer:

Explanation

Step 1: $k$ ki value nikalna:

Sari frequencies ka sum 62 hai:

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

3k^2 + 4k - 2 = 62 \implies 3k^2 + 4k - 64 = 0

Is quadratic equation ko solve karne par:

(3k + 16)(k - 4) = 0

Kyunki frequency negative nahi ho sakti, isliye $k = 4$ .

Step 2: Frequencies table:

$f_i$ values: $6, 8, 15, 15, 17, 1$ . (Total $\sum f_i = 62$ )

Step 3: Mean ( $\mu$ ) aur Variance ( $\sigma^2$ ) nikalna:

Formula: $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i}$

\sum f_i x_i^2 = (6 \cdot 0^2) + (8 \cdot 1^2) + (15 \cdot 2^2) + (15 \cdot 3^2) + (17 \cdot 4^2) + (1 \cdot 5^2)

\sum f_i x_i^2 = 0 + 8 + 60 + 135 + 272 + 25 = 500

\mu^2 + \sigma^2 = \frac{500}{62} \approx 8.064

Step 4: Greatest Integer Function:

[\mu^2 + \sigma^2] = [8.064] = 8

Choose the correct answer:

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