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JEE Main 20262 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2026 · 2 April · Evening Shift · Mathematics · Statistics

The mean and variance of n observations are 8 and 16 ,...

The mean and variance of nn observations are 88 and 1616, respectively. If the sum of the first (n−1)(n-1) observations is 4848 and the sum of squares of the first (n−1)(n-1) observations is 496496, then the value of nn is:

Options

  1. A2121
  2. B1616
  3. C1313
  4. D77

Correct answer

D. 77

Step-by-step solution

Let the nn-th observation be xnx_n.

Given mean xˉ=8\bar{x} = 8, we have:

∑i=1n−1xi+xnn=8\dfrac{\sum_{i=1}^{n-1} x_i + x_n}{n} = 8

48+xnn=8⇒xn=8n−48\dfrac{48 + x_n}{n} = 8 \Rightarrow x_n = 8n - 48

Given variance σ2=16\sigma^2 = 16, we have:

∑i=1n−1xi2+xn2n−(xˉ)2=16\dfrac{\sum_{i=1}^{n-1} x_i^2 + x_n^2}{n} - (\bar{x})^2 = 16

496+xn2n−64=16\dfrac{496 + x_n^2}{n} - 64 = 16

496+xn2n=80\dfrac{496 + x_n^2}{n} = 80

496+xn2=80n496 + x_n^2 = 80n

Substituting xn=8n−48x_n = 8n - 48:

496+(8n−48)2=80n496 + (8n - 48)^2 = 80n

496+64n2−768n+2304=80n496 + 64n^2 - 768n + 2304 = 80n

64n2−848n+2800=064n^2 - 848n + 2800 = 0

Dividing by 1616:

4n2−53n+175=04n^2 - 53n + 175 = 0

4n2−28n−25n+175=04n^2 - 28n - 25n + 175 = 0

4n(n−7)−25(n−7)=04n(n - 7) - 25(n - 7) = 0

(4n−25)(n−7)=0(4n - 25)(n - 7) = 0

Since nn must be an integer, n=7n = 7.

Answer: 77
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