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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Statistics

NTA Abhyas 2020 · Mathematics · Statistics

The mean of n observations is ¯ X . If the first observation is...

The mean of nn observations is X¯.\overset{¯}{X}. If the first observation is increased by 12,{1}^{2}, second by 22{2}^{2} and so on, then the new mean is

Options

  1. AX¯+n(n+1)(2n+1)6\overset{¯}{X} + n\dfrac{\left(n + 1\right)\left(2n + 1\right)}{6}
  2. BX¯+(n+1)(2n+1)6\overset{¯}{X} + \dfrac{\left(n + 1\right)\left(2n + 1\right)}{6}
  3. CX¯+n+12\overset{¯}{X} + \dfrac{n + 1}{2}
  4. D\overset{¯}{X} + \dfrac{\left(n + 1\right)}{4}

Correct answer

B. X¯+(n+1)(2n+1)6\overset{¯}{X} + \dfrac{\left(n + 1\right)\left(2n + 1\right)}{6}

Step-by-step solution

Let,& x1,x2,.....xn{x}_{1}, {x}_{2}, . . . . . {x}_{n} be nn observations.
Then,& X¯=1nΣxi.<br/>Let,\overset{¯}{X} = \dfrac{1}{n}Σ{x}_{i} .<br />Let,{y}_{1} = {x}_{1} + {1}^{2}, {y}_{2} = {x}_{2} + {2}^{2},<br/><br />{y}_{3} = {x}_{3} + {3}^{2}, . . . . , {y}_{n} = {x}_{n} + n^{2}
Then, the meanofthenewseriesis<br/>1nmean of the new series is<br />\dfrac{1}{n}Σ{y}_{i} = 1nΣ(xi+i2)=1nΣxi+1n(12+22+32+....+n2)\dfrac{1}{n}Σ\left({x}_{i} + i^{2}\right) = \dfrac{1}{n}Σ{x}_{i} + \dfrac{1}{n}\left({1}^{2} + {2}^{2} + {3}^{2} + . . . . +n^{2}\right)
=X¯+1n= \overset{¯}{X} + \dfrac{1}{n}⋅ n(n+1)(2n+1)6=X¯+(n+1)(2n+1)6\dfrac{n \left(n + 1\right) \left(2n + 1\right)}{6} = \overset{¯}{X} + \dfrac{\left(n + 1\right) \left(2n + 1\right)}{6}
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