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

NTA Abhyas 2020 · Mathematics · Statistics

The mean square deviation of a set of observations& x _ 1 , x _ 2 ....

The mean square deviation of a set of observations& x1,x2......xn{x}_{1}, {x}_{2} . . . . . . {x}_{n} about a point mm is defined as 1n∑i=1n(xi−m)2\dfrac{1}{n}\mathop{∑}_{i = 1}^{n}\left({x}_{i} - m\right)2. If the mean square deviations about –1– 1 and 11 of a set of observations are 77 and 33 respectively. The standard deviation of those observations is

Options

  1. A2\sqrt{2}
  2. B22
  3. C55
  4. D3\sqrt{3}

Correct answer

D. 3\sqrt{3}

Step-by-step solution

∑i=1n(xi+1)2n=7\dfrac{\mathop{∑}_{i = 1}^{n}{\left({x}_{i} + 1\right)}^{2}}{n} = 7(given)
⇒∑xi2+2∑xi+n=7n⇒ ∑ {x}_i^{2} + 2 ∑ {x}_{i} + n = 7n
⇒∑xi2+2∑xi=6n⇒ ∑ {x}_i^{2} + 2 ∑ {x}_{i} = 6n … (1)
Also, ∑(xi−1)2n=2\dfrac{∑ {\left({x}_{i} - 1\right)}^{2}}{n} = 2 (given)
⇒∑xi2−2∑xi+n=3n⇒ ∑ {x}_i^{2} - 2 ∑ {x}_{i} + n = 3n
⇒∑xi2−2∑xi=2n⇒ ∑ {x}_i^{2} - 2 ∑ {x}_{i} = 2n … (2)
From (1)( 1 ) and (2)( 2 )
∑xi2=4n,∑xi=n∑ {x}_i^{2} = 4n, ∑ {x}_{i} = n
Standard deviation=∑xi2n−(∑xin)2= \sqrt{\dfrac{∑ {x}_i^{2}}{n} - \left(\dfrac{∑ xi}{n}\right)2}
=4nn−1=3= \sqrt{\dfrac{4n}{n} - 1} = \sqrt{3}
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