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

NTA Abhyas 2020 · Mathematics · Statistics

If both the mean and the standard deviation of 50 observations x _...

If both the mean and the standard deviation of 5050 observations x1,x2,…,x50{x}_{1}, {x}_{2}, \ldots , {x}_{50} are equal to 1616, then the mean of ${\left({x}_{1} - 4\right)}^{2}, {\left({x}_{2} - 4\right)}^{2}, \ldots , {\left({x}_{50} - 4\right)}^{2} is

Options

  1. A525525
  2. B480480
  3. C400400
  4. D380380

Correct answer

C. 400400

Step-by-step solution

For observations x1,x2,...........x50{x}_{1}, {x}_{2},........... {x}_{50}
Mean, x−=∑xi50=16\overset{-}{x} = \dfrac{∑ {x}_i^{}}{50} = 16..... (i)
Variance, σ2=∑xi250−(x−)2=162{\sigma}^{2} = \dfrac{∑ {x}_i^{2}}{50} - {( \overset{-}{x} )}^{2} = {16}^{2}
⇒∑xi250=162+(x−)2=162+162=512⇒ \dfrac{∑ {x}_i^{2}}{50} = {16}^{2} + {( \overset{-}{x} )}^{2} = {16}^{2} + {16}^{2} = 512..... (ii)
So, the mean value of (x1−4)2,(x2−4)2,.......(x50−4)2{( {x}_{1} - 4 )}^{2}, {( {x}_{2} - 4 )}^{2},....... {( {x}_{50} - 4 )}^{2} will be
=∑(xi−4)250=∑xi2−8∑xi+16×5050= \dfrac{∑ {( {x}_{i} - 4 )}^{2}}{50} = \dfrac{∑ {x}_i^{2} - 8 ∑ {x}_{i} + 16 \times 50}{50}
=∑xi250−8∑xi50+16=512−8×16+16=400= \dfrac{∑ {x}_i^{2}}{50} - 8\dfrac{∑ {x}_{i}}{50} + 16 = 512 - 8 \times 16 + 16 = 400& (using (i) and (ii))
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