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

NTA Abhyas 2020 · Mathematics · Statistics

The mean and variance of 20 observations are found to be 10 and 4...

The mean and variance of 2020 observations are found to be 1010 and 44 respectively. On rechecking, it was found that an observation 88 is incorrect. If the wrong observation is omitted, then the correct variance is

Options

  1. A77
  2. B10019\dfrac{100}{19}
  3. C1400361\dfrac{1400}{361}
  4. D1440361\dfrac{1440}{361}

Correct answer

D. 1440361\dfrac{1440}{361}

Step-by-step solution

We have n=20,x¯old=10,Varold=4n = 20, {\overset{¯}{x}}_{old} = 10, Va{r}_{old} = 4
⇒x¯=Σxi20⇒Σxi=200⇒ \overset{¯}{x} = \dfrac{Σ{x}_{i}}{20} ⇒ Σ{x}_{i} = 200
⇒(Σxi)new=192,X¯new=1929⇒ {\left(Σ{x}_{i}\right)}_{new} = 192,{\overset{¯}{X}}_{new} = \dfrac{192}{9}
Var(old)=Σxi220−(x¯old)2Va{r}_{\left(old\right)} = \dfrac{Σ{x}_i^{2}}{20} - {\left({\overset{¯}{x}}_{old}\right)}^{2}
4=Σxi220−1004 = \dfrac{Σ{x}_i^{2}}{20} - 100
Σxi2(old)=2080Σ{{x}_i^{2}}_{\left(old\right)} = 2080
Σxi2new=2080−64=2016Σ{{x}_i^{2}}_{new} = 2080 - 64 = 2016
Var(new)=201619−(19219)2=1440361Va{r}_{\left(new\right)} = \dfrac{2016}{19} - {\left(\dfrac{192}{19}\right)}^{2} = \dfrac{1440}{361}
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