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Statistics question

2020 · 3 Sep · Shift 2 · Q36
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Statistics question

2020 · 3 Sep · Shift 2 · Q36

JEE MainMathematicsStatisticsMCQ+4 / −1
Let xi (1 ≤\le≤ i ≤\le≤ 10) be ten observations of a random variable X. If ∑i=110(xi−p)=3\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3i=1∑10​(xi​−p)=3 and ∑i=110(xi−p)2=9\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9i=1∑10​(xi​−p)2=9 where 0 eee p ∈\in∈ R, then the standard deviation of these observations is :
  1. A
    710{7 \over {10}}107​
  2. B
    910{9 \over {10}}109​
  3. C
    45{4 \over 5}54​
  4. D
    35\sqrt {{3 \over 5}}53​​
View written solutionFree

Correct answer: B

  1. Let the mean of the 101010 observations be xˉ\bar xxˉ.

Given, ∑i=110(xi−p)=3\sum_{i=1}^{10}(x_i-p)=3∑i=110​(xi​−p)=3 So, ∑i=110xi−10p=3\sum_{i=1}^{10}x_i-10p=3∑i=110​xi​−10p=3 Hence, 10xˉ−10p=310\bar x-10p=310xˉ−10p=3 xˉ−p=310\bar x-p=\frac{3}{10}xˉ−p=103​

  1. Also given, ∑i=110(xi−p)2=9\sum_{i=1}^{10}(x_i-p)^2=9∑i=110​(xi​−p)2=9

Use the identity ∑i=1n(xi−a)2=∑i=1n(xi−xˉ)2+n(xˉ−a)2\sum_{i=1}^{n}(x_i-a)^2=\sum_{i=1}^{n}(x_i-\bar x)^2+n(\bar x-a)^2∑i=1n​(xi​−a)2=∑i=1n​(xi​−xˉ)2+n(xˉ−a)2 with n=10n=10n=10 and a=pa=pa=p.

Thus, 9=∑i=110(xi−xˉ)2+10(xˉ−p)29=\sum_{i=1}^{10}(x_i-\bar x)^2+10\left(\bar x-p\right)^29=∑i=110​(xi​−xˉ)2+10(xˉ−p)2 Substitute xˉ−p=310\bar x-p=\frac{3}{10}xˉ−p=103​: 9=∑i=110(xi−xˉ)2+10(310)29=\sum_{i=1}^{10}(x_i-\bar x)^2+10\left(\frac{3}{10}\right)^29=∑i=110​(xi​−xˉ)2+10(103​)2 9=∑i=110(xi−xˉ)2+10⋅91009=\sum_{i=1}^{10}(x_i-\bar x)^2+10\cdot \frac{9}{100}9=∑i=110​(xi​−xˉ)2+10⋅1009​ 9=∑i=110(xi−xˉ)2+9109=\sum_{i=1}^{10}(x_i-\bar x)^2+\frac{9}{10}9=∑i=110​(xi​−xˉ)2+109​ So, ∑i=110(xi−xˉ)2=9−910=8110\sum_{i=1}^{10}(x_i-\bar x)^2=9-\frac{9}{10}=\frac{81}{10}∑i=110​(xi​−xˉ)2=9−109​=1081​

  1. Variance of the observations is σ2=110∑i=110(xi−xˉ)2\sigma^2=\frac{1}{10}\sum_{i=1}^{10}(x_i-\bar x)^2σ2=101​∑i=110​(xi​−xˉ)2 σ2=110⋅8110=81100\sigma^2=\frac{1}{10}\cdot \frac{81}{10}=\frac{81}{100}σ2=101​⋅1081​=10081​

Therefore, standard deviation is σ=81100=910\sigma=\sqrt{\frac{81}{100}}=\frac{9}{10}σ=10081​​=109​

  1. Checking options:
  • A: 710\frac{7}{10}107​ ❌
  • B: 910\frac{9}{10}109​ ✅
  • C: 45\frac{4}{5}54​ ❌
  • D: 35\sqrt{\frac{3}{5}}53​​ ❌

Hence, the correct answer is B.

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