Let x _ i (1 ≤ i ≤ 10) be ten observation of a random variable X . If ∑ _ i = 1 ^ 10 ( x _ i - p) = 3 and ∑ _ i = 1 ^ 10 ( x _ i - p) ^ 2 = 9 where 0 ≠ p ∈ R , then the standard deviation of these observations is:

Correct answer: C. (9)/(10) Standard Deviation (S . D .) = ∑ _ i = 1 ^ 10 ( x _ i - p) ^ 2 10 - ( ∑ _ i = 1 ^ 10 ( x _ i - p) 10 ) ^ 2 = (9)/(10) - ((3)/(10)) ^ 2 = (9)/(10)

JEE Main 20203 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Statistics

JEE Main 2020 · 3 September · Evening Shift · Mathematics · Statistics

Let x _ i (1 ≤ i ≤ 10) be ten observation of a random variable X ....

Let

{x}_{i}

\left(1 \le i \le 10\right)

be ten observation of a random variable

. If

\mathop{∑}_{i = 1}^{10}\left({x}_{i} - p\right) = 3 and

\mathop{∑}_{i = 1}^{10}{\left({x}_{i} - p\right)}^{2} = 9

where

0 \ne p \in R

, then the standard deviation of these observations is:

Options

A $\dfrac{4}{5}$
B $\sqrt{\dfrac{3}{5}}$
C $\dfrac{9}{10}$
D $\dfrac{7}{10}$

Correct answer

C. $\dfrac{9}{10}$

Step-by-step solution

Standard Deviation $\left(S . D .\right) = \sqrt{\dfrac{\mathop{∑}_{i = 1}^{10}{\left({x}_{i} - p\right)}^{2}}{10} - {\left(\dfrac{\mathop{∑}_{i = 1}^{10}\left({x}_{i} - p\right)}{10}\right)}^{2}}$
$= \sqrt{\dfrac{9}{10} - {\left(\dfrac{3}{10}\right)}^{2}} = \dfrac{9}{10}$

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