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Statistics question

2020 · 6 Sep · Shift 1 · Q31
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  5. /2020 · 6 Sep · Shift 1 · Q31

Statistics question

2020 · 6 Sep · Shift 1 · Q31

JEE MainMathematicsStatisticsMCQ+4 / −1
If ∑i=1n(xi−a)=n\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = ni=1∑n​(xi​−a)=n and ∑i=1n(xi−a)2=na\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = nai=1∑n​(xi​−a)2=na (n, a > 1) then the standard deviation of n observations x1 , x2 , ..., xn is :
  1. A
    aaa – 1
  2. B
    na−1n\sqrt {a - 1}na−1​
  3. C
    n(a−1)\sqrt {n\left( {a - 1} \right)}n(a−1)​
  4. D
    a−1\sqrt {a - 1}a−1​
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