If i=1∑n(xi+1)2=9n and i=1∑n(xi−1)2=5n, then standard deviation of these 'n' observations (x1) is:

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If {_{{{i}={1}}}^{{n}}}{({x}_{{i}}+{1})}^{{2}}={9}{n} and {_{{

Question

If $i = 1 \sum n (x_{i} + 1)^{2} = 9 n$ and $i = 1 \sum n (x_{i} - 1)^{2} = 5 n$ , then standard deviation of these 'n' observations $(x_{1})$ is:

23

(3)

(2) ∣ ∣

(2)

3

(3)

5

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Theodore Discussed

i = 1 \sum n (x_{i} + 1)^{2} = 9 n

and

i = 1 \sum n (x_{i} - 1)^{2} = 5 n

, then standard deviation of these 'n' observations

(x_{1})

is:

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Question Text	If $i = 1 \sum n (x_{i} + 1)^{2} = 9 n$ and $i = 1 \sum n (x_{i} - 1)^{2} = 5 n$ , then standard deviation of these 'n' observations $(x_{1})$ is:
Answer Type	Video solution: 1
Upvotes	150