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In a group of data, there are n observations, `x,x_(2), ..., x_(n)." If "sum_(i=1)^(n)(x_(i)+1)^(2)=9n and sum_(i=1)^(n)(x_(i)-1)^(2)=5n`, the standard deviation of the data isA. 2B. `sqrt(7)`C. 5D. `sqrt(5)` |
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Answer» Correct Answer - D We have, `sum_(i=1)^(n)(x_(i)+1)^(2)=9n " ...(i)" ` ` and sum_(i=1)^(n)(x_(i)-1)^(2)=5n " ...(ii)" ` On substracting Eq. (ii) from Eq. (i) is, we get `rArr sum_(i=1)^(n){(x_(i)+1)^(2)-(x_(i)-1)^(2)}=4n` `rArr sum_(i=1)^(n)4x_(i) =4n rArr sum_(i=1)^(n)x_(i)=n rArr (sum_(i=1)^(n)x_(i))/(n)=1` ` therefore " mean " (bar(x))=1` Now, standard deviation `=sqrt((sum_(i=1)^(n)(x_(i)-bar(x))^(2))/(n))=sqrt((sum_(i=1)^(n)(x_(i)-1)^(2))/(n))` `=sqrt((5n)/(n))=sqrt(5)` |
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