If underset i=1 oversetn Sigma(xi-a)=n and underset i=1 oversetn Sigma(xi-a)2=n a,(n, a1) then the standard deviation of n observations x 1, x 2, ldots ., x n is

Question

If underset i=1 oversetn Sigma(xi-a)=n and underset i=1 oversetn Sigma(xi-a)2=n a,(n, a>1) then the standard deviation of n observations x 1, x 2, ldots ., x n is (A) n √ a -1 (B) √a-1 (C) a-1 (D) √ n ( a -1)

Tardigrade Admin · Accepted Answer

S.D =√( displaystyle∑i=1n(xi-a)/n)-(( displaystyle∑i=1n(xi-a)/n))2 =√(n a/n)-((n/n))2 textGiven displaystyle∑i=1n(xi-a)=n displaystyle∑i=1n(xi-a)2=n a =√a-1

Q. If $i = 1 Σ n (x_{i} - a) = n$ and $i = 1 Σ n (x_{i} - a)^{2} = na, (n, a > 1)$ then the standard deviation of n observations $x_{1}, x_{2}, \dots ., x_{n}$ is

$n a - 1$

$a - 1$

$a - 1$

$n (a - 1)$

$S . D = \frac{i = 1 \sum n ( x _{i} - a )}{n} - ⎝ ⎛ \frac{i = 1 \sum n ( x _{i} - a )}{n} ⎠ ⎞^{2}$
$= \frac{na}{n} - (\frac{n}{n})^{2}$
${Given i = 1 \sum n (x_{i} - a) = n i = 1 \sum n (x_{i} - a)^{2} = na}$
$= a - 1$

Q. If i=1Σn​(xi​−a)=n and i=1Σn​(xi​−a)2=na,(n,a>1) then the standard deviation of n observations x1​,x2​,….,xn​ is

na−1​

a−1​

a−1

n(a−1)​

S.D=ni=1∑n​(xi​−a)​−⎝⎛​ni=1∑n​(xi​−a)​⎠⎞​2​ =nna​−(nn​)2​ {Giveni=1∑n​(xi​−a)=ni=1∑n​(xi​−a)2=na} =a−1​

Q. If $i = 1 Σ n (x_{i} - a) = n$ and $i = 1 Σ n (x_{i} - a)^{2} = na, (n, a > 1)$ then the standard deviation of n observations $x_{1}, x_{2}, \dots ., x_{n}$ is

$n a - 1$

$a - 1$

$a - 1$

$n (a - 1)$

$S . D = \frac{i = 1 \sum n ( x _{i} - a )}{n} - ⎝ ⎛ \frac{i = 1 \sum n ( x _{i} - a )}{n} ⎠ ⎞^{2}$
$= \frac{na}{n} - (\frac{n}{n})^{2}$
${Given i = 1 \sum n (x_{i} - a) = n i = 1 \sum n (x_{i} - a)^{2} = na}$
$= a - 1$