If x1,x2,......x18 are observations such, that ∑ limitsj=118(xj-8)=9 and ∑ limitsj=118(xj-8)2=45, then the standard deviation of these observations is

Question

If x1,x2,......x18 are observations such, that ∑ limitsj=118(xj-8)=9 and ∑ limitsj=118(xj-8)2=45, then the standard deviation of these observations is (A)  √(81/34)  (B)  5  (C)  √5  (D)  (3/2)

Tardigrade Admin · Accepted Answer

Standard deviation =√( undersetj=1 mathop overset18 mathop Sigma (xj-8)2/n)-( ( undersetj=1 mathop overset18 mathop Sigma (xk-8)/n) )2 =√(45/18)-( (9/18) )2 =√(45/18)-(1/4)=√(81/36)=(9/6)=(3/2)

Q. If $x_{1}, x_{2}, ...... x_{18}$ are observations such, that $j = 1 \sum 18 (x_{j} - 8) = 9$ and $j = 1 \sum 18 (x_{j} - 8)^{2} = 45,$ then the standard deviation of these observations is

$\frac{81}{34}$

$5$

$5$

$\frac{3}{2}$

Standard deviation $= \frac{j = 1 Σ 18 ( x _{j} - 8 ) ^{2}}{n} - ⎝ ⎛ \frac{j = 1 Σ 18 ( x _{k} - 8 )}{n} ⎠ ⎞^{2}$
$= \frac{45}{18} - (\frac{9}{18})^{2}$
$= \frac{45}{18} - \frac{1}{4} = \frac{81}{36} = \frac{9}{6} = \frac{3}{2}$

Q. If x1​,x2​,......x18​ are observations such, that j=1∑18​(xj​−8)=9 and j=1∑18​(xj​−8)2=45, then the standard deviation of these observations is

3481​​

5

5​

23​

Standard deviation =nj=1Σ18​(xj​−8)2​−⎝⎛​nj=1Σ18​(xk​−8)​⎠⎞​2​ =1845​−(189​)2​ =1845​−41​​=3681​​=69​=23​

Q. If $x_{1}, x_{2}, ...... x_{18}$ are observations such, that $j = 1 \sum 18 (x_{j} - 8) = 9$ and $j = 1 \sum 18 (x_{j} - 8)^{2} = 45,$ then the standard deviation of these observations is

$\frac{81}{34}$

$5$

$5$

$\frac{3}{2}$

Standard deviation $= \frac{j = 1 Σ 18 ( x _{j} - 8 ) ^{2}}{n} - ⎝ ⎛ \frac{j = 1 Σ 18 ( x _{k} - 8 )}{n} ⎠ ⎞^{2}$
$= \frac{45}{18} - (\frac{9}{18})^{2}$
$= \frac{45}{18} - \frac{1}{4} = \frac{81}{36} = \frac{9}{6} = \frac{3}{2}$