If the variance of 1, 2, 3, 4, 5, ..., 10 is (99/12), then the standard deviation of 3, 6, 9, 12, ...,30 is

Question

If the variance of 1, 2, 3, 4, 5, ..., 10 is (99/12), then the standard deviation of 3, 6, 9, 12, ...,30 is (A)  (297/4)  (B)  (3/2)√33  (C)  (3/2)√99  (D)  √(99/12)

Tardigrade Admin · Accepted Answer

Given, σ 102=(99/12)=(33/4) ⇒ σ 10=(√33/2) SD of required series =3σ 10 =(3√33/2)

Q. If the variance of $1, 2, 3, 4, 5, ..., 10$ is $\frac{99}{12},$ then the standard deviation of $3, 6, 9, 12, ..., 30$ is

$\frac{297}{4}$

$\frac{3}{2} 33$

$\frac{3}{2} 99$

$\frac{99}{12}$

$\frac{3 3}{2}$

Given, $σ_{10}^{2} = \frac{99}{12} = \frac{33}{4}$
$\Rightarrow$ $σ_{10} = \frac{33}{2}$
SD of required series $= 3 σ_{10}$ $= \frac{3 33}{2}$

Q. If the variance of 1,2,3,4,5,...,10 is 1299​, then the standard deviation of 3,6,9,12,...,30 is

4297​

23​33​

23​99​

1299​​

233​​

Given, σ102​=1299​=433​ ⇒ σ10​=233​​ SD of required series =3σ10​ =2333​​