Let -12 be the variance of the dataset 1-4-7-10- - 301 and -22 be the variance of the dataset 7-13-19-25- - 607- Then the value of -22-12-12

S_1={1,4,7,10,…,301} S_2={7,13,19,25,…,607} Let n^th term of S_1 is denoted by T_n. Then nbsp;n^th term of S_2 will be denoted as 2T_n+5. We know Var(aX+b)=a ...

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Standard XI

Mathematics

Standard Deviation

Let σ12 be th...

Question

Let $σ_{1}^{2}$ be the variance of the dataset ${1, 4, 7, 10, \dots, 301}$ and $σ_{2}^{2}$ be the variance of the dataset ${7, 13, 19, 25, \dots, 607}$ . Then the value of $\frac{σ_{2}^{2} - σ_{1}^{2}}{σ_{1}^{2}}$ is

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Solution

$S_{1} = {1, 4, 7, 10, \dots, 301}$
$S_{2} = {7, 13, 19, 25, \dots, 607}$
Let $n^{th}$ term of $S_{1}$ is denoted by $T_{n}$ .
Then $n^{th}$ term of $S_{2}$ will be denoted as $2 T_{n} + 5$ .

We know $V a r (a X + b) = a^{2} V a r (X)$
So, $σ_{2}^{2} = 2^{2} σ_{1}^{2} = 4 σ_{1}^{2}$
$\frac{σ_{2}^{2} - σ_{1}^{2}}{σ_{1}^{2}} = \frac{σ_{2}^{2}}{σ_{1}^{2}} - 1$
$= \frac{4 σ_{1}^{2}}{σ_{1}^{2}} - 1$
$= 4 - 1$
$= 3$

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Let σ21 be the variance of the dataset {1,4,7,10,…,301} and σ22 be the variance of the dataset {7,13,19,25,…,607}. Then the value of σ22−σ21σ21 is

S1={1,4,7,10,…,301} S2={7,13,19,25,…,607} Let nth term of S1 is denoted by Tn. Then nth term of S2 will be denoted as 2Tn+5. We know Var(aX+b)=a2Var(X) So, σ22=22σ21=4σ21 σ22−σ21σ21=σ22σ21−1 =4σ21σ21−1 =4−1 =3

Let $σ_{1}^{2}$ be the variance of the dataset ${1, 4, 7, 10, \dots, 301}$ and $σ_{2}^{2}$ be the variance of the dataset ${7, 13, 19, 25, \dots, 607}$ . Then the value of $\frac{σ_{2}^{2} - σ_{1}^{2}}{σ_{1}^{2}}$ is