A set of four observations has mean 1 and variance 13.

Question

A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1 . Then, the variance of all these 10 observations is equal to :

Accepted Answer

$\begin{gathered} {x}_{1} \quad  {x}_{2} \quad  {x}_{3} \quad  {x}_{4} \quad  n \quad  \quad  {y}_{1} \quad  {y}_{2} \quad  {y}_{3} \quad  {y}_{4} \quad  {y}_{5} \quad  {y}_{6} \quad  n \ 1 \quad  1 \quad  1 \quad  1 \quad  4 \quad  \quad  2 \quad  2 \quad  2 \quad  2 \quad  2 \quad  2 \quad  6 \end{gathered}$

$1\Rightarrow {\mu }_{x}=\frac{\sum _{i=1}^{4}{x}_{i}}{4}=1\ldots .(i)$

${\sigma }_{2}^{x}=[|((|\frac{\sum _{i=1}^{4}{x}_{2}^{i}}{4}))|-{\mu }_{2}^{x}]|=13\ldots \ldots (ii)$

Set $2\Rightarrow {\mu }_{y}=\frac{\sum _{j=1}^{6}{y}_{j}}{6}=2\ldots \ldots (iii)$

${\sigma }_{2}^{y}=[[|((|\frac{\sum _{j=1}^{6}{y}_{2}^{j}}{6}))|-{\mu }_{2}^{y}]]|=1\ldots \ldots (iv)$

We mathematically extract from equations $(i)$ and $(ii)$ :

$\sum _{i=1}^{4}{x}_{i}=4\ldots .(v)\, \sum _{i=1}^{4}{x}_{2}^{i}=4({\mu }_{2}^{x}+13)=4(1+13)=56\ldots \ldots (vi)$

We mathematically extract from equations (iii) and (iv) :

$\sum _{j=1}^{6}{y}_{j}=12\ldots .(vii)\, \sum _{j=1}^{6}{y}_{2}^{j}=6({\mu }_{2}^{y}+1)=6(4+1)=30\ldots \ldots (viii)$

We have mathematically compiled a summary of all the information in equations (v) through (viii) to compute the mean and the variance for the 10 datapoints in set $1\cup$ set 2 :

${\mu }_{t}=\frac{\sum _{i=1}^{4}{x}_{i}+\sum _{j=1}^{6}{y}_{j}}{4+6}=\frac{4+12}{10}=1.6\ldots ..(ix)$

${\sigma }_{2}^{t}=[(\frac{\sum _{4}^{i=1}{x}_{2}^{i}+\sum _{6}^{j=1}{y}_{2}^{j}}{4+6})-{\mu }_{2}^{t}]$

$=\frac{56+30}{4+6}-\begin{gathered} (1.6{)}^{2} \end{gathered}$

$=8.6-2.56=6.04$

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