Mathematics - Statistics Question with Solution | TestHub

Given,

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$k^2-1$	$k^2-1$	$k^2+1$	$k-3$

Now we know that,

$\text{Mean}=\frac{\sum X_i{f}_i}{\sum f_i}$

$\Rightarrow \mu =\frac{2k+ 2k^2 –2 + 3k^2–3 + 4k^2+4 +5k – 15}{62}$

$\Rightarrow \mu =\frac{9k^2+7k-16}{62} .......\left(1\right)$

And $\sum f_i=62$

$\Rightarrow 3k^2+4k-2=62$

$\Rightarrow 3{\mathrm{k}}^2+4\mathrm{k}-64=0$

$\Rightarrow (3\mathrm{k}+16)(\mathrm{k}-4)=0$

$\Rightarrow k=4$

Now putting the value of $k$ in above equation $\left(1\right)$ we get,

$\mu =\frac{156}{62}$

Now finding variance we get,

${\sigma }^2=\frac{\sum {\left(X_i\right)}^2{f}_i}{\sum f_i}-{\left(\text{Mean}\right)}^2$

$\Rightarrow {\sigma }^2=\frac{1·\left(2k\right)+4\left(k^2-1\right)+9\left(k^2-1\right)+16\left(k^2+1\right)+25\left(k-3\right)}{62}-{\left(\frac{156}{62}\right)}^2$

$\Rightarrow {\sigma }^2=\frac{29{\mathrm{k}}^2+27\mathrm{k}-72}{62}-{\mathrm{\mu }}^2$

Now putting the value of $k$ in above equation we get,

$\Rightarrow {\sigma }^2=\frac{500}{62}-{\mathrm{\mu }}^2$

$\Rightarrow {\mathrm{\sigma }}^2+{\mathrm{\mu }}^2=\frac{500}{62}$

$\Rightarrow \left[{\mathrm{\sigma }}^2+{\mathrm{\mu }}^2\right]=8$