Let μ be the mean and σ be the standard deviation of the...

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,

$\text{Mean} = \dfrac{∑ {X}_{i}{f}_{i}}{∑ {f}_{i}}$

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

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

And $∑ {f}_{i} = 62$

$⇒ 3{k}^{2} + 4k - 2 = 62$

$⇒ 3{k}^{2} + 4k - 64 = 0$

$⇒ ( 3k + 16 ) ( k - 4 ) = 0$

$⇒ k = 4$

Now putting the value of $k$ in above equation \left(1\right)$we get,</p><p>$\mu = \dfrac{156}{62}$</p><p>Now finding variance we get,</p><p>${\sigma}^{2} = \dfrac{∑ {\left({X}_{i}\right)}^{2}{f}_{i}}{∑ {f}_{i}} - {\left(\text{Mean}\right)}^{2}$</p><p>$⇒ {\sigma}^{2} = \dfrac{1 \cdot \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(\dfrac{156}{62}\right)}^{2}$</p><p>$⇒ {\sigma}^{2} = \dfrac{29{k}^{2} + 27k - 72}{62} - {\mu}^{2}$</p><p>Now putting the value of$k$in above equation we get,</p><p>$⇒ {\sigma}^{2} = \dfrac{500}{62} - {\mu}^{2}$</p><p>$⇒ {\sigma}^{2} + {\mu}^{2} = \dfrac{500}{62}$</p><p>$⇒ \left[{\sigma}^{2} + {\mu}^{2}\right] = 8