Let be the mean and be the standard deviation of the distribution x_i 0 1 2 3 4 5 f_i k+2 …

Question

Let $\mu $ be the mean and $\sigma $ be the standard deviation of the distribution

${x}_{i}$
 0
 1
 2
 3
 4
 5

${f}_{i}$
 $k+2$
 $2k$
 ${k}^{2}-1$
 ${k}^{2}-1$
 ${k}^{2}+1$
 $k-3$

where $\sum {f}_{i}=62$. If $[x]$ denotes the greatest integer $\le x$, then $[{\mu }^{2}+{\sigma }^{2}]$ is equal to :

Accepted Answer

We have, $\Sigma {f}_{i}=62$

$\begin{gathered} (K+2)+2K+({K}^{2}-1))+({K}^{2}-1)+({K}^{2}+1)+(K-3)=62 \ \Rightarrow 3{K}^{2}+4K-64=0 \ \Rightarrow (3K+16)(K-4)=0 \ \Rightarrow K=4\, \end{gathered}$

$(\because k=\frac{-16}{3})$

$\begin{gathered} {x}_{i} \quad  {f}_{i} \quad  {f}_{i}{x}_{i} \quad  {f}_{i}{x}_{2}^{i} \ 0 \quad  6 \quad  0 \quad  0 \ 1 \quad  8 \quad  8 \quad  8 \ 2 \quad  15 \quad  30 \quad  60 \ 3 \quad  15 \quad  45 \quad  135 \ 4 \quad  17 \quad  68 \quad  272 \ 5 \quad  1 \quad  5 \quad  25 \ 62 \quad  156 \quad  500 \end{gathered}$

$\mu =\frac{\Sigma {f}_{i}{x}_{i}}{\Sigma {f}_{i}}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62}$

$\begin{gathered} {\sigma }^{2}= \frac{\Sigma {f}_{i}{x}_{2}^{i}}{\Sigma {f}_{i}}-{(\frac{\Sigma {f}_{i}{x}_{i}}{\Sigma {f}_{i}})}^{2} \ \ = \frac{1	imes 8+4	imes 15+9	imes 15+16	imes 17+25	imes 1}{62}-{(\frac{156}{62})}^{2} \ \ =\frac{500}{62}-{(\frac{156}{62})}^{2}=\frac{500}{62}-{\mu }^{2} \end{gathered}$

$\begin{gathered} 	herefore {\sigma }^{2}+{\mu }^{2}=\frac{500}{62} \ [{\sigma }^{2}+{\mu }^{2}]=8 \end{gathered}$

Solutions · 1