Let the mean and the standard deviation of the probability distribution .tg tdborder-color…

Question

Let the mean and the standard deviation of the probability distribution

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Answer: 5

Accepted Answer

Mean $(\mu )=\Sigma {x}_{i}P({x}_{i})$

Standard deviation $(\sigma )=\sqrt{(\Sigma {x}_{2}^{i}P({x}_{i}))-{\mu }^{2}}$

$\begin{gathered} \Rightarrow \, \mu =\frac{1}{3}\alpha +K-\frac{3}{4} \ \sigma =\sqrt{(\frac{1}{3}{\alpha }^{2}+K+0+\frac{9}{4})-{(\frac{1}{3}\alpha +K-\frac{3}{4})}^{2}} \ \because \Sigma {P}_{i}=1\Rightarrow \frac{1}{3}+K+\frac{1}{6}+\frac{1}{4}=1 \ \Rightarrow \, K=\frac{1}{4}\Rightarrow \mu =\frac{1}{3}\alpha -\frac{1}{2} \ \because \sigma -\mu =2 \ {\sigma }^{2}=(\mu +2{)}^{2} \end{gathered}$

$\begin{gathered} \frac{1}{3}{\alpha }^{2}+\frac{5}{2}-{\mu }^{2}=(\mu +2{)}^{2} \ \frac{1}{3}{\alpha }^{2}+\frac{5}{2}={(\frac{1}{3}\alpha -\frac{1}{2})}^{2}+{(\frac{1}{3}\alpha +\frac{3}{2})}^{2} \ \Rightarrow \alpha =0,6 \end{gathered}$

$\begin{gathered} \alpha =0,K=\frac{1}{4} \alpha =6,K=\frac{1}{4} \ \mu =-\frac{1}{2},\sigma =\frac{3}{2} \mu =\frac{3}{2},\sigma =\frac{7}{2} \ \sigma +\mu =1 \sigma +\mu =5 \end{gathered}$

Both (1) and (5) are correct but according to NTA
(5) is correct

Solutions · 1