If the mean of the following probability distribution of a radam variable [X] : .tg tdbord…

Question

If the mean of the following probability distribution of a radam variable $$\mathrm{X}$$ :

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Accepted Answer

$X$
 0
 2
 4
 6
 8

$P(X)$
 $a$
 $2a$
 $a+b$
 $2b$
 $3b$

$\begin{gathered} =\sum {x}_{i}P({x}_{i}) \ \frac{46}{9}=4a+4a+4b+12b+24b \ \frac{46}{9}=8a+40b \ \frac{23}{9}=4a+20b \ 36a+180b=23\, \end{gathered}$

Sum of probability is 1

$\Rightarrow 4a+6b=1\,$

$\begin{gathered} a=\frac{1}{12},b=\frac{1}{9} \ {\sigma }^{2}=\sum {x}_{2}^{i}P({x}_{i})-{(\sum {x}_{i}P({x}_{i}))}^{2} \ =4	imes 2a+16(a+b)+36(2b)+64(3b)-{(\frac{46}{9})}^{2} \ =8(a+2(a+b)+9b+24b)-{(\frac{46}{9})}^{2} \ =8(3a+35b)-{(\frac{46}{9})}^{2} \ =8(\frac{3}{12}+\frac{35}{9})-{(\frac{46}{9})}^{2} \ =8(\frac{149}{36})-{(\frac{46}{9})}^{2}=\frac{566}{81} \end{gathered}$

Solutions · 1