Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: Dec 2014

For poisons distributions variance =m=?

1)$${\mkern1mu\raise1pt\hbox{.}\mkern1mu\raise4pt\hbox{.}\mkern1mu\raise1pt\hbox{.}\mkern1mu} \, \, p(X=x)=\frac{e^{-m} m^{x} }{x!} =\frac{e^{-3} 3^{x} }{x!} $$ $${\mkern1mu\raise1pt\hbox{.}\mkern1mu\raise4pt\hbox{.}\mkern1mu\raise1pt\hbox{.}\mkern1mu} \, \, p\left(X=2\right)=\frac{e^{3} 3^{2} }{2!} =0.2240$$

2) $p\left(x \ge 4\right)=1-p\left(x \lt 4\right)$ $$\begin{array}{l} {=1-\left[p\left(x=0\right)+p\left(x=3\right)+p\left(x=2\right)+p\left(x=3\right)\right]} \\ {=1-\left[\frac{e^{-3} 3^{0} }{0!} +\frac{e^{-3} 3^{1} }{1!} +\frac{e^{-3} 3^{2} }{2!} +\frac{e^{-3} 3^{3} }{3!} \right]} \\ {=1-e^{-3} \left[1+3+\frac{9}{2} +\frac{9}{2} \right]} \\ {=1-13e^{-3} } \\ {=0.3528} \\ {Hence\, \, p\left(x=2\right)=0.2240\, \, \, \, \& } \\ {p\left(x\ge 4\right)=0.3528} \end{array}$$