Answers

2014-04-29T15:20:43+05:30
P(x=3)>p(x=2)

(e^-lamda d^2)/3! > (e^-lamda lamda^2)/ 2!

henece d= 3 

mean = 3
0
2014-05-02T22:58:46+05:30
Let α be the mean
 \frac{e^{- \alpha } \alpha^3 }{3!}> \frac{e^{- \alpha } \alpha^2}{2!} \\ a>3
so mean value of α=3(minimum value)


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