ii. If x denotes the outcome when a fair die is tossed. find the MGE of X about the origin. Hence find first two moments about the origin.

i. Given,

$ p=\frac{1}{2400} \\ n=200$

$ m= n \times p = 200 \times \frac{1}{2400} \\ m = \frac{200}{2400}$

$m=\frac{1}{12}$

Since, Give p is very small, hence we use poisson distribution

$P(x=r) = \frac{e^{-m} \times m^r}{r!}$

For at least one fatal accident in a year. Here, r = 1,2,3,----200, Therefore for simplification, we first calculate No accident in a year i.e r=0

$\begin{align} P(x=0) &= \frac{e^{-(\frac{1}{12})} \times -(\frac{1}{12})^0}{0!}\\ &= \frac{0.9200 \times 1}{1}\\ P(x=0)& = 0.9200\\ \end{align}$

\

$\begin{equation} \therefore \text{for at least one fatal accident in a year}\\ = 1- P(x=0)\\ = 1-0.9200 \\ = 0.08 \\ \end{equation}$

ii. We have here X taking values 1,2,3,4,5,6

$\therefore Probability, p = \frac16$

$\begin{align} M_0(t) &= E(e^{txi} = \sum {pie^{txi}} \\ &= \frac1\sigma e^{t}+frac1\sigma e^{2t}+frac1\sigma e^{3t}+frac1\sigma e^{4t}+frac1\sigma e^{5t}+frac1\sigma e^{6t}\\ &= frac1\sigma [e^{t}+e^{2t}+e^{3t}+e^{4t}+e^{5t}+e^{6t}] - - - - 1\\ Mean(\mu _1 ')&= [\frac{d}{dt} M_{0(t)}]_{t=0} \\ \\ \text{diff 1 w.r.t 't'}\\ (\mu _1 ') &= \frac1\sigma [e^{t}+e^{2t}+e^{3t}+e^{4t}+e^{5t}+e^{6t}]_{t=0} \\ (\mu _1 ') &= \frac1\sigma [1+2+3+4+5+6] \\ (\mu _1 ') &= \frac{21}{6} \\ (\mu _1 ') &= \frac72 \\ Variance \ (\mu _2) &= \mu _2 '- (\mu _2 ')^2 \\ \text{To find $\mu _2 '$ }\\ \mu _2 '&= [\frac{d^2}{dt^2} M_0(t)]_{t=0}\\ \text{diff 1 w.r.t 't'}\\ \mu _2 '&= \frac{d}{dt} \Big\{ \frac1\sigma [e^{t}+e^{2t}+e^{3t}+e^{4t}+e^{5t}+e^{6t}]_{t=0} \Big\} \\ \text{diff again w.r.t 't'}\\ \mu _2 '&= \frac1\sigma [e^{t}+4e^{2t}+9e^{3t}+16e^{4t}+25e^{5t}+36e^{6t}]_{t=0} \\ \mu _2 '&= \frac1\sigma[1+4+9+16+25+36]\\ \mu _2 '&=\frac{91}{6}\\ \\ \therefore Variance \ (\mu _2) & = \mu _2 ' - (\mu _1 ')^2 \\ &= \frac{91}{6} - (\frac72)^2\\ &= \frac{91}{6} - \frac{49}{4}\\ Variance (\mu _2)&= \frac{35}{12}\\ \end{align}$