written 5.0 years ago by | • modified 3.9 years ago |
1)no accident in a year
2)more than 3 accidents in a year.
($e^{-1}=0.3679,e^{-2} = 0.1353,e^{-3}= 0.0498$
written 5.0 years ago by | • modified 3.9 years ago |
1)no accident in a year
2)more than 3 accidents in a year.
($e^{-1}=0.3679,e^{-2} = 0.1353,e^{-3}= 0.0498$
written 5.0 years ago by | • modified 4.9 years ago |
For poison distribution $P(x=r) = \frac{ e^{-m}.m^{r}}{r!}$ where x =0,1,2
Given m=3
i) No.accidents in a year, i.e.$r=0$
$P(x=0) = \frac {e^{-3} \times (3)^0 }{0!}$
$P(x=0)=\frac {0.0498\times 1}{1}$
$P(x=0)=0.0498$
therefore expected number of drivers with no accidents
=$N\times P(x=0)$
=$1000\times 0.0498$
=$49.8 \approx 50$
ii)More than 3 accidents in a year
$P(more \ than \ 3 \ accident) = 1-p[x=0,1,2]$
=$1-[\frac{e^{-3}.3^{0}}{0!}+\frac{e^{-3}.3^1}{1!}+\frac{e^{-3}.(3)^{2}}{2!}]$
= $1-[0.0498+0.1494+0.2241]$
=$1-0.4233$
$P(more \ than \ 3 \ accident)=0.5767$
Expected number of drivers with more then 3 accidents
= $N\times p(more \ than \ 3 \ accidents)$
=$1000 (0.5767)$
$=576.7 \approx =577$