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The number of accidents in a year involving taxi drivers in a city follows a poisson distribution with mean equal to $3$ out of $1000$ taxi drivers find approximately the number of driver with more than $3$ accidents in a year $[e^{-3}=0.0498].$

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Toolbox:
  • A random variable $X$ is said to have a poisson distribution if the probability mass function of $X$ is
  • $P(X=x)=\large\frac{e^{\Large -\lambda }\lambda^x}{x!}$$\qquad (x=0,1,2........$ for some $\lambda > 0)$
  • Constants of a poisson distribution :
  • Mean=Variance=$\lambda$
  • The parameter of the Poisson distribution is $\lambda$
  • A Poisson random variable corresponds to rare events.
Step 1:
Let $X$ be the random variable denoting the number of accidents involving taxi drivers in a year.
$X\sim P(3)$
$P(X=x)=\large\frac{e{-3}3^x}{x!}\qquad$$ x=0,1,2........$
Step 2:
Probability that a taxi driver is involved in more than 3 accidents in a year.
$P(X > 3)=1-P(X\leq 3)$
$\qquad\quad\;\;=1-[P(X=0)+P(X=1)+P(X=2)+P(X=1)]$
$\qquad\quad\;\;=1-e^{-3}[\large\frac{3^0}{0!}+\frac{3^1}{1!}+\frac{3^2}{2!}+\frac{3^3}{3!}$
$\qquad\quad\;\;=1-0.0498[1+3+\large\frac{9}{2}+\frac{9}{2}]$
$\qquad\quad\;\;=1-0.0498\times 13$
$\qquad\quad\;\;=1-0.6474$
$\qquad\quad\;\;=0.3526$
No of taxi drivers out of 1000 who are expect ion to be involved in more than 3 accidents =1000$\times 0.3526=353$(approx)
answered Sep 19, 2013 by sreemathi.v
 

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