Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 $[e^{-3}=0.0498].$

Can you answer this question?

1 Answer

0 votes
  • 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.
$P(X > 3)=1-P(X\leq 3)$
$\qquad\quad\;\;=1-0.0498\times 13$
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App