logo

Ask Questions, Get Answers

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

Let $X$ have a poisson distribution with mean $4$.Find $(i) \;P(X\leq $3$)\qquad[e^{-4} = 0.0183].$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • A random variable $X$ is said to have a poisson distribution of the probability mass function of $X$ is
  • $P(X=x)=\large\frac{e^{-\lambda }\lambda^x}{x!}$$\qquad (x=0,1,2........$ for some $\lambda > 0)$
  • Constants of a poisson distribution : Mean=Variance=$\lambda$
  • A continuous random variable $X$ is said to follow a normal distribution with parameter $\mu$ and $\sigma$ (or $\mu$ and $\sigma^2$) if the probability density function is
  • $f(x)=\large\frac{1}{\sigma \sqrt{2\pi}}$$e^{-\large\frac{1}{2}(\frac{x-\mu}{\sigma})^2};-\infty < x < \infty,-\infty< \mu <0$ and $\sigma > 0$
  • $X\sim N(\mu,\sigma)$
  • Constants of a normal distribution :
  • Mean =$\mu$,variance =$\sigma^2$,standard deviation =$\sigma$
Step 1:
$X\sim P(4)$
$\therefore P(X=x)=\large\frac{e^{\Large -4}4^{\Large x}}{x!}$$\qquad x=0,1,2....$
Step 2:
$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$
$\qquad\qquad=e^{\Large -4}\big[\large\frac{4^0}{0!}+\frac{4^1}{1!}+\frac{4^2}{2!}+\frac{4^3}{3!}\big]$
$\qquad\qquad=0.0183\big[\large\frac{(1+4+8)3+32)}{3}\big]$
$\qquad\qquad=0.0061[71]$
$\qquad\qquad=0.4331$
answered Sep 18, 2013 by sreemathi.v
 

Related questions

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