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Probability Distribution
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Let $x$ have a poisson distribution with mean $4$.Find$P(2\leq$X$<$5$)[e^{-4}=0.0183].$
tnstate
class12
bookproblem
ch10
sec-1
exercise10-4
p218
q1
q1-2
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asked
Apr 21, 2013
by
poojasapani_1
edited
Sep 18, 2013
by
sreemathi.v
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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^{\Large -\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:
$P(2 \leq X <5)=P(X=2)+P(X=3)+P(X=4)$
$\qquad\qquad\quad\;\;=e^{-4}\big[\large\frac{4^2}{2!}+\frac{4^3}{3!}+\frac{4^4}{4!}\big]$
$\qquad\qquad\quad\;\;=0.0183\big[8+\large\frac{32}{3}+\frac{32}{3}\big]$
$\qquad\qquad\quad\;\;=0.0061[88]$
$\qquad\qquad\quad\;\;=0.5368$
answered
Sep 18, 2013
by
sreemathi.v
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