Tag: business maths

The mean of P.D. is $5$. 
In P.D,  the variance and mean of the same distribution is same.
So, variance = $5$

For a Poisson distribution, mean ($\mu$) and variance ($\sigma^2$ )are equal .
i.e.$\mu = \sigma^2$

Mean = $ \lambda$
In Poisson distribution, the expected value of a Poisson-distributed random variable is equal to that is mean and is equal to its variance. 
Variance =  $ \lambda$
S.D = $ \sqrt {variance} =\sqrt { \lambda  } $

$P(X=3)/P(X=2)$
$=\dfrac{\lambda^{3}(2!)}{3!(\lambda^{2})}$
$=\dfrac{\lambda}{3}$
$=\dfrac{1}{2(3)}$
$=\dfrac{1}{6}$
The ratio is $1:6$

Given $P(X=0)=.1$
$\Rightarrow e^{-\lambda}=\cfrac{1}{10}\therefore \lambda = \log _e 10$

In Poisson distribution the  parameter $(  > 0)$, it is always greater than zero or a finite positive value.

X is a Poisson variable with parameter 0.09.
In Poisson distribution,  Poisson variable(X) is equal to  expected value of X and also to its variance. 
 S.D =  $\sqrt {variance}  =  \sqrt { X } =\sqrt { 0.09 } =0.3$

Given $\sigma =1.5$
We know that for a Poisson's distribution,mean and variance are equal.
$\mu=\sigma^2$
$\Rightarrow \mu =2.25$

$S.D= \sqrt { \lambda  }$

Therefore, the probability of getting $6$ heads with $6$ coins $ = {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{64}} = P\left( { > {a _y}} \right)$
Then, 
$\eta P = 6400 \times \frac{1}{{64}} = 100 = m\left( {{a _y}} \right)$
So, by poison's law, $P\left( {x = n} \right) = \frac{{{e^{ - m}}{m^x}}}{{x!}}$
$ = \frac{{{e^{ - 100}}{{100}^x}}}{{x!}}$