Tag: probability distributions
Questions Related to probability distributions
For a poission distribution variable $X$ is such that $P(X = 2) = 9 P(X= 4) + 90 P(X= 6)$ the mean is
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$2$
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$3$
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$1$
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None of these
For a Poission distribution, which of the following is true
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$Mean = Mode$
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$Median = S.D.$
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$Mean = Variance$
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$Median = Variance$
At a telephone enquiry system the number of phone calls regarding relevant enquiry follow Poisson distribution with a average of 5 phone calls during IO-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is
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$\displaystyle \frac{6}{5^{e}}$
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$\displaystyle \frac{5}{6}$
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$\displaystyle \frac{6}{55}$
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$\displaystyle \frac{6}{e^{5}}$
According to Poisson distribution
$\displaystyle P(X=r)=\frac{e^{-m}m^{r}}{r!}$
$\displaystyle \therefore P(X\leq 1)=P(X=0)+P(X=1)$
$\displaystyle =e^{-m}+\frac{e^{-m} m}{1!}$
Given m $\displaystyle =$mean$ = 5 $
$\displaystyle \therefore P(x \leq 1)=e^{-5}+5\times e^{-5}=e^{-5}(1+5)=\frac{6}{e^{5}}$
The probability of r successes in case of poissons distrbution is
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$\dfrac{e^{\gamma }m}{\angle \gamma }$
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$\dfrac{\gamma ^{m}e^{m}}{\angle \gamma }$
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$\dfrac{e^{m}\gamma }{\angle \gamma }$
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$\dfrac{e^{-m}m^{r}}{\angle \gamma }$
in Poisson distribution, probability $ P(x; m )= \dfrac { { e }^{ - m }{ m }^{ x } }{ \angle x } $
$m$ = mean ; $x$= number of success $ \angle x =$ factorial $ x $
For $r$ success $x= r$;
$ P(r; m )= \dfrac { { e }^{ -m }{ m }^{ r } }{ \angle r } $
A random variable $X$ has Poisson distribution with mean $2$. Then $P(X > 1.5)$ equals
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$2/e^{2}$
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$0$
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$1-\dfrac{3}{e^{2}}$
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$\dfrac{3}{e^{2}}$
For Poisson distribution of mean = $\mu = 2 $
P$(x ; \mu) = \frac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$
$P(X>1.5) = 1 - [P(X=0) + P(X=1)]$
P$(1; 2) = \dfrac { { e }^{-2}{2}^{} } {1!}=2{e}^{-2}$
$P(X>1.5) = 1 - [P(X=0) + P(X=1)] $
$ = 1 - [{e}^{-2}+ 2{e}^{-2} ] = 1 - 3 {e}^{-2} = 1 - \frac { 3 }{ { e }^{ 2 } } $
If $X$ is a random poisson variate such that $\alpha =p(X=1)=p(X=2)$, then $p(X=4)=$
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$2\alpha $
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$\dfrac{\alpha }{3}$
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$\alpha e^{-2}$
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$\alpha e^{2}$
In Poisson distribution such that $ \alpha = p(X=1)=p(X=2) $
$ p(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $
=> $ \alpha = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} $ => $ \alpha = { e }^{-\mu}{\mu} $
=> $ { e }^{-\mu}{\mu} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2} $
=> $ \mu = 2 $
$ p(4; \mu) = \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} = \dfrac { { e }^{ -\mu }{ \mu \times { \mu }^{ 3 } } }{ 24 } = \dfrac { \alpha {\times { 2 }^{ 3 } } }{ 24 } = \dfrac { \alpha }{ 3 } $
The variance of P.D. with parameter $\lambda $ is
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$\lambda $
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$\sqrt{\lambda }$
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$\dfrac{1}{\lambda}$
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$\dfrac{1}{\sqrt {\lambda}}$
$E[{ X }^{ 2 }]= \sum _{ k=0 }^{ \infty }{ k^{ 2 } } \sum _{ }^{ }{ } \dfrac { 1 }{ k! } \lambda ^{ k }e^{ -\lambda }\ \ = ^{ k }e^{ -\lambda }\sum _{ 1 }^{ \infty }{ k\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } \= ^{ k }e^{ -\lambda }(\sum _{ 1 }^{ \infty }{ (k-1)\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } ) +\sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\ =^{ k }e^{ -\lambda }(\sum _{ 2 }^{ \infty }{ (\lambda \dfrac { 1 }{ (k2)! } \lambda ^{ k-2 } } )+ \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\ = ^{ k }e^{ -\lambda }(\sum _{ i=0 }^{ \infty }{ (\lambda \dfrac { 1 }{ i! } \lambda ^{ i } } )+ \sum _{ j=0 }^{ \infty }{ \dfrac { 1 }{ j! } \lambda ^{ j } } )\ = ^{ k }e^{ -\lambda }(\lambda e^{ \lambda }+e^{ \lambda })\ \ = { \lambda }^{ 2 }+ \lambda \ \ Variance= E[{ X }^{ 2 }]- { E[X] }^{ 2 }\ = { \lambda }^{ 2 }+ \lambda - { \lambda }^{ 2 }\ = \lambda \ \ \ $
If a random variable $X$ has a poisson distributionsuch that $P(X=1)=P(X=2)$, its mean and varianceare
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$1,1$
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$2, 2$
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$2, 3$
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$2,4$
In Poisson distribution such that $P(X=1)=P(X=2)$
$ P(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!}$
=> $ P(1; \mu) = P(2; \mu) $
=> $ \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} $
=> $ 1 =$ $ \dfrac {\mu}{2} $
=> $ \mu = 2 $
In Poisson distribution Variance $(m)$ is equal to mean.
Mean = Variance = $2 $
If m is the variance of P.D., then the ratio of sum of the terms in even places to the sum of the terms in odd places is
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$e^{-m}\cosh m$
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$e^{-m}\sinh m$
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$\coth m$
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$\tanh m$
Sum of the terms in Even places will be
$=e^{-m}[m+\dfrac{m^{3}}{3!}+\dfrac{m^{5}}{5!}+...]$
$=e^{-m}\dfrac{1}{2}[e^{m}-e^{-m}]$ ...(i)
Sum of the terms in Odd places will be
$=e^{-m}[1+\dfrac{m^{2}}{2!}+\dfrac{m^{4}}{4!}+...]$
$=e^{-m}\dfrac{1}{2}[e^{m}+e^{-m}]$ ...(ii)
Taking the ratio of i and ii, we get
$\dfrac{i}{ii}$
$=\dfrac{e^{-m}\dfrac{1}{2}[e^{m}-e^{-m}]}{e^{-m}\dfrac{1}{2}[e^{m}+e^{-m}]}$
$=\dfrac{e^{m}-e^{-m}}{e^{m}+e^{-m}}$
$=\dfrac{2sinhm}{2coshm}$
$=tanh(m)$.
If ${m}$ is the variance of Poisson distribution, then sum of the terms in even places is
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$e^{-m}$
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$e^{-m}\cosh m$
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$e^{-m}\sinh m$
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$e^{-m}\coth m$
If m is the variance of P. D, then P$(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $
Sum of the terms in even places = [ P$(1; \mu) + P(3; \mu) + P(5; \mu) + .........$ ]
= [$\dfrac { { e }^{-\mu}{\mu}^{1}}{1!} + \dfrac { { e }^{-\mu}{\mu}^{3}}{3!} + \dfrac { { e }^{-\mu}{\mu}^{5}}{5!} + ....... $]
= $ { e }^{-\mu} [ \mu + \dfrac {{\mu}^{3}}{3!} + \dfrac {{\mu}^{5}}{5!} + .......]$ --------------------- (1)
Since ${ e }^{ x }=1+\dfrac { x }{ 1! } +\dfrac { { x }^{ 2 } }{ 2! } +\dfrac { { x }^{ 3 } }{ 3! } +\dfrac { { x }^{4} }{ 4! } +....$
${ e }^{ -x }=1+\dfrac {( -x )}{ 1! } +\dfrac { { (-x) }^{ 2 } }{ 2! } +\dfrac { {( -x) }^{ 3 } }{ 3! } +\dfrac { { (-x) }^{4} }{ 4! } +....$
on subtracting , we get
${e}^{x} - { e }^{-x} = 2[ x + \dfrac {{x}^{3}}{3!} + \dfrac {{x}^{5}}{5!} + .....] $
$ \dfrac {{e}^{x} - { e }^{-x}}{2} = [ x +\dfrac { { x }^{ 3 } }{ 3! } + \dfrac {{x}^{5}}{5!} + .....] $
So,
$ \dfrac {{e}^{\mu} - { e }^{-\mu}}{2} = [ x +\dfrac { { \mu }^{ 3 } }{ 3! } + \dfrac {{\mu}^{5}}{5!} + .....] $
put this value in equation (1),
Sum of the terms in even places = $ { e }^{-\mu} [ \mu + \dfrac {{\mu}^{3}}{3!} + \dfrac {{\mu}^{5}}{5!} + .......]$
= $ { e }^{-\mu} (\dfrac {{e}^{\mu} - { e }^{-\mu}}{2}) $
= $ { e }^{-\mu} \sinh { \mu } $
= $ { e }^{-m} \sinh {m} $ [ Variance (m) is equal to mean ($\mu$) in Poisson distribution ]