Tag: probability distributions

Questions Related to probability distributions

If m is the variance of P.D., then the ratio of sum of the terms in odd places to the sum of the terms in even places is

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  1. $e^{-m}\cosh m$

  2. $e^{-m}\sinh m$

  3. $\coth m$

  4. $\tanh m$

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Correct Option: C
Explanation:

If m is the variance of P. D, then  P$(x; \mu) = \frac { { e }^{-\mu}{\mu}^{x}}{x!} $
Sum of the terms in odd places = [ P$(0; \mu) + P(2; \mu)  + P(4; \mu) + .........$ ]


                                       =  [$\dfrac { { e }^{-\mu}{\mu}^{0}}{0!} + \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} + \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} + ....... $]

                                      =  $  { e }^{-\mu} [ 1 + \dfrac {{\mu}^{2}}{2!} + \dfrac {{\mu}^{4}}{4!} + .......]$       --------------------- (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 adding , we get  
 ${e}^{x} + { e }^{-x} =  2[ 1 + \dfrac {{x}^{2}}{2!} + \dfrac {{x}^{4}}{4!} + .....] $
$ \dfrac {{e}^{x} + { e }^{-x}}{2} =  [ 1 +\dfrac { { x }^{ 2} }{ 2! } + \dfrac {{x}^{4}}{4!} + .....] $
So, 
$ \dfrac {{e}^{\mu} + { e }^{-\mu}}{2} =  [ 1+\dfrac { { \mu }^{ 2 } }{ 2! } + \dfrac {{\mu}^{4}}{4!} + .....] $

put this value in equation (1),
Sum of the terms in odd places = $  { e }^{-\mu} [ 1 + \dfrac {{\mu}^{2}}{2!} + \dfrac {{\mu}^{4}}{4!} + .......]$
                    =  $  { e }^{-\mu} (\dfrac {{e}^{\mu} + { e }^{-\mu}}{2}) $
                    = $ { e }^{-\mu} \cosh { \mu  } $
                    = $ { e }^{-m} \cosh {m} $       [ Variance (m) is equal to mean ($\mu$) in Poisson distribution ]
Similarly Sum of the terms in even places =  $ { e }^{-m} \sinh {m} $ 
The ratio of sum of the terms in odd places to the sum of the terms in even places is = $ \dfrac { { e }^{ -m }\cosh { m }  }{ { e }^{ -m }\sinh { m }  } =\coth { m }   $ 

A : the sum of the times in odd places in a P.D is $e^{-\lambda }$ cosh $\lambda$ 
R : cosh $\lambda =\frac{\lambda ^{1}}{1!}+\frac{\lambda ^{3}}{3!}+\frac{\lambda ^{5}}{5!}+......$

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  1. Both A and R are true and R is the correct

    explanation of A

  2. Both A and R are true but R is not correct

    explanation of A

  3. A is true but R is false

  4. A is false but R is true

Show answer
Correct Option: C
Explanation:

Considering all odd places in Poisson's Distribution, we get 
$e^{-\lambda}+\dfrac{e^{-\lambda}.\lambda^{2}}{2!}+\dfrac{e^{-\lambda}.\lambda^{4}}{4!}....$

$=e^{-\lambda}[1+\dfrac{\lambda^{2}}{2!}+\dfrac{\lambda^{4}}{4!}+....]$

$=e^{-\lambda}[\dfrac{e^{-\lambda}+e^{\lambda}}{2}]$

$=e^{-\lambda}(\cosh\lambda)$.
Hence reason is false.

If $X$ is a poisson variate with $P(X=0)=P(X=1)$, then $P(X=2)$ is

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\dfrac{e}{2}$

  2. $\dfrac{e}{6}$

  3. $\dfrac{1}{6e}$

  4. $\dfrac{1}{2e}$

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Correct Option: D
Explanation:

Here,  $P(X=0)=P(X=1)$
$\Rightarrow \displaystyle \frac{e^{-\lambda}\lambda^0}{0!}=\frac{e^{-\lambda}\lambda^1}{1!}\Rightarrow \lambda = 1$
$\displaystyle \therefore P(X = 2) = \frac{e^{-\lambda}1^2}{2!}=\frac{1}{2e}$

If $X$ is a random poisson variate such that $E(X^{2})=6$, then $E(X)=$

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  1. $-3$

  2. $2$

  3. $-3&2$

  4. $-2$

Show answer
Correct Option: B
Explanation:

Variance  = $E(X^{ 2 }) - E[X]^{ 2 }$
Mean = Variance for P.D.
Let mean = $m$
Therefore
$m = $ $ 6 - m*m $
Hence
$m = 2$

For a Poisson variate $X$ if $P(X=2)=3P(X=3)$, then the mean of $X$ is

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  1. $1$

  2. $1/2$

  3. $1/3$

  4. $1/4$

Show answer
Correct Option: A
Explanation:

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=2)=3P(x=3)$
$3\dfrac{\lambda^{3}e^{-\lambda}}{3!}=\dfrac{\lambda^{2}e^{-\lambda}}{2!}$
$3\lambda=3$
$\lambda=1$
Hence mean=variance=$1$

If $X$ is a poisson variate such that $P(X=0)=\dfrac{1}{2}$, the variance of $X$ is

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  1. $\dfrac{1}{2}$

  2. $2$

  3. $\log _{e}2$

  4. $3$

Show answer
Correct Option: C
Explanation:

Poisson's distribution implies
$P(X=n)=\frac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence substituting $x=0$ 

we get
$e^{-\lambda}=0.5$
$e^{\lambda}=2$
$\lambda=ln(2)$
Hence mean=variance=$log _{e}(5)$

If in a poisson frequency distribution, the frequency of $3$ successes is $\displaystyle \frac{2}{3}$ times the frequency of $4$ successes, the mean of the distribution is

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  1. $\displaystyle \frac{2}{3}$

  2. $\displaystyle \frac{1}{3}$

  3. $6$

  4. $\sqrt{6}$

Show answer
Correct Option: C
Explanation:

According to question
$P(X=3; \mu )= \dfrac {2}{3}P(X=4; \mu) $
$\dfrac { { e }^{-\mu}{\mu}^{3}}{3!}  = \dfrac {2}{3} \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} $
=> $\dfrac { { e }^{-\mu}{\mu}^{3}}{6}  = \dfrac { { e }^{-\mu}{\mu}^{4}}{36} $
=> ${ e }^{ -\mu  }{ \mu  }^{ 3 }\left[ \dfrac { 1 }{ 6 } -\dfrac { \mu  }{ 36 }  \right] $ 
=> $ \mu = 0 \  or\   \mu = \dfrac{36}{6} = 6 $

If X is a poisson variate such that $P(X=2)=9p(X=4)+90p(X=6)$ , then the mean of x is

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  1. $3$

  2. $2$

  3. $1$

  4. $0$

Show answer
Correct Option: C
Explanation:

For Poisson's distribution, $P(X)$ is given by 
$\displaystyle P(X)=\frac { { e }^{ -\lambda  }.{ \lambda  }^{ x } }{ x! } $
It is given that $P(X=2)=9P(X=4)+  90P(X=6)$
$\Rightarrow \displaystyle \frac { { e }^{ -\lambda  }.{ \lambda  }^{ 2 } }{ 2! } =9.\frac { { e }^{ -\lambda  }.{ \lambda  }^{ 4 } }{ 4! } +90.\frac { { e }^{ -\lambda  }.{ \lambda  }^{ 6 } }{ 6! } $
Cancelling ${ e }^{ -\lambda  }$ from both sides, we get
$\displaystyle \frac { { \lambda  }^{ 2 } }{ 2! } =9.\frac { { \lambda  }^{ 4 } }{ 4! } +90.\frac { { \lambda  }^{ 6 } }{ 6! } $
Since $\lambda$ can not be zero, cancel out $\lambda^2$ from both sides to obtain
${ \lambda  }^{ 4 }+3{ \lambda  }^{ 2 }-4=0$.
Solving , we get $\lambda = 1$ as a valid solution.

If $X$ is a poisson variate such that $P(X=0)=0.1,P(X=2)=0.2$, then the parameter $\lambda $

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  1. $2$

  2. $4$

  3. $5$

  4. $3$

Show answer
Correct Option: A
Explanation:

$P(X=0)=0.1$ and $P(X=2)=0.2$
$P$$(0; \mu) = \dfrac{ { e }^{-\mu}{\mu}^{0}}{0!} =  { e }^{-\mu}$  =  $0.1 $
$P$$(2; \mu) = \dfrac{ { e }^{-\mu}{\mu}^{2}}{2!} $  $=  0.2 $
                    =>  $\dfrac {{0.1}{\mu}^{2}}{2} $ $= 0.2$
                    =>  $ \mu $ $= 2 $
The parameter  is mean ($ \mu $) in Poisson distribution, so $ =$ $2  $

If $X$ is a poisson variate with $P(X=0) = 0.8,$ then the variance of $X$ is

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  1. $log _{e}20$

  2. $log _{10}20$

  3. $log _{e}(5/4)$

  4. $0$

Show answer
Correct Option: C
Explanation:

In Poisson distribution,   $P(X=0) = 0.8,$
 $ p(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $ 
=>  $ p(0; \mu) = \dfrac { { e }^{-\mu}{\mu}^{0}}{0!} $
=>    0.8 = $ { e }^{-\mu} $ 
=>    $ \dfrac {4}{5} = { e }^{-\mu} $
=>    ${ e }^{\mu}  = \dfrac {5}{4} $ 
taking log to both sides 
 =>  $  \log _{ e }{ { e }^{ \mu  } }  = \log _{e} (\dfrac {5}{4}) $ 
 =>  $ \mu =  \log _{e} (\dfrac {5}{4}) $