Tag: business maths

Questions Related to business maths

If a random variable $X$ has a poisson distributionsuch that $P(X=1)=P(X=2)$, its mean and varianceare

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

  1. $1,1$

  2. $2, 2$

  3. $2, 3$

  4. $2,4$

Show answer
Correct Option: B
Explanation:

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

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

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

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

  3. $\coth m$

  4. $\tanh m$

Show answer
Correct Option: D
Explanation:

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

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

  1. $e^{-m}$

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

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

  4. $e^{-m}\coth m$

Show answer
Correct Option: C
Explanation:

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 ]

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

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

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

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

  3. $\coth m$

  4. $\tanh m$

Show answer
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 $\lambda =\frac{\lambda ^{1}}{1!}+\frac{\lambda ^{3}}{3!}+\frac{\lambda ^{5}}{5!}+......$

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

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

Show answer
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 $E(X)=$

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

  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

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

  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

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

  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

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

  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 $