Tag: random variables and probability distribution

Questions Related to random variables and probability distribution

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 $

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

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

  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 $

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

  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

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

  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}) $

If in a poisson distribution $P(X=1)=P(X=2)$; the mean of the distribution $f(x)=e^{-x}\dfrac{\lambda ^{x}}{\angle x}$ is

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

  1. $1$

  2. $2$

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

  4. $\dfrac{3}{2}$

Show answer
Correct Option: B
Explanation:

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

If for a poisson distribution $P(X=0)=0.2$, then the variance of the distribution is

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

  1. $5$

  2. $log _{10}5$

  3. $log _{e}5$

  4. $log _{5}e$

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
$e^{-\lambda}=0.2$
$e^{\lambda}=5$
$\lambda=ln(5)$
Hence mean=variance=$log _{e}(5)$

In a Poisson distribution, the probability $P(X=0)$ is twice the probability $P(X=1)$. The mean of the distribution is

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

  1. $\displaystyle \frac{1}{4}$

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

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

  4. $\displaystyle \frac{3}{4}$

Show answer
Correct Option: C
Explanation:

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=0)=2P(x=1)$
$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=2\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$2\lambda=1$
$\lambda=\dfrac{1}{2}$
Hence mean=variance=$0.5$

Suppose $X$ is a poisson variable such that $P(X=2)=\frac{2}{3}P(X=1)$, then $P(x=0)$ is

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

  1. $\dfrac{3}{4}$

  2. $e^{\dfrac{4}{3}}$

  3. $e^{\dfrac{-4}{3}}$

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

Show answer
Correct Option: C
Explanation:

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$

Where mean=variance=$\lambda $
Hence, $P(X=2)=\dfrac{2}{3}P(x=1)$
$\dfrac{\lambda^{2}e^{-\lambda}}{2!}=\dfrac{2\lambda^{1}e^{-\lambda}}{3.1!}$
$\dfrac{\lambda}{2}=\dfrac{2}{3}$
$\lambda=\dfrac{4}{3}$
Hence, $P(X=0)=e^{\dfrac{-4}{3}}$

If $X$ is a poisson variable such that $P(X=2)=\frac{2}{3}P(X=1)$, then $P(x=3)$ is

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

  1. $e^{\frac{-4}{3}}$

  2. $\frac{64}{162}e^{\frac{-4}{3}}$

  3. $e^{\frac{-3}{4}}$

  4. $e^{\frac{3}{4}}$

Show answer
Correct Option: B
Explanation:

$P(X=2)= \cfrac {2}{3}P(X=1) $

$P$$(2; \mu) = \cfrac { { e }^{-\mu}{\mu}^{2}}{2!} = \cfrac {2}{3} P(1; \mu) $=$ \cfrac {2}{3} \cfrac { { e }^{-\mu}{\mu}^{1}}{1!}$

$ { e }^{-\mu} \mu (\cfrac {\mu}{2} - \cfrac {2}{3})=0$

either $\mu = 0\  or \mu = \cfrac{4}{3}$

but here $ \mu = \cfrac{4}{3}$$P(X=3) =$ $ \cfrac { { e }^{-\mu}{\mu}^{3}}{3!} = \cfrac { { e }^{ \frac { -4 }{ 3 }  }{( \frac { 4 }{ 3 }  })^{ 3 } }{ 6 } = \cfrac { 64 }{ 162 } { e }^{ \frac{-4}{3} } $