Tag: probability distributions

Questions Related to probability distributions

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. $\dfrac{3}{2}$

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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

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  1. $5$

  2. $log _{10}5$

  3. $log _{e}5$

  4. $log _{5}e$

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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} } $

If $X$ is a Poisson variate with parameter $1.5$, then $P(X>1)$ is

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

  1. $1-e^{-1.5}$

  2. $e^{-1.5}(2.5)$

  3. $1-e^{-1.5}(2.5)$

  4. $1-e^{-1.5}(3.5)$

Show answer
Correct Option: C
Explanation:

$P(X>1)$
$=1-[P(X=0)+P(X=1)]$
$=1-[e^{-1.5}+1.5e^{-1.5}]$
$=1-e^{1.5}(1+1.5)$
$=1-e^{1.5}(2.5)$

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

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:

$P(X=0)=P(X=1) $
P$(0; \mu) = \dfrac { { e }^{-\mu}{\mu}^{0}}{0!} =  P(1; \mu) = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!}$
$ { e }^{-\mu}  (1 - \mu )$ = 0 
 $\mu = 1 $
P(X=2) = $ \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} = \dfrac{{ e }^{-1}}{2} = \dfrac {1}{2e} $

A random variable $X$ follows poisson distribution such that $P(X=k)=P(X=k+1)$ then the parameter of the distribution $\lambda =$

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

  1. $K$

  2. $K+1$

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

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

Show answer
Correct Option: B
Explanation:

$P(X=k)=P(X=k+1)$
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $


$ P(k;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ k } }{ k! } =P(k+1;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ k+1 } }{ (k+1)! }$

$ 1 = \dfrac {\mu}{k+1} $

$ \mu = k+1 $

In a poisson distribution $P(X=0)=P(X=1)=k$, then the value of $k$ is

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

  1. $1$

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

  3. $e$

  4. $\sqrt{2}$

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Correct Option: B
Explanation:

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

Given,$P(X=0)=P(X=1)=k $

$ \dfrac { { e }^{ -\mu }{ \mu }^{ 0 } }{ 0! } =\dfrac { { e }^{ -\mu }{ \mu }^{ 1 } }{ 1! }  $
$ { e }^{ -\mu } ={ e }^{ -\mu } \mu $
$ { e }^{ -\mu } (1 - \mu) = 0 $ 
Since $ { e }^{ -\mu } \neq 0$ , => $ \mu = 1 $
$ P(X=0) = \dfrac { { e }^{ -\mu }{ \mu }^{ 0 } }{ 0! } = { e }^{ -\mu } ={ e }^{ - 1} = \dfrac {1}{e} $

If for a poisson variable $ X$, $P(X=1)=2.\ P(X=2)$, then the parameter $\lambda $ is

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

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: B
Explanation:

For Poisson's distribution, $P(X)$ is given by 
$\displaystyle P(X)=\frac { { e }^{ -\lambda  }.{ \lambda  }^{ x } }{ x! } $
Hence $\displaystyle P(X=1) =\displaystyle  \frac { { e }^{ -\lambda }.{ \lambda  }^{ 1 } }{ 1! } $.
Similarly $\displaystyle P(X=2)=\frac { { e }^{ -\lambda }.{ \lambda  }^{ 2 } }{ 2! } $.
Given that, $P(X=1) = 2 P(X=2)$
$\Rightarrow \displaystyle \frac { { e }^{ -\lambda }.{ \lambda  }^{ 1 } }{ 1! } =2.\frac { { e }^{ -\lambda }.{ \lambda  }^{ 2 } }{ 2! } $
Simplifying we get $\lambda = 1$