Mathematics

Probability Distributions

457 Questions

Probability distributions describe the likelihood of different outcomes in a statistical experiment. Key topics include calculating confidence intervals, understanding Type II errors, and working with the standard normal distribution. These concepts are frequently tested in mathematics and statistics sections of various competitive exams.

Standard normal distributionType II error probabilityConfidence interval interpretationPoisson distributionProbability density function

Probability Distributions Questions

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

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

  1. $\displaystyle \frac{6}{5^{e}}$
  2. $\displaystyle \frac{5}{6}$
  3. $\displaystyle \frac{6}{55}$
  4. $\displaystyle \frac{6}{e^{5}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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 $
$\displaystyle \therefore P(x \leq 1)=e^{-5}+5\times e^{-5}=e^{-5}(1+5)=\frac{6}{e^{5}}$

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

The probability of r successes in case of poissons distrbution is

  1. $\dfrac{e^{\gamma }m}{\angle \gamma }$
  2. $\dfrac{\gamma ^{m}e^{m}}{\angle \gamma }$
  3. $\dfrac{e^{m}\gamma }{\angle \gamma }$
  4. $\dfrac{e^{-m}m^{r}}{\angle \gamma }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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 } $ 

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

A random variable $X$ has Poisson distribution with mean $2$. Then $P(X > 1.5)$ equals

  1. $2/e^{2}$
  2. $0$
  3. $1-\dfrac{3}{e^{2}}$
  4. $\dfrac{3}{e^{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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$(0; 2) = \dfrac { { e }^{-2}{2}^{0}}{0!}={e}^{-2}$
 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 } } $

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

If $X$ is a random poisson variate such that $\alpha =p(X=1)=p(X=2)$, then $p(X=4)=$

  1. $2\alpha $
  2. $\dfrac{\alpha }{3}$
  3. $\alpha e^{-2}$
  4. $\alpha e^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 } $  

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

The variance of P.D. with parameter $\lambda $ is

  1. $\lambda $
  2. $\sqrt{\lambda }$
  3. $\dfrac{1}{\lambda}$
  4. $\dfrac{1}{\sqrt {\lambda}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$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 \ \ \ $

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

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

  1. $1,1$
  2. $2, 2$
  3. $2, 3$
  4. $2,4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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


If ${m}$ is the variance of Poisson distribution, then sum of the terms in even places is

  1. $e^{-m}$
  2. $e^{-m}\cosh m$
  3. $e^{-m}\sinh m$
  4. $e^{-m}\coth m$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 ]

Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution 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

  1. $e^{-m}\cosh m$
  2. $e^{-m}\sinh m$
  3. $\coth m$
  4. $\tanh m$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 }   $ 

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

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

  1. $1$
  2. $1/2$
  3. $1/3$
  4. $1/4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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=$1$

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

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

  1. $5$
  2. $log _{10}5$
  3. $log _{e}5$
  4. $log _{5}e$
Reveal answer Fill a bubble to check yourself
C Correct answer
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=$log _{e}(5)$

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

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

  1. $\displaystyle \frac{1}{4}$
  2. $\displaystyle \frac{1}{3}$
  3. $\displaystyle \frac{1}{2}$
  4. $\displaystyle \frac{3}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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=$0.5$

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

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

  1. $K$
  2. $K+1$
  3. $\dfrac{K}{2}$
  4. $\dfrac{K+1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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

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

  1. $1$
  2. $\displaystyle\frac{1}{e}$
  3. $e$
  4. $\sqrt{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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} $

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

If the first two terms of a Poisson distribution are equal to $k$, find $k$.

  1. $e$
  2. $\displaystyle \frac{1}{e}$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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