Mathematics · Quantitative Aptitude

Statistics and Dispersion

515 Questions

Statistics and dispersion involve the calculation of mean, standard deviation, variance, and coefficient of variation for data sets. These questions also cover probability distributions and cumulative frequency analysis. Such quantitative aptitude topics are heavily featured in banking and SSC examinations.

Standard deviationNormal distributionMean calculationCumulative frequencyCoefficient of variation

Statistics and Dispersion Questions

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

For a poission distribution variable $X$ is such that $P(X = 2) = 9 P(X= 4) + 90 P(X= 6)$ the mean is

  1. $2$
  2. $3$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For $P.D. P(X=r)=\displaystyle \dfrac{e^{-\lambda}\lambda ^{r}}{r!},r=0,1,2,\cdots$
$\therefore \displaystyle \dfrac{e^{-\lambda}\lambda ^{2}}{2!}=\dfrac{9e^{-\lambda}\lambda ^{4}}{4!}+90 \dfrac{e^{-\lambda}\lambda ^{6}}{6!}$        (given)
$\Rightarrow \displaystyle\dfrac{\lambda ^{2}}{2}=\dfrac{9}{24}\lambda ^{4}+\dfrac{90}{720}\lambda ^{6}$ 
$\Rightarrow \lambda ^{4}+3\lambda ^{2}-4=0$
$\Rightarrow \left ( \lambda^{2}+4 \right )\left ( \lambda^{2}-1 \right )=0=>\lambda= \pm 1$
$\Rightarrow \lambda=1$ as $\lambda>0$ and $(\lambda^{2}+4=0$ impossible$)$
$\therefore$ mean$=\lambda=1$
Multiple choice business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

For a Poission distribution, which of the following is true

  1. $Mean = Mode$
  2. $Median = S.D.$
  3. $Mean = Variance$
  4. $Median = Variance$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For Poission distribution we have $\displaystyle P\left ( r \right )=\dfrac{e^{-\pi }\lambda ^{r}}{r^{i}}\left ( r=0, 1, 2,...\infty  \right)$ 
$\displaystyle$ mean $\displaystyle =\dfrac{\sum f _{i}x _{i}}{\sum f _{i}}=\sum _{r=0}^{\infty }rP\left ( r \right ),\sum f _{i}P\left ( r\right )=1$$\displaystyle =0+\lambda e^{-\lambda }+2\dfrac{\lambda ^{2}e^{-\lambda }}{2!}+3\dfrac{\lambda ^{3}e^{-\lambda }}{3!}+.....\infty$ $\displaystyle =\lambda e^{-\lambda }\left ( 1+\dfrac{\lambda }{11}+\dfrac{\lambda ^{2}}{2!}+\dfrac{\lambda ^{3}}{3!}+.....\infty  \right )$$\displaystyle =\lambda e^{\lambda }.e^{\lambda }=\lambda $
similarly, $\displaystyle o^{2}=$ Variance $=\sum _{r=0}^{\infty }r^{2}P\left ( r \right )-\left ( \sum _{r=0}^{\infty } rP\left ( r \right ) \right )^{2}$ $=\displaystyle \lambda e^{\lambda }\left ( e^{-\lambda }+\lambda e^{-\lambda } \right )-\lambda ^{2}=\lambda =$ mean 
$\therefore \text{mean}=\text{variance}$
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

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

  1. $\dfrac{1}{2}$
  2. $2$
  3. $\log _{e}2$
  4. $3$
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
$e^{-\lambda}=0.5$
$e^{\lambda}=2$
$\lambda=ln(2)$
Hence mean=variance=$log _{e}(5)$

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

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

  1. $\displaystyle \frac{2}{3}$
  2. $\displaystyle \frac{1}{3}$
  3. $6$
  4. $\sqrt{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

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

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

  1. $3$
  2. $2$
  3. $1$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $log _{e}20$
  2. $log _{10}20$
  3. $log _{e}(5/4)$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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}) $

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

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

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

If $X$ is a Poisson variate with parameter $\displaystyle \frac{3}{2}$, find $P(X\geq 2)$

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

Here $\lambda = \cfrac{3}{2}$
$\therefore P(X\geq 2) = 1-P(X=0)-P(X=1)$
$\displaystyle =1-\cfrac{e^{-3/2}(3/2)^0}{0!}-\cfrac{e^{-3/2}(3/2)^1}{1!}=1-\cfrac{5}{2}e^{-3/2}$