Mathematics · Quantitative Aptitude

Statistics and Dispersion

559 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 statistics data handling analysis bar charts frequency table and diagrams bar diagram

The following frequency distribution is classified as 
X : 0   1    2   3   4   5   6  
f:    2   5   8   10   9   6   3    

  1. continuous distribution

  2. discrete distribution

  3. cumulative frequency distribution

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A discrete distribution represents data where variables take on specific, distinct values (like 0, 1, 2, etc.) rather than a continuous range. Since the variable X is listed as distinct integers, this is a discrete distribution.

Multiple choice statistics data handling analysis bar charts frequency table and diagrams bar diagram

For the mid-values 4,8,12,16,20 the first class of the distribution is _________.

  1. 0-8

  2. 1-7

  3. 2-6

  4. none of them
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mid-values are calculated as (lower limit + upper limit) / 2. For the first class, if the mid-value is 4 and the interval width is 4 (based on the difference between consecutive mid-values), the class is 2-6, where (2+6)/2 = 4.

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

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

If ${ \mu  } _{ 2 }=20,{ \mu  } _{ 2 }^{ 1 }=276$ for a discrete random variable $X$, then the mean of the random variable $X$ is

  1. $16$
  2. $5$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\mu _2=E(X^2)-[E(X)]^2......(i)$

$\mu _2'=E(X)^2.....(ii)$
Given: $\mu _2=20, \mu _2'=276$
from equation (ii) $\implies E(x^2)=276$
and from equation (i), $\implies 20=276-[E(X)]^2$
$\implies [E(X)]^2=276-20=250\\implies E(X)=16$
is the mean of random variable.

Multiple choice maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

If average of $6$ consecutive numbers is $48$, what is the difference between the smallest and the largest numbers ?

  1. $5$
  2. $10$
  3. $9$

  4. data inadequate
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the $6$ numbers are $x, x+1, x+2, x+3, x+4, x+5$.

Since,
$average=48$
$\dfrac{x+x+1+x+2+x+3+x+4+x+5}{6}=48$
$6x+15=288$
$6x=273$
$x=45.5$
So,
The smallest number $=45.5$
The largest number $=x+5=45.5+5=50.5$
Difference $=50.5-45.5=5$

Multiple choice maths average arithmetic mean of ap introduction to averages means

The mean of ${1^{2,}}{2^2},{3^2},{4^2},{5^2},{6^2},{7^2}$ is:

  1. 10

  2. 20

  3. 30

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The sum of squares of the first 7 natural numbers is 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 = 1 + 4 + 9 + 16 + 25 + 36 + 49 = 140. The mean is 140 / 7 = 20.

Multiple choice maths average arithmetic mean of ap introduction to averages means

Mean of the first $n$ terms of the A.P. $a, (a + d), (a + 2d), ........$ is

  1. $\displaystyle a + \frac{nd}{2}$
  2. $\displaystyle a + \frac{(n - 1)d}{2}$
  3. $a + (n - 1) d$
  4. $a + nd$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Required mean $= \displaystyle \frac{a + (a + d) + (a + 2d) + ....... + { a + (n - 1) d }}{n}$
$\displaystyle = \frac{\displaystyle \frac{n}{2} [a + a + (n - 1) d]}{n} = a + \frac{(n - 1)d}{2}$