Quantitative Aptitude

Statistics

606 Questions
All Ungrouped data dispersion and consistency 209 Unknown terms from given average 124 Grouped data central tendencies 101 Merging means and variances of groups 53 Replacement change in average 27 Alligation and weighted average 21 Data interpretation from tables and charts 20 Net successive percentage change 19 Grouped data variance via step deviation 18 Combination of multiple ratios 1

Statistics Questions

Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

Using step deviation method find the mean.

X 20-40 40-60 60-80 80-100
frequency 4 8 12 16


  1. $40$
  2. $50$
  3. $60$
  4. $70$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 X Mid point (X)  Frequency (F)  i=class interval width  A=50 Assumed Means  d'=$\dfrac{x-A}{i}$  fd' 
 20-40 30  4 20  50  -1  -4 
40-60  50=A  8  20  50  0  0 
60-80  70  12  20  50  1  12 
80-100  90  16  20  50  2  32 
    $\Sigma f=40$        $\Sigma f d'=40$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \frac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =50 +\dfrac{40}{40}\times 20$
$= 50 + 20$
$= 70$

Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

 Marks 0-10  10-20  20-30  30-40  40-50  50-60  60-70  70- 80 80-90  90-100 
 Frequency 3  5  6  7  8   9  10  12  6 4 


Find the mean mark using step deviation method:

  1. $54$
  2. $55$
  3. $56$
  4. $57$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 X Mid Point (X)  Frequency (F)  i=class interval width  A=45 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
 0-10 5  3  10  45  -4  -12 
10-20  15  5  10  45  -3  -15
20-30  25  6  10  45  -2  -12 
30-40  35  7  10  45  -1  -7 
40-50  45=A 8  10  45  0  0 
50-60  55  9  10  45  1  9 
60-70  65  10  10  45  2  20 
70-80  75  12  10  45  3  36 
80-90  85  6  10  45  4  24 
90-100   95 4  10  45  5  20 
    $\Sigma f=70$        $\Sigma fd'=63$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =45 +\dfrac{63}{70}\times 10$
$= 45 + 9$
$= 54$

Multiple choice maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

 Marks 0-5  5-10  10-15  15-20  20-25  25-30  30-35  35-40  40-45  45-50 
Frequency  3  5  7  8  10  11  14  19  15  13 

For the following distribution, find the mean using step deviation method. (Round off your answer to the nearest whole number)

  1. $29$
  2. $31$
  3. $35$
  4. $37$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 X Mid point (x)  Frequency (F)  i=class  interval width  A=22.5 Assumed mean  d'=$\dfrac{x-A}{i}$  fd' 
0-5  2.5  3  5  22.5  -4  -12 
5-10  7.5  5  5  22.5  -3  -15 
10 -15 12.5  7  5  22.5  -2  -14 
15-20  17.5  8  5  22.5  -1  -8 
20-25  22.5=A  10  5  22.5  0  0 
25-30  27.5  11  5  22.5  1  11 
30-35  32.5  14  5  22.5  2  28 
35-40  37.5  19  5  22.5  3  57 
40-45  42.5  15  5  22.5  4  60 
45-50  47.5  13  5  22.5  5  65 
    $\Sigma f=105$        $\Sigma fd'=172$ 


The formula used for arithmetic mean of grouped data by step deviation method is, $\overline {X} =A + \dfrac{\sum fd'}{\sum f} \times i$ 
$A =$ Assumed mean of the given data
$\sum$ = Summation of the frequencies given in the grouped data
$\sum fd'$ = Summation of the frequencies and deviation of a given mean data
$d' = \dfrac{(x - A)}{i}$
$i =$ Class interval width
$\overline {X}$ = arithmetic mean
$\overline {X} =22.5 +\dfrac{172}{105}\times 5$
$= 22.5 + 8.19$
$= 30.69$ $\approx$ $31$

Multiple choice statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

$n=25,\sum x=125,\sum x^2=650, \sum y=100,\sum y^2=460,\sum xy=508$. It was observed that two pair of values of $(x,y)$ were copied as $ (6,14)$ and $(8,6) $ instead of $(8,12),(6,8).$ The correct correlation coefficient is 

  1. $0.667$
  2. $0.87$
  3. $-0.25$
  4. $0.356$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Corrected $\sum x=125-6-8+8+6=125$

Corrected $\sum y=100-14-6+12+8=100$
Corrected $\sum x^2=650-(6)^2-(8)^2+(6)^2+(8)^2=650$
Corrected $\sum y^2=460-(14)^2-(6)^2+(12)^2+(8)^2=436$
Corrected $\sum xy=508-6\times 18-8\times 6+8\times 12+6\times 8=520$
The formula for Pearson product moment correlation is
$r=\dfrac{n\sum xy-(\sum x)(\sum y)}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}$

    $=\dfrac{25\times 520-125\times 100}{\sqrt{[25\times 650-(125)^2][25\times 436-(100)^2]}}$

   $=\dfrac{13000-12500}{\sqrt{[16250-15625][10900-10000]}}$

   $=\dfrac{500}{\sqrt{[625][900]}}$

   $=\dfrac{500}{(25)(30)}$

   $=\dfrac{2}{3}$

   $=0.667$

Multiple choice economics measures of dispersion coefficient of variance variance and standard deviation statistics and probability range and mean deviation

If $\sum\limits _{i = 1}^9 {\left( {{x _i} - 5} \right) = 9}$ and $\sum\limits _{i = 1}^9 {{{\left( {{x _i} - 5} \right)}^2}}  = 45$, then the standard deviation of the $9$ items ${x _1},{x _2},.....,{x _9}$ is

  1. $2$
  2. $3$
  3. $9$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
S.D of $xi-5$ is

$\sigma =\sqrt{\dfrac{\sum _{i=1}^{9}(xi-5)^2}{9}-\left [ \dfrac{\sum _{i=1}^{9}(xi-5)^2}{9} \right ]^2}$

$\sigma =\sqrt{5-1}=2$
Multiple choice maths measures of dispersion coefficient of variance variance and standard deviation statistics and probability range and mean deviation

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250 $. Then Co-efficient of variation is

  1. $10\%$
  2. $40\%$
  3. $50\%$

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\sum(x-\overline{x})^2=250$, $\overline{x}=50$
$\Rightarrow$  Standard deviation $(\sigma)=\sqrt{\dfrac{250}{10}}=\sqrt{25}=5$
$\Rightarrow$  Coefficient of variation $=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}}$
                                             $=\dfrac{\sigma}{Mean}\times 100$

                                             $=\dfrac{5}{50}\times 100$

                                             $=10\%$
Multiple choice maths measures of dispersion coefficient of variance variance and standard deviation statistics and probability range and mean deviation

The sum of the squares of deviation of 10 observations from their mean 50 is 250, then coefficient of varition is

  1. 10%

  2. 40%

  3. 50%

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

Given $\displaystyle \Sigma \left ( x _{i}-\overline{x} \right )^{2}=250$,$n=10,\overline{x}=50$

Now, $\sigma=\sqrt{\dfrac{1}{n}\Sigma \left ( x _{i}-\overline{x} \right )^{2}}$

$= \sqrt{\dfrac{1}{10}\times 250}=5$ 
Hence coefficient of variation $\displaystyle =\dfrac{\sigma }{\overline{x}}\times 100=\dfrac{5}{50}\times 100=10$%

Multiple choice maths measures of dispersion coefficient of variance variance and standard deviation statistics and probability range and mean deviation

The sum of the squares of deviation of 10 observations from their mean 50 is 250, then coefficient of variation is

  1. 10%

  2. 40%

  3. 50%

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,   $\sum (x-\bar{x})^2 = 250, n = 10, \bar{x} =50$
Thus standard deviation $ = \sqrt{\cfrac{\sum (x-\bar{x})^2}{n}}=\sqrt{25}=5$
$\therefore$ Coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100 =\cfrac{5}{50}\times 100$ % $= 10$%

Multiple choice maths measures of dispersion coefficient of variance variance and standard deviation statistics and probability range and mean deviation

The mean of a distribution is 4. If its coefficient of variation is 58%. Then the S.D. of the distribution is

  1. 2.23

  2. 3.23

  3. 2.32

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

Given,  mean $\bar{x} = 4,$ and coefficient of variation $=58$ %
If S.D of the given distribution is $\sigma$ then we know that,
Coefficient of variation $=\cfrac{\sigma}{\bar{x}}\times 100$ %
$\Rightarrow 58 = \cfrac{\sigma}{4}\times 100\Rightarrow \sigma = \cfrac{58\times 4}{100}=2.32$

Multiple choice maths measures of dispersion coefficient of variance variance and standard deviation statistics and probability range and mean deviation

The mean of a distribution is $14$ and standard deviation is $5$. What is the value of the coefficient of variation?

  1. $57.7\%$
  2. $45.7\%$
  3. $35.7\%$

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

Coefficient of variation is given by $CV = \dfrac{SD}{Mean}\times 100 $
$\Rightarrow \dfrac{5}{14}\times 100 = 35.7\%$

Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean
Class-intervals 0-10 10-20 20-30 30-40 40-50
Frequency   12    11    14    10    13

Find the arithmetic mean for the given grouped frequency distribution.

  1. $\displaystyle 15\frac{1}{6}$
  2. $\displaystyle 25\frac{1}{6}$
  3. $\displaystyle 35\frac{1}{6}$
  4. $\displaystyle 45\frac{1}{6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider the following table, to calculate mean:

$ci$  $f _i$ $x _i$  $f _ix _i$
 0-10  12  5  60
 10-20  11 15  165
 20-30  14  25  350
 30-40  10  35  350
 40-50  13  45  585
 $N=\Sigma f _i=60$          
 $\Sigma f _ix _i=1510$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{1510}{60}=25\dfrac16$
Hence, option $B$ is correct.
Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

  1. $8.2$ plants
  2. $6.5$ plants
  3. $5.7$ plants

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

Answer: Using direct method to calculate mean :


No. of plants   No. of houses   $x _i$ $f _i.x _i$ 
0-2  1   1 1 
 2-4 2  3  6 
4-6   1 5  5 
6-8  5  7  35 
8-10  6  9  54 
10-12  2  11  22 
12-14  3  13  39 
  $\Sigma f _i=20$    $\Sigma f _i.x _i=162$ 

Mean=$\cfrac{\Sigma f _i.x _i}{\Sigma f _i}=\cfrac{162}{20}=8.1\;plants$

We used direct method for finding the mean as the width of the class is very small and also the frequency for each class have very small values. Thus, it will be easy to calculate.

Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

Compute the missing frequencies $'f _1'$ and $'f _2'$ in the following data, if the mean is $166\frac {9}{26}$ and the sum of the observation is 52.

Classes Frequency
140-150 5
150-160 $f _1$
160-170 20
170-180 $f _2$
180-190 6
190-200 2
Total 52
  1. $f _1=7, f _2=3$
  2. $f _1=10, f _2=6$
  3. $f _1=9, f _2=8$

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

Given:- $\Sigma f = 52 $

 classes frequency$ (f _i) $  $ x _i = \cfrac{\text{lower limit + upper limit}}{2} $  $ x _i f _i $ 
140-150  5   145  725
150-160  $ f _1 $  155  $ 155f _1 $ 
 160-170 20  165  3300 
170-180  $ f _2 $  175  $175f _2$ 
180-190  6  185  1110 
190-200  2 195  390 
  $ \Sigma f _i = 52 $    $ \Sigma x _i f _i = 5525 + 155f _1 + 175f _2 $ 

Also $ \Sigma f _i = 33  f _1 + f _2 = 52 $

$ \Rightarrow f _1 + f _2 = 19\longrightarrow eq.(i) $
Now Mean = $ \cfrac{\Sigma x _i f _i}{\Sigma f _i} = \cfrac{5525 + 155f _1 + 175f _2}{52} $ 
Given:- Mean = $ \cfrac{4325}{26} $
$ \Rightarrow \cfrac { 5525+155f _{ 1 }+175f _{ 2 } }{ 52 } =\cfrac { 4325 }{ 26 } $
$ \Rightarrow 5525 + 155f _1 + 175f _2 = 8650 $
$ \Rightarrow 155f _1 + 175f _2 = 8650 - 5525 = 3125$
$ 31f _1 + 35f _2 = 625 \longrightarrow eq.(ii) $
from eq. (i) $ & $ (ii), we get
$ f _2 = 9 \Rightarrow f _1 = 10 $
D) None of these

Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

If the mean of four observations is $20$ and when a constant  is added to each observation the mean becomes $22$ The value of $c$ is?

  1. $-2$
  2. $2$
  3. $4$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle\text{Let } x _1,x _2,x _3,x _4\text{ be four observations.}$
$\displaystyle\text{According to question}$
$\displaystyle \frac{ x _1+x _2+x _3+x _4}{4}=20$
$\Rightarrow \displaystyle { x _1+x _2+x _3+x _4}=80$
$\displaystyle\text{After adding 'c' to each observation the new A.M becomes 22.}$
$\Rightarrow \displaystyle \frac{ (x _1+c)+(x _2+c)+(x _3+c)+(x _4+c)}{4}=22$
$\Rightarrow \displaystyle  (x _1+x _2+x _3+x _4)+4c=88$
$\Rightarrow \displaystyle  80+4c=88$
$\Rightarrow \displaystyle  4c=8$
$\Rightarrow \displaystyle  c=2$
Options B is correct.

Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency $\displaystyle f _{1}$ and $\displaystyle f _{2}$.

Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 5 $\displaystyle f _{1}$ 10 $\displaystyle f _{2}$ 7 8
  1. $5, 8$
  2. $6, 12$
  3. $8, 11$
  4. $8, 12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Class       Frequency(f)  ClassMark (x)         fx
0-20         5      10         50
20-40     ${f} _{1} $      30       $ 30{f} _{1} $
40-60         10       50          500
60-80      ${f} _{2} $        70        $ 70{f} _{2} $
80-100          7         90          630
100-120           8         110           880
Total $30 + {f} _{1} +{f} _{2} $   $ 2060  + 30{f} _{1}+70{f} _{2} $

Given $ 30 + {f} _{1} +{f} _{2} = 50 $
$ => {f} _{1} +{f} _{2} = 20 $   -- (1)

Given, Mean $ = \cfrac { \sum { fx }  }{ \sum { f }} =62.8 $
$ => \cfrac { 2060  + 30{f} _{1}+70{f} _{2}}{30 + {f} _{1} +{f} _{2}} = 62.8 $

$ =>  2060  + 30{f} _{1}+70{f} _{2} = 1884 + 62.8{f} _{1} + 62.8{f} _{2} $ 

$ 32.8{f} _{1} - 7.2{f} _{2} =176 $

=> $ 8.2{f} _{1} - 1.8{f} _{2} = 44 $

=> $ 4.1{f} _{1} - 0.9{f} _{2} = 22 $ -- (2)

Solving both equations 1, 2, we get
$ {f} _{1} = 8, {f} _{2} = 12 $