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

If $n=10, \bar{x}=12$ and $\sum x^2=1530$, then calculate the coefficient of variation.

  1. $20$
  2. $25$
  3. $30$
  4. $35$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\sigma=\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$

   
   $=\sqrt{\dfrac{1530}{10}-(12)^2}$

   $=\sqrt{153-144}$
   $=\sqrt{9}$
   $=3$

Coefficient of variation $=\dfrac{\sigma}{\overline{x}}\times 100$

                                       $=\dfrac{3}{12}\times 100$

                                       $=\dfrac{1}{4}\times 100$
                                       $=25$

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

$\bar{x} = A + \dfrac{\sum fd}{N}$ is the formula of

  1. Median

  2. Mode

  3. Arithmetic mean

  4. Mean deviation
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\text{Arithmetic mean for ungrouped data:-}$ Arithmetic mean for ungrouped data is the average of the values in the data set. i.e., sum of the values in the data set divided by the total number of values in the data set.

for example, if the data set is $x _1,x _2,x _3,...,x _n$ then the mean is $\bar x =\dfrac{x _1+x _2+x _3+...+x _n}n=\dfrac{\sum x _i}n$

$\text{Arithmetic mean for grouped data:-}$ If $x _1,x _2,x _3,...,x _n$ are the values with corresponding frequencies $f _1,f _2,f _3,...,f _n$ then the arithmetic mean is $\bar x =\dfrac{f _1x _1+f _2x _2+f _3x _3+...+f _nx _n}{f _1+f _2+...+f _n}=\dfrac{\sum f _ix _i}{\sum f _i}$

Alternate formula for arthmetic mean of grouped data is $\bar x =A+\dfrac{\sum f _id _i}{\sum f _i}=A+\dfrac{\sum f _id _i}N$

where $d _i=(x _i-A)$,
$N=\sum f _i$ and
A is assumed mean (usually middle term)


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

Find the mean marks from the following data:

Marks Number of students
Below 10 5
Below 20 9
Below 30 17
Below 40 29
Below 50 45
Below 60 60
Below 70 70
Below 80 78
Below 90 83
Below 100 85
  1. $37.12$ marks
  2. $41.5$ marks
  3. $44.26$ marks
  4. $48.4$ marks
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Answer:- Taking continuous class interval with width $= 10$

$ \therefore $ For $"90-100" = Below\ 100 - Below\ 90 = 85 - 83 = 2$
$ \therefore $ For $"80-90" = Below\ 90 - Below\ 80 = 83 - 78 = 5$
Similarly for other classes also.
Thus our frequency distribution table is 

 Marks Students$ f _i $  $ x _i = \cfrac{\text{lower limit + upper limit}}{2} $  $ f _ix _i $ 
0-10  5  5  25
10-20  4  15  60
20-30 8  25 200
30-40  12   35 420 
 40-50 16   45 720 
50-60 15   55 825 
60-70 10   65 650 
70-80  8  75  600
80-90  5   85  425
90-100  2   95  190
  $ \Sigma f _i = 85 $    $ \Sigma f _ix _i = 4115 $ 

$ \therefore $ Mean $= 48.4$

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

In a frequency dist. if $\displaystyle d _{i}$ is deviation of variates from a number e and mean = $\displaystyle e+\frac{\Sigma f _{i}d _{i}}{\Sigma f _{i}}$, then e is

  1. Lower limit

  2. Assumed mean

  3. Number of observation

  4. Class interval
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Formula of finding Mean using step deviation method is

mean = $\displaystyle e+\frac{\Sigma f _{i}d _{i}}{\Sigma f _{i}}$
where,
$e=$Assumed Mean
$\Sigma f _id _i=$Sum of all $frequency(f _i)\times deviation(d _i)$
$\Sigma f _i=$ Sum of all frequencies
Hence the correct answer is assumed mean.

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 geometric mean of $10$ observations on a certain variable was calculated as $16.2$. It was later discovered that one of the observations was wrongly recorded as $12.9$; infact it was $21.9$. The correct geometric mean is:

  1. $\left (\dfrac {(16.2)^{9}\times 21.9}{21.9}\right )^{1/10}$
  2. $\left (\dfrac {(16.2)^{10}\times 21.9}{21.9}\right )^{1/10}$
  3. $\left (\dfrac {(16.2)^{10}\times 21.9}{12.9}\right )^{1/10}$
  4. $\left (\dfrac {(16.2)^{11}\times 21.9}{21.9}\right )^{1/11}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Geometric mean of $n$ numbers $=(\prod _{i=1}^{n} x _{i})^{1/n}$

Here, $(\prod _{i=1}^{10} x _{i})^{1/10}=16.2$
$\Rightarrow (\prod _{i=1}^{10} x _{i})=(16.2)^{10}$
Now suppose $x _{10}$ was wrongly recorded, so we rewrite above relation as $(\prod _{i=1}^{9} x _{i})\times x _{10}=(16.2)^{10}$
$\Rightarrow (\prod _{i=1}^{9} x _{i})=\dfrac{(16.2)^{10}}{x _{10}}$
Now, the correct value is $21.9$, so multiply both sides by $21.9$ and also put value of $x _{10}=12.9$ in above equation
$(\prod _{i=1}^{9} x _{i})\times 21.9=\dfrac{(16.2)^{10}}{12.9}\times 21.9$
$=$ Correct Geometric mean=$((\prod _{i=1}^{9} x _{i})\times 21.9)^{1/10}$
$=\left(\dfrac{(16.2)^{10}}{12.9}\times 21.9\right)^{1/10}$
$=\left(\dfrac{(16.2)^{10}\times 21.9}{12.9}\right)^{1/10}$
Hence, $(C)$ 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 $

Multiple choice business mathematics and statistics applied statistics weighted methods to calculate index numbers construction of index numbers index numbers

The most appropriate average in averaging the price relatives is:

  1. Median

  2. Harmonic mean

  3. Arithmetic mean

  4. Geometric mean
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Rightarrow$  The most appropriate average in averaging the price relative is : $Geometric\,\, mean.$

$\Rightarrow$  The geometric mean is the average of a set of products, the calculation of which is commonly used to determine the performance results of an investment or portfolio. It is technically defined as "the 'n'th root product of 'n' numbers.
$\Rightarrow$  The geometric mean must be used when working with percentages, which are derived from values.

Multiple choice business mathematics and statistics index numbers weighted methods to calculate index numbers construction of index numbers applied statistics

Construct a composite index number as a weighted mean from the following data:

Index Number 122 145 101 98 137 116
Weight 7 2 4 1 6 5

  1. 120

  2. 122

  3. 130

  4. 132
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
 $Index\,number$        $I$ $Weight$$w$  $I.w$ 
 $122$ $7$  $854$ 
$145$  $2$  $290$ 
$101$  $4$  $404$ 
$98$  $1$  $98$ 
$137$  $6$  $822$ 
$116$  $5$  $580$ 
$Total$  $\sum w=25$  $\sum Iw=3048$ 

$\Rightarrow$  Composite index number = $\dfrac{\sum Iw}{\sum w}=\dfrac{3048}{25}=121.92\approx 122$.

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

For a symmetrical distribution lower quartitl is 20 and upper quartile is 40.The value of 50th percentile is

  1. 20

  2. 40

  3. 30

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

First quartile also called the lower quartile or the 25th percentile(splits off the lowest 25% of data from the highest 75%)
Second quartile also called the median or the 50th percentile (cuts data set in half)
Third quartile  also called the upper quartile or the 75th percentile (splits off the highest 25% of data from the lowest 75%)
Since its a symmetrical distribution therefore the median will be 30