Tag: mean

Questions Related to mean

The harmonic mean of $\dfrac { a }{ 1-ab } and \dfrac { a }{ 1+ab }$ is:

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

  1. $a$

  2. $\dfrac { a }{ 1-{ a }^{ 2 }b^{ 2 } }$

  3. $\dfrac { 1 }{ 1-{ a }^{ 2 }b^{ 2 } }$

  4. $\dfrac { a }{ 1+{ a }^{ 2 }b^{ 2 } }$

Show answer
Correct Option: A
Explanation:

Here, $n=2$

$x _1=\dfrac{a}{1-ab}$ and $x _2=\dfrac{a}{1+ab}$
Then, $\dfrac{1}{x _1}=\dfrac{1-ab}{a}$ and $\dfrac{1}{x _2}=\dfrac{1+ab}{a}$

$\Rightarrow$  Harmonic mean $=\dfrac{n}{\dfrac{1}{x _1}+\dfrac{1}{x _2}}$

                                  $=\dfrac{2}{\dfrac{1-ab}{a}+\dfrac{1+ab}{a}}$

                                  $=\dfrac{2a}{2}$

                                  $=a$

The relation among AM, GM and HM is

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

  1. $AM\times GM=HM$

  2. $HM=\sqrt{AM\times GM}$

  3. $GM^2=AM\times HM$

  4. $AM^2=GM\times HM$

Show answer
Correct Option: C
Explanation:

For any given numbers, their means, Arithmetic mean (AM), Geometric mean, (GM) and Harmonic mean are related by the relation
$ GM^2 = AM \times HM $

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

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

  1. Median

  2. Mode

  3. Arithmetic mean

  4. Mean deviation

Show answer
Correct Option: C
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)


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.

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

  1. $\displaystyle 15\frac{1}{6}$

  2. $\displaystyle 25\frac{1}{6}$

  3. $\displaystyle 35\frac{1}{6}$

  4. $\displaystyle 45\frac{1}{6}$

Show answer
Correct Option: B
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.

Below is given the distribution of money (in Rs.) collected by students for flood relief fund. Find mean of money (in Rs.) collected by a student

Money (in Rs.) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
No. of students   5    7    5    2    6
statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. Rs.$ 22.80$

  2. Rs.$ 23.80$

  3. Rs. $24.80$

  4. Rs.$ 25.80$

Show answer
Correct Option: B
Explanation:
Consider the following table, to calculate mean:
|  $ci$   |  $f _i$ |    $x _i$ |  $f _ix _i$ | | --- | --- | --- | --- | |  $0-10$    |   5 |  5 |  25 | |  $10-20$  |   7 | 15 |  105 | |  $20-30$  |  5 |  25 |  125 | |  $30-40$ |  2 |  35 |  70 | |  $40-50$ |  6 |  45 |  270  |
 $N=\Sigma f _i=25$          
 $\Sigma f _ix _i=595$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{595}{25}=23.80$
Hence, option $B$ is correct.

The measurements (in mm) of the diameters of the head of screws are given below :
Calculate mean diameter of head of a screw.

Diameter (in mm) 33 - 35 36 - 38 39 - 41 42 - 44 45 - 47
No. of screw   10   19   23   21   27
statistics measures of central tendency geometric and harmonic mean geometric mean mean

  1. $38.08$ mm

  2. $40.08$ mm

  3. $41.08$ mm

  4. $45.08 $ mm

Show answer
Correct Option: C
Explanation:

Consider the following table, to calculate mean:

 $ci$  $f _i$  $x _i$  $f _ix _i$
 32.5-35.5  10  34  340
 35.5-38.5  19  37  703
 38.5-41.5  23  40  920
41.5-44.5  21  43  903
44.5-47.5  27  46  1242
 $N=\Sigma f _i=100$          
 $\Sigma f _ix _i=4108$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{4108}{100}=41.08$
Mean diameter of head of a screw is $41.08$mm
Hence, option $C$ is correct.

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

  1. $37.12$ marks

  2. $41.5$ marks

  3. $44.26$ marks

  4. $48.4$ marks

Show answer
Correct Option: D
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 $= 48.4$

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?

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

  1. $8.2$ plants

  2. $6.5$ plants

  3. $5.7$ plants

  4. None of these

Show answer
Correct Option: D
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.

Record of no. of days of medical leave taken by $ 30$ employees within a year is given below.

No. of days 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
No. of employees 5 7 11 4 3

Find mean number of days of medical leave taken by an employee in a year.

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

  1. $29.67$ days

  2. $25.67$ days

  3. $22.67$ days

  4. $20.6$ days

Show answer
Correct Option: C
Explanation:

Consider the following table, to calculate mean:

 $ci$  $f _i$  $x _i$ $f _ix _i$ 
 0-10  5  5  25
 10-20  7  15  105
 20-30  11  25  275
 30-40  4  35  140
 40-50  3  45  135
$N=\Sigma f _i=30$          
 $\Sigma f _ix _i=680$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{680}{30}=22.67$
Mean number of days of medical leave taken by an employee in a year is $22.67$ days.
Hence, option $C$ is correct.

To find the concentration of $SO _2$ in the air (in parts, per

million), the data was collected for 30 localities, in a certain city

and is presented below:

Concentration of $SO _2$ (in ppm) Frequency
0.00-0.04 4
0.04-0.08 9
0.08-0.12 9
0.12-0.16 2
0.16-0.20 4
0.20-0.24 2

Find the mean concentrations of $SO _2$ in the air.

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

  1. $0.099$ ppm

  2. $0.09$ ppm

  3. $0.99$ ppm

  4. $0.0909$ ppm

Show answer
Correct Option: A
Explanation:

Consider the following table, to calculate mean:

$ci$ $f _i$  $x _i$  $f _ix _i$ 
$0.00-0.04$ $4$  $0.02$  $0.08$ 
$0.04-0.08$  $9$  $0.06$  $0.54$ 
$0.08-0.12$  $9$  $0.10$  $0.90$ 
$0.12-0.16$  $2$  $0.14$  $0.28$ 
$0.16-020$  $4$  $0.18$  $0.72$ 
$0.20-0.24$  $2$  $0.22$  $0.44$ 
$N=\Sigma f _i=30$          
 $\Sigma f _ix _i=2.96$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{2.96}{30}=0.0986667 \approx 0.099$
mean concentration of $SO _2$ in air is $0.099ppm$
Hence, option $A$ is correct.