Tag: measures of dispersion

Questions Related to measures of dispersion

The mean weight of 9 students is 24 kg. If one more student is joined in the group the mean is unaltered, then the weight of the 10th student is

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

  1. 25 kg

  2. 24 kg

  3. 26 kg

  4. 23 kg

Show answer
Correct Option: A
Explanation:
The sum of weights of the 9 students = $25\times 9=225$ kg
If one more student is joined in the group, then total number of students is 10, and the mean isn 25.
The sum of weights of the 10 students is $25\times 10=250$ kg
The weight of the tenth student is $250-225=25$ kg

The arithmetic mean in a measure of central tedency and is popularlyknown as mean. Arithmetic mean is obtained by dividing the sum of the values of all items of a series by the number of items of that series. Normally, arithmetic mean is denoted by $\bar X$ which is red as  '$X$ bar'. It can be computed for unclassified or ungrouped data or individual series as well as classified or grouped data or discrete or continuous series.

From the following data calculate arithmentic mean.

Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of students 10 20 30 50 40 30


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

  1. $25$

  2. $40$

  3. $15$

  4. $35$

Show answer
Correct Option: D
Explanation:

b'Let us take assumed mean $=$ 45
Calculation of deviations from assumed mean 

Marks No. of (X-45)/10
   X m students  f      d   fd
0-10 5 10 -4 -40
10-20 15 20 -3 -60
20-30 25 30 -2 -60
30-40 35 50 -2 -50
40-50 45 40 0 0
50-60 55 30 +1 30

                              N $=$180         +9              $\sum fd = $-180
Mean $A \displaystyle + \frac{\sum fd}{N} \times c = 45+ \frac{-180 \times 10}{180} = 35$'

The standard deviation and mean diameter of the circles respectively are

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

  1. $5.55,44,5$

  2. $5.25,43,5$

  3. $5.55,44,5$

  4. $5.25,44,5$

Show answer
Correct Option: A

Following table gives frequency distribution of trees planted by different housing societies in a particular locality.
Find mean number of trees planted by housing society by using 'step deviation method'.

No. of tress 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40
No. of Societies 2 7 9 8 6 4
maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $25.42$ trees

  2. $27.42$ trees

  3. $29.42$ trees

  4. $31.42$  trees

Show answer
Correct Option: A
Explanation:

Consider the following table, to calculate mean by "step deviation method":

$x _i$=mid value of class interval
Assumed mean $a=27.5$
class interval $c=5$

 $ci$  $f _i$  $x _i$  $d _i=\dfrac{x _i-a}{c}$ $f _id _i$
 10-15 2   12.5  $\dfrac{12.5-27.5}{5}=$-3  -6
 15-20 7  17.5  -2  -14
 20-25 9   22.5  -1  -9
 25-30 8  27.5  0  0
 30-35 6   32.5  1  6
 35-40 4   37.5  2  8
 $N=\Sigma f _i=36$          
 $\Sigma f _id _i=-15$

Mean $\overline x=a +\dfrac {\Sigma f _id _i}{N}\times c$

$\therefore \overline x=27.5 + \dfrac{-15}{36} \times 5=25.42$

Hence, option $A$ is corect.

Following table gives age groupwise distribution of people suffering from 'Asthama' due to air pollution in certain city. Find mean age of person suffering from 'Asthama' .

Age (in years) 7 - 11 11 - 15 15 - 19 19 - 23 23 - 27 27 - 31 31 - 35 35 - 39
No. of People  5  9  13  21  16  15  12   9
maths measures of dispersion mean of grouped data : step deviation method step deviation method of finding mean step deviation method

  1. $23.88$years

  2. $24.88$ years

  3. $25.88 $years

  4. $26.88$ years

Show answer
Correct Option: A
Explanation:

Consider the following table, to calculate mean:

 $ci$ $ f _i$  $x _i$  $f _ix _i$
 7-11  5  9  45
 11-15  9  13  117
 15-19  13  17  221
 19-23  21  21  441
 23-27  16  25  400
 27-31  15  29  435
 31-35  12 33   396
 35-39  9  37  333
$N=\Sigma f _i=100$          
 $\Sigma f _ix _i=2388$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{2388}{100}=23.88$
Mean age of people suffering from Asthama is $23.88$years
Hence, option $A$ is correct.

Find the arithmetic mean for the following grouped frequency distribution:

Class-intervals$ $6-10$ $10-14$ $14-18$ $18-22$  $22-26$  $26-30$
Frequency  $ 4$   $ 6$   $9$   $ 12$    $ 7 $    $2$


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

  1. $15.8$

  2. $16.8$

  3. $17.8$

  4. None of these

Show answer
Correct Option: C
Explanation:
Consider the following table, to calculate mean:
|  $ci$ |  $f _i$ | $x _i$  |  $f _ix _i$ | | --- | --- | --- | --- | |  $6-10$ |  $4$ | $ 8$ | $ 32$ | | $ 10-14$ |  $6$ |  $12$ |  $72$ | |  $14-18$ |  $9$ |  $16$ | $ 144$ | |  $18-22$ | $12$ | $ 20$ |  $240$ | | $ 22-26$ |  $7$ | $ 24$ |  $168$ | |  $26-30$ | $ 2$ |  $28$ |  $56$ |
$N=\Sigma f _i=40$          
$\Sigma f _ix _i=712$

Mean $\overline x=\dfrac {\Sigma f _ix _i}{N}$
$\therefore \overline x=\dfrac{712}{40}=17.8$
Hence, option $C$ is correct.

Following table shows frequency distribution of advertisement on T.V. by shift of origin ans scale method.

Duration (in sec.) 25 - 30 30 - 35 35 - 40 40 - 45 45 - 50 50 - 55
No. of advertisements 10 32 15 9 7 2

Obtain mean duration of advertisement on T.V. by shift of origin and scale method.

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

  1. $31.97$ seconds

  2. $32.97$ seconds

  3. $34.97$ seconds

  4. $35.97$ seconds

Show answer
Correct Option: D
Explanation:

Consider the following table, to calculate mean by "shift of origin method":

$x _i$=mid value of class interval
Assumed mean $a=42.5$

 $ci$  $f _i$  $x _i$ $d _i=x _i-a$   $f _id _i$
 $25-30$  $10$  $27.5$  $-15$  $-150$
 $30-35$  $32$  $32.5$  $-10$  $-320$
 $35-40$  $15$  $37.5$  $-5$  $-75$
 $40-45$  $9$  $42.5$  $0$  $0$
 $45-50$  $7$  $47.5$  $5$  $35$
 $50-55$  $2$  $52.5$  $10$  $20$
 $N=\Sigma f _i=75$          
 $\Sigma f _id _i=-490$
Mean $\overline x=a +\cfrac {\Sigma f _id _i}{N}$
$\therefore \overline x=42.5 + \cfrac{-490}{75}=35.966667 \Rightarrow 35.97$

Mean duration of advertisement on T.V is $35.97$ seconds 
Hence, option $D$ is correct.

Number of calories (in' 00) consumed daily by a sample of 15 years old boys are given below.

Calories 1000 - 1500 1500 - 2000 2000 - 2500 2500 - 3000 3000 - 3500 3500 - 4000 4000 - 4500
No. of boys 5 13 16 18 27 10 4

Obtain mean calories consumed daily by a boy by step deviation method.

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

  1. $2222$ calories

  2. $2548$ calories

  3. $2563$ calories

  4. $2761$ calories

Show answer
Correct Option: D
Explanation:
Consider the following table, to calculate mean by "step deviation method":
$x _i$=mid value of class interval
Assumed mean $a=2750$
class interval $c=500$

 $ci$  $f _i$  $x _i$  $d _i=\dfrac{x _i-a}{c}$  $f _id _i$ 
 $1000-1500$  $5$  $1250$  $-3$  $-15$
 $1500-2000$  $13$  $1750$  $-2$  $-26$
 $2000-2500$  $16$  $2250$  $-1$  $-16$
 $2500-3000$  $18$  $2750$  $0$  $0$
 $3000-3500$  $27$  $3250$  $1$  $27$
 $3500-4000$  $10$  $3750$  $2$  $20$
 $4000-4500$  $4$  $4250$  $3$  $12$

$N=\Sigma f _i=93$          
$\Sigma f _id _i=2$

Mean $\overline x=a +\dfrac {\Sigma f _id _i}{N}\times c$

$\therefore \overline x=2750 + \dfrac{2}{93} \times 500=2760.75 \approx 2761$
mean calories consumed daily by a boy is $2761$ calories. 

Hence, option $D$ is correct.

Consider the table given below

Marks 0-10 10-20 20-30 30-40 40-50 50-60
Number of Students 12 18 27 20 17 6

The arithmetic mean of the marks given above is

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

  1. $18$

  2. $28$

  3. $27$

  4. $6$

Show answer
Correct Option: B
Explanation:
$ x _i = \cfrac{\text{upper limit + lower limit}}{2} $

| Marks | No. of students $ \left( f _i \right) $ | $ x _i $  | $ x _i f _i $  | | --- | --- | --- | --- | | 0-10  | 12 | 5  | 60  | | 10-20  | 18  | 15  | 270  | | 20-30  | 27  | 25  | 675  | | 30-40  | 20  | 35  | 700  | | 40-50  | 17  | 45  | 765  | | 50-60  | 6  | 55  | 330  | |   | $ \Sigma{f _i} = 100 $  |   | $ \Sigma{x _i f _i} = 2800 $  | $ \therefore \; Mean=\cfrac { \Sigma { x _{ i }f _{ i } } }{ \Sigma { f _{ i } } } = \cfrac{2800}{100} = 28$
B) 28

The marks obtained by $20$ students of Class $X$ of a certain school in a English paper consisting of $100$ marks are presented in table below. Find the mean of the marks obtained by the students using step deviation method.

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

  1. $61$

  2. $62$

  3. $63$

  4. $64$

Show answer
Correct Option: C
Explanation:

Answer:- Using Shortcut Method

Class interval width (i) = $50-40 = 10$

Marks  X  F  $d=\cfrac{X-A}{i}$  Fd 
40-50  45  5  -2  -10 
50-60  55   4  -1 -4 
60-70  65 = A  3  0 0 
70-80  75   6  1 6 
80-90  85   2  2 4 
    $\Sigma F = 20$    $\Sigma Fd = -4$ 

Mean ( By step deviation formula) = $A+\cfrac{\Sigma Fd}{\Sigma F} \times i = 65+\left(\cfrac{-4}{20} \times 10 \right) = 65=2 = 63$

C) Mean = $63$