25. Chapter 25. Measures of Central Tendency NIOS Class 10 Mathematics Textbook Solutions

Chapter 25. Measures of Central Tendency NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation

Chapter 25. MEASURES OF CENTRAL TENDENCY

CHECK YOUR PROGRESS 25.1

1. Write formula for calculating mean of n observations  .

Solution: We have,   

2. Find the mean of first ten natural numbers.

Solution: We have,   Mean

3. The daily sale of sugar for 6 days in a certain grocery shop is given below. Calculate the mean daily sale of sugar.

Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
74 kg	121 kg	40 kg	82 kg	70.5 kg	130.5 kg

Solution: We have,

Mean  

4. The heights of 10 girls were measured in cm and the results were as follows:

142  ,  149  ,  135  , 150  , 128  , 140  , 149  , 152  , 138  , 145

Find the mean height.

Solution: We have,

Mean   

5. The maximum daily temperature (in °C) of a city on 12 consecutive days are given below:

32.4       29.5       26.6       25.7      23.5      24.6    24.2     22.4     24.2     23.0     23.2     28.8

Calcualte the mean daily temperature.

Solution: We have,

Mean  

6. The enrolment in a school in last five years was 605, 710, 745, 835 and 910. Verify that the sum of deviations of  from their mean () is 0.

Solution: Here,

We have,  

 

Now,     ,    ,   ,    , 

   

Thus, the sum of deviations of  from their mean () is 0.      Verified.

7. Mean of 9 observatrions was found to be 35. Later on, it was detected that an observation which was 81, was taken as 18 by mistake. Find the correct mean of the observations.

Solution: Given,  Number of observations = 9 , Incorrect mean = 35 ,

Incorrect observation recorded = 18 ,  Correct observation should have been = 81

Incorrect sum= 35 × 9 = 315
The recorded value was 18, but it should have been 81.

Correct sum = Incorrect sum –(18 + 81)

Correct sum = 315 – (18 + 81) = 378

Correct mean  

8. The mean marks obtained by 25 students in a class is 35 and that of 35 students is 25. Find the mean marks obtained by all the students.

Solution:  Given,  25 students, mean marks = 35 and also 35 students, mean marks = 25

Combined total marks = 35 × 25 + 25 × 35 = 875 + 875 =1750

Total students = 25 + 35 = 60

The combined mean =  

CHECK YOUR PROGRESS 25.2

1. Find the mean marks of the following distribution:

Marks	1	2	3	4	5	6	7	8	9	10
Frequency	1	3	5	9	14	18	16	9	3	2

Solution:  We have,

Marks	Frequency
1 2 3 4 5 6 7 8 9 10	1 3 5 9 14 18 16 9 3 2	1 6 15 36 70 108 112 72 27 20
Total

Mean 

2. Calculate the mean for each of the following distributions:

(i)

x	6	10	15	18	22	27	30
f	12	36	54	72	62	42	22

(ii)

x	5	5.4	6.2	7.2	7.6	8.4	9.4
f	3	14	28	23	8	3	1

Solution: (i) We have,

6 10 15 18 22 27 30	12 36 54 72 62 42 22	72 360 810 1296 1364 1134 660
Total

Mean  

(ii) We have,

5 5.4 6.2 7.2 7.6 8.4 9.4	3 14 28 23 8 3 1	15 75.6 173.6 165.6 60.8 25.2 9.4
Total

Mean

3. The wieghts (in kg) of 70 workers in a factory are given below. Find the mean weight of a worker.

Weight (in kg)	60	61	62	63	64	65
Number of workers	10	8	14	16	15	7

Solution: We have,

Weights (in kg)  ()	Number of workers
60 61 62 63 64 65	10 8 14 16 15 7	600 488 868 1008 960 455
Total

Mean  

4. If the mean of following data is 17.45 determine the value of p:

	15	16	17	18	19	20
	3	8	10	P	5	4

Solution: We have,

15 16 17 18 19 20	3 8 10 P 5 4	45 128 170 18p 95 80
Total

Mean 

 

   

The value of p is 10 .

CHECK YOUR PROGRESS 25.3

1. Following table shows marks obtained by 100 students in a mathematics test

Marks	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Number of students	12	15	25	25	17	6

Calculate mean marks of the students by using Direct Method.

Solution: We have,

Marks	Class mark	Number of students
0 – 10 10 – 20 20 – 30 30 – 40  40 – 50  50 – 60	5 15 25 35 45 55	12 15 25 25 17 6	60 225 625 875 765 330
Total

Mean  

2. The following is the distribution of bulbs kept in boxes:

Number of bulbs	50 – 52	52 – 54	54 – 56	56 – 58	58 – 60
Number of boxes	15	100	126	105	30

Find the mean number of bulbs kept in a box. Which method of finding the mean did you choose?

Solution: We construct the frequency distribution table :

Class	Frequency	Class marks
50 – 52 52 – 54 54 – 56 56 – 58 58 – 60	15 100 126 105 30	51 53 55 57 59	– 4  – 2 0 2 4	– 60 – 200 0 210 120
Total

Here,

Using assumed mean method , we have 

Mean 

3. The weekly observations on cost of living index in a certain city for a particular year are given below:

Cost of living index	140 – 150	150 – 160	160 – 170	170 – 180	180 – 190	190 – 200
Number of weeks	5	8	20	9	6	4

Calculate mean weekly cost of living index by using Step deviation Method.

Solution: We construct the frequency distribution table :

Cost of living index	Class mark	Number of weeks
140 – 150 150 – 160 160 – 170 170 – 180 180 – 190 190 – 200	145 155 165 175 185 195	5 8 20 9 6 4	– 30 – 20 – 10 0 10 20	– 3 – 2 – 1 0 1 2	– 15 – 16 – 20 0 6 8
Total

Here,

Using step deviation method , we have

Mean  

4. Find the mean of the following data by using (i) Assumed Mean Method and (ii) Step deviation Method.

Class	150 – 200	200 – 250	250 – 300	300 – 350	350 – 400
Frequency	48	32	35	20	10

Solution: We construct the frequency distribution table :

Class	Frequency	Class marks
150 – 200 200 – 250 250 – 300 300 – 350 350 – 400	48 32 35 20 10	175 225 275 325 375	– 100 – 50 0 50 100	– 4800 – 1600 0 1000 1000
Total

Here,

Using assumed mean method , we have 

Mean   

(ii)  We construct the frequency distribution table :

Class	Class mark	Frequency
150 – 200 200 – 250 250 – 300 300 – 350 350 – 400	175 225 275 325 375	48 32 35 20 10	– 100  – 50 0 50 100	– 2 – 1 0 1 2	– 96  – 32 0 20 20
Total

Here,

Using step deviation method , we have

Mean   

CHECK YOUR PROGRESS 25.4

1. Following are the goals scored by a team in a series of 11 matches

1, 0 , 3, 2, 4, 5, 2, 4, 4, 2, 5

Determine the median score.

Solution: We arrange the data in the ascending order: 0 , 1 , 2 , 2, 2 , 3, 4 , 4, 4 , 5 ,5

Here,  (odd)

The median 

2. In a diagnostic test in mathematics given to 12 students, the following marks (out of 100) are recorded

46, 52, 48, 39, 41, 62, 55, 53, 96, 39, 45, 99

Calculate the median for this data.

Solution: We arrange the data in the ascending order: 39 , 39 , 41 , 45 , 46 , 48 , 52 , 53 , 55 ,62 , 96 , 99  

Here,  (even)

The median

 

3. A fair die is thrown 100 times and its outcomes are recorded as shown below:

Outcome	1	2	3	4	5	6
Frequency	17	15	16	18	16	18

Find the median outcome of the distributions.

Solution:  We have,

Outcome	Frequency	Cumulative frequency
1 2 3 4 5 6	17 15 16 18 16 18	17 32 48 66 82 100

Here, n = 100 (even)

Since n is even, so the median will be the mean of    observations,

i.e., 50th and 51th observations.  Thus, we look at the cumulative frequency column to find the first cumulative frequency greater than or equal to 50 and 51, which is 66. Therefore, both the 50th and 51st observations are 4. Hence, the median is 4.

4. For each of the following frequency distributions, find the median:

(a)

	2	3	4	5	6	7
	4	9	16	14	11	6

(b)

	5	10	15	20	25	30	35	40
	3	7	12	20	28	31	28	26

(c)

	2.3	3	5.1	5.8	7.4	6.7	4.3
	5	8	14	21	13	5	7

Solution: (a) We have,

		Cumulative frequency
2 3 4 5 6 7	4 9 16 14 11 6	4 13 29 43 54 60

Here, n= 60 (even)

Since n is even, so the median will be the mean of    observations,

i.e., 30th and 31th observations.  Thus, we look at the cumulative frequency column to find the first cumulative frequency greater than or equal to 30 and 31, which is 43. Therefore, both the 30th and 31st observations are 45. Hence, the median is 5.

 (b) We have,

		Cumulative frequency
5 10 15 20 25 30 35 40	3 7 12 20 28 31 28 26	3 10 22 42 70 101 129 155

Here n = 155 (odd)

Median  

(c) We have,

		Cumulative frequency
2.3 3 5.1 5.8 7.4 6.7 4.3	5 8 14 21 13 5 7	5 13 27 48 61 66 73

Here n = 73 (odd)

Median 

CHECK YOUR PROGRESS 25.5

1. Find the mode of the data:

         5, 10, 3, 7, 2, 9, 6, 2, 11, 2

Solution: Arranging the data in increasing order, we have

2 ,2 ,2 , 3, 5 , 6 , 7 , 9 , 10 , 11

We find the frequency of 2 is 3 times and is more than the frequency of all other scores.

So, mode of the data is 2 .

2. The number of TV sets in each of 15 households are found as given below:

      2, 2, 4, 2, 1, 1, 1, 2, 1, 1, 3, 3, 1, 3, 0

What is the mode of this data?

Solution: Arranging the data in increasing order, we have

0 , 1 ,1 , 1, 1, 1, 1, 2 , 2 ,2 , 2, 3 , 3 , 3 , 4

We find the frequency of 1 is 6 times and is more than the frequency of all other scores.

So, mode of the data is 1 .

3. A die is thrown 100 times, giving the following results

Outcome	1	2	3	4	5	6
Frequency	15	16	16	15	17	20

 Find the modal outcome from this distribution.

Solution:  The highest frequency in the table is 20 and the corresponding outcome is 6. Therefore, the mode is 6.

4. Following are the marks (out of 10) obtained by 80 students in a mathematics test:

Marks obtained	0	1	2	3	4	5	6	7	8	9	10
Number of students	5	2	3	5	9	11	15	16	9	3	2

Determine the modal marks.

Solution: The highest frequency in the “number of students” row is 16 and the corresponding score is 7. Hence, the mode is 7.

TERMINAL EXERCISE

1. Find the mean of first five prime numbers.

Solution: We have,

Mean 

2. If the mean of 5, 7, 9, x, 11 and 12 is 9, find the value of x.

Solution: We have,  

    

3. Following are the marks obtained by 9 students in a class

51, 36, 63, 46, 38, 43, 52, 42 and 43

(i) Find the mean marks of the students.

(ii) What will be the mean marks if a student scoring 75 marks is also included in the class.

Solution:  The marks obtained by 9 students are: 51, 36, 63, 46, 38, 43, 52, 42, 43

(i)  Total marks = 51 + 36 + 63 + 46 + 38 + 43 + 52 + 42 + 43 = 414

Number of students = 9

  

Therefore, the mean marks of the students is 46 .

(ii) Total marks =  414 + 75 = 489

Number of students = 9+1=10

  

New mean marks = 48.9

4. The mean marks of 10 students in a class is 70. The students are divided into two groups of 6 and 4 respectively. If the mean marks of the first group is 60, what will be the mean marks of the second group?

Solution: Given,  total number of students = 10  , Mean marks of all 10 = 70

Total marks of all students = 10 × 70 = 700

Total marks of first group = 6 × 60 = 360

Second group: Number of students = 4

Let the mean marks = x

Total marks of second group  

A/Q,    

  

Therefore, the  mean marks of the second group = 85

5. If the mean of the observations  is  , show that   

Solution: The mean of n observations  is

        and     

Where,  is a constant value .

LHS :     RHS          Proved.

6. There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Determine the mean of the given numbers.

Solution:  Let, the numbers are : .

We have,  

     

  

Therefore,  the mean of the given numbers is 56.5.

7. Find the mean of the following data:

(a)

	5	9	13	17	22	25
	3	5	12	8	7	5

(b)

	16	18	28	22	24	26
	1	3	5	7	5	4

Solution: (a)  We have,

5 9 13 17 22 25	3 5 12 8 7 5	15 45 156 136 154 125
Total

Mean 

(b)  We have,

16 18 28 22 24 26	1 3 5 7 5 4	16 54 140 154 120 104
Total

Mean 

8. Find the mean of the following data

(a)

Classes	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Frequencies	2	3	5	7	5	3

(b)

Classes	100 – 200	200 – 300	300 – 400	400 – 500	500 – 600	600 – 700
Frequencies	3	5	8	6	5	3

(c) The ages (in months) of a group of 50 students are as follows. Find the mean age.

Age	156 – 158	158 – 160	160 – 162	162 – 164	164 – 166	166 – 168
Number of students	2	4	8	16	14	6

Solution: (a) We have,

Classes	Class mark	Frequencies
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70	15 25 35 45 55 65	2 3 5 7 5 3	30 75 175 315 275 195

Mean  

(b) We have,

Classes	Class mark	Frequencies
100 – 200 200 – 300 300 – 400 400 – 500 500 – 600 600 – 700	150 250 350 450 550 650	3 5 8 6 5 3	450 1250 2800 2700 2750 1950
Total

Mean 

(c) We have,

Age	Class mark	Frequencies
156 – 158 158 – 160 160 – 162 162 – 164 164 – 166 166 – 168	157 159 161 163 165 167	2 4 8 16 14 6	314 636 1288 2608 2310 1002
Total

Mean  

9. Find the median of the following data:

(a) 5, 12, 16, 18, 20, 25, 10            (b) 6, 12, 9, 10, 16, 28, 25, 13, 15, 17            (c) 15, 13, 8, 22, 29, 12, 14, 17, 6

Solution: (a)  We arranging the data in increasing order :

    5 , 10 , 12 , 16 , 18 , 20 , 25

Here n = 7 (odd)

Median

(b) We arranging the data in increasing order :

    6 , 9 , 10 , 12 , 13 , 15 , 16 , 17 , 25 , 28

Here, n= 10 (even)

Median 

(c)  We arranging the data in increasing order :

     6 , 8 , 12 , 13 , 14 , 15 , 17 , 22 , 29

Here n = 9 (odd)

Median  

10. The following data are arranged in ascending order and the median of the data is 60.Find the value of x.

    26, 29, 42, 53, x, x + 2, 70, 75, 82, 93

Solution:   We have,

 

 

Therefore, the value of x is 59 .

11. Find the median of the following data:

(a)

	25	30	35	45	50	55	65	70	85
	5	14	12	21	11	13	14	7	3

(b)

	35	36	37	38	39	40	41	42
	2	3	5	4	7	6	4	2

Solution:   We have,

		Cumulative frequency
25 30 35 45 50 55 65 70 85	5 14 12 21 11 13 14 7 3	5 19 31 52 63 76 90 97 100

Here, n = 100

Since n is even, so the median will be the mean of    observations,

i.e., 50th and 51th observations.  Thus, we look at the cumulative frequency column to find the first cumulative frequency greater than or equal to 50 and 51, which is 52. Therefore, both the 50th and 51st observations are 45. Hence, the median is 45.

(b) We have,

		Cumulative frequency
35 36 37 38 39 40 41 42	2 3 5 4 7 6 4 2	2 5 10 14 21 27 31 33

Here, n= 33

Median

The 17^th term falls within the cumulative frequency of 21 . So, the median is 39 .

12. Find the mode of the following data:

(a) 8, 5, 2, 5, 3, 5, 3, 1

(b) 19, 18, 17, 16, 17, 15, 14, 15, 17, 9

Solution: (a)  We arranging the data in increasing order :

     1 , 2, 3 ,3 , 5 , 5, 5, 8

So, the mode is 5 ( 3 times) ]

(b) We arranging the data in increasing order :

    9 , 14 , 15 , 15 , 16 , 17 , 17 , 17 , 18 ,19

So, the mode is 17 ( 3 times) ]

13. Find the mode of the following data which gives life time (in hours) of 80 bulbs selected at random from a lot.

Life time (in hours)	300	500	700	900	1100
Number of bulbs	10	12	20	27	11

Solution: The highest frequency in the table is 27 , which corresponding to a lifetime of 900 hours. So, the mode is 900 .

14. In the mean of the following data is 7, find the value of p:

	4	P	6	7	9	11
	2	4	6	10	6	2

Solution: We have,

4 P 6 7 9 11	2 4 6 10 6 2	8 4p 36 70 54 22
Total

Mean  

 

Therefore, the value of p is 5 .

15. For a selected group of people, an insurance company recorded the following data:

Age (in years)	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80
Number of deaths	2	12	55	95	71	42	16	7

Determine the mean of the data.

Solution:  We construct the frequency distribution table :

Age (in years)	Frequency	Class marks
0 – 10 10 – 20 20 – 30 30 – 40  40 – 50 50 – 60 60 – 70 70 – 80	2 12 55 95 71 42 16 7	5 15 25 35 45 55 65 75	– 40 – 30 – 20 – 10 0 10 20 30	– 80  – 360  – 1100  – 950 0 420 320 210
Total

Here,  

Using assumed mean method , we have 

Mean   

Therefore, the mean of the data is 39.87 years .

16. If the mean of the observations: x + 1, x + 4, x + 5, x + 8, x + 11 is 10, the mean of the last three observations is

(A) 12.5        (B) 12.2          (C) 13.5         (D) 14.2

Answer: (B)  12.2

[ A/Q,    

 

The sum of the last three observations

  ]

17. If each observation in the data is increased by 2, than their mean

(A) remains the same      (B) becomes 2 times the original mean      (C) is decreased by 2      (D) is increased by 2

Answer: (D) is increased by 2 .

18. Mode of the data: 15, 14, 19, 20, 14, 15, 14, 18, 14, 15, 17, 14, 18 is

(A) 20        (B) 18         (C) 15        (D) 14

Answer:  (D) 14

[Arranging the data in increasing order, we have

14,14,14,14,14,15,15,15,17,18,18,19,20

So, the mode is 14 (5^th times) ]

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