CBSE NCERT Solutions For Class 10th Maths Chapter 14 : Statistics. NCERT Solutins For Class 10 Mathematics. Exercise 14.1, Exercise 14.2, Exercise 14.3, Exercise 14.4.

---

NCERT Solutions for Class X Maths Chapter 14 Statistics – Mathematics CBSE

Exercise 14.1

1. 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?

Solution:

Class Interval	fi	xi	fixi
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
	Sum fi = 20		Sum fixi = 162

Mean can be calculated as follows:

In this case, the values of fi and xi are small hence direct method has been used.

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution: In this case, value of xi is quite large and hence we should select the assumed mean method.

Let us take assumed mean a = 150

Class Interval	fi	xi	di = xi – a	fidi
100-120	12	110	-40	-480
120-140	14	130	-20	-280
140-160	8	150
160-180	6	170	20	120
180-200	10	190	40	400
	Sum fi = 50			Sum fidi = -240

Now, mean of deviations can be calculated as follows:

Mean can be calculated as follows:

x = d + a = -4.8 + 150 = 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Daily pocket allowance (in Rs)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	7	6	9	13	f	5	4

Solution:

Class Interval	fi	xi	fixi
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18	234
19-21	f	20	20f
21-23	5	22	110
23-25	4	24	96
	Sum fi = 44 + f		Sum fixi = 752 + 20f

We have;

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per min	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Solution:

Class Interval	fi	xi	di = xi – a	fidi
65-68	2	66.5	-9	-18
68-71	4	69.5	-6	-24
71-74	3	72.5	-3	-9
74-77	8	75.5
77-80	7	78.5	3	21
80-83	4	81.5	6	24
83-86	2	84.5	9	18
	Sum fi = 30			Sum fidi = 12

Now, mean can be calculated as follows:

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	50-52	53-55	56-58	89-61	62-64
Number of boxes	15	110	135	115	25

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

Solution:

Class Interval	fi	xi	di = x – a	fidi
50-52	15	51	-6	90
53-55	110	54	-3	-330
56-58	135	57
59-61	115	60	3	345
62-64	25	63	6	150
	Sum fi = 400			Sum fidi = 75

Mean can be calculated as follows:

In this case, there are wide variations in fi and hence assumed mean method is used.

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Solution:

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
100-150	4	125	-100	-2	-8
150-200	5	175	-50	-1	-5
200-250	12	225
250-300	2	275	50	1	2
300-350	2	325	100	2	4
	Sum fi = 25				Sum fiui = -7

Mean can be calculated as follows:

7. To find out the concentration of SO₂ in the air (in parts per million, i.e. ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO2 (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 concentration of SO₂ in the air.

Solution:

Class Interval	fi	xi	fixi
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
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.22	0.44
	Sum fi = 30		Sum fixi = 2.96

Mean can be calculated as follows:

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of students	11	10	7	4	4	3	1

Solution:

Class Interval	fi	xi	fixi
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39
	Sum fi = 40		Sum fixi = 499

Mean can be calculated as follows:

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-98
Number of cities	3	10	11	8	3

Solution:

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
45-55	3	50	-20	-2	-6
55-65	10	60	-10	-1	-10
65-75	11	70
75-85	8	80	10	1	8
85-95	3	90	20	2	6
	Sum fi = 35				Sum fiui = -2

Mean can be calculated as follows:

Exercise 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year.

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution: Solution: Modal class = 35 – 45, l = 35, h = 10, f1 = 23, f0 = 21 and f2 = 14

Calculations for Mean:

Class Interval	fi	xi	fixi
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
	Sum fi = 80		Sum fixi = 2830

The mode of the data shows that maximum number of patients is in the age group of 26.8, while average age of all the patients is 35.37.

2. The following data gives the information on the observed lifetime (in hours) of 225 electrical components:

Lifetime (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Solution: Modal class = 60-80, l = 60, f1 = 61, f0 = 52, f2 = 38 and h = 20

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

Expenditure	Number of families
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

Solution: Modal class = 1500-2000, l = 1500, f1 = 40, f0 = 24, f2 = 33 and h = 500

Calculations for mean:

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
1000-1500	24	1250	-1500	-3	-72
1500-2000	40	1750	-1000	-2	-80
2000-2500	33	2250	-500	-1	-33
2500-3000	28	2750
3000-3500	30	3250	500	1	30
3500-4000	22	3750	1000	2	44
4000-4500	16	4250	1500	3	48
4500-5000	7	4750	2000	4	28
	fi = 200				fiui = -35

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher	Number of states/UT
15-20	3
20-25	8
25-30	9
30-35	10
35-40	3
40-45
45-50
50-55	2

Solution: Modal class = 30-35, l = 30, f1 = 10, f0 = 9, f2 = 3 and h = 5

Calculation for mean:

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
15-20	3	17.5	-15	-3	-9
20-25	8	22.5	-10	-2	-16
25-30	9	27.5	-5	-1	-9
30-35	10	32.5
35-40	3	37.5	5	1	3
40-45		42.5	10	2
45-50		47.5	15	3
50-55	2	52.5	20	4	8
	Sum fi = 35				Sum fiui = -23

The mode shows that maximum number of states has 30-35 students per teacher. The mean shows that average ratio of students per teacher is 29.22

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored	Number of batsmen
3000-4000	4
4000-5000	18
5000-6000	9
6000-7000	7
7000-8000	6
8000-9000	3
9000-1000	1
10000-11000	1

Find the mode of the data.

Solution: Modal class = 4000-5000, l = 4000, f1 = 18, f0 = 4, f2 = 9 and h = 1000

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data.

Number of cars	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	7	14	13	12	20	11	15	8

Solution: Modal class = 40 – 50, l = 40, f1 = 20, f0 = 12, f2 = 11 and h = 10

Exercise 14.3(NCERT)

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)	Number of customers
65-85	4
85-105	5
105-125	13
125-145	20
145-165	14
165-185	8
185-205	4

Solution:

Class Interval	Frequency	Cumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	8	64
185-205	4	68
	N = 68

Here; n = 68 and hence n/2 = 34

So, median class is 125-145 with cumulative frequency = 42

now, l = 125, n = 68, cf = 22, f = 20, h = 20

Median can be calculated as follows:

Calculations for Mode:

Modal class = 125-145, f1 = 20, f0 = 13, f2 = 14 and h = 20

Calculations for Mean:

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
65-85	4	75	-60	-3	-12
85-105	5	95	-40	-2	-10
105-125	13	115	-20	-1	-13
125-145	20	135
145-165	14	155	20	1	14
165-185	8	175	40	2	16
185-205	4	195	60	3	12
	Sum fi = 68				Sum fiui = 7

Mean, median and mode are more or less equal in this distribution.

2. If the median of the distribution given below is 28.5, find the value of x and y.

Class Interval	Frequency
0-10	5
10-20	x
20-30	20
30-40	15
40-50	y
50-60	5
Total	60

Solution: n = 60 and hence n/2 = 30

Median class is 20 – 30 with cumulative frequency = 25 + x

lower limit of median class = 20, cf = 5 + x , f = 20 and h = 10

Now, from cumulative frequency, we can find the value of x + y as follows:

Hence, x = 8 and y = 7

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)	Number of policyt hodlers
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Solution:

Class interval	Frequency	Cumulative frequency
15-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

Here; n = 100 and n/2 = 50, hence median class = 35-45

In this case; l = 35, cf = 45, f = 33 and h = 5

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	5
163-171	4
172-180	2

Find the median length of leaves.

Solution:

Class Interval	Frequency	Cumulative frequency
117.5-126.5	3	3
126.5-135.5	5	8
135.5-144.5	9	17
144.5-153.5	12	29
153.5-162.5	5	34
162.5-171.5	4	38
171.5-180.5	2	40

We have; n = 40 and n/2 = 20 hence median class = 144.5-153.5

Thus, l = 144.5, cf = 17, f = 12 and h = 9

5. The following table gives distribution of the life time of 400 neon lamps.

Lifetime (in hours)	Number of lamps
1500-2000	14
2000-2500	56
2500-3000	60
3000-3500	86
3500-4000	74
4000-4500	62
4500-5000	48

Find the median life time of a lamp.

Solution:

Class Interval	Frequency	Cumulative Frequency
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400

We have; n = 400 and n/2 = 200 hence median class = 3000 – 3500

So, l = 3000, cf = 130, f = 86 and h = 500

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Solution: Calculations for median:

Class Interval	Frequency	Cumulative Frequency
1-4	6	6
4-7	30	36
7-10	40	76
10-13	16	92
13-16	4	96
16-19	4	100

Here; n = 100 and n/2 = 50 hence median class = 7-10

So, l = 7, cf = 36, f = 40 and h = 3

Calculations for Mode:

Modal class = 7-10,

Here; l = 7, f1 = 40, f0 = 30, f2 = 16 and h = 3

Calculations for Mean:

Class interval	fi	xi	fixi
1-4	6	2.5	15
4-7	30	5.5	165
7-10	40	8.5	340
10-13	16	11.5	184
13-16	4	14.5	51
16-19	4	17.5	70
	Sum fi = 100		Sum fixi = 825

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75
Number of students	2	3	8	6	6	3	2

Solution:

Class Interval	Frequency	Cumulative frequency
40-45	2	2
45-50	3	5
50-55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

 

We have; n = 30 and n/2 = 15 hence median class = 55-60

So, l = 55, cf = 13, f = 6 and h = 5

Exercise 14.4(NCERT)

1. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution:

Daily income	Cumulative frequency
Less than 120	12
Less than 140	26
Less than 160	34
Less than 180	40
Less than 200	50

2. During the medical checkup of 35 students of a class, their weights were recorded as follows:

Weight (in kg)	Number of students
Less than 38
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

Weight (in kg)	Frequency	Cumulative Frequency
36-38
38-40	3	3
40-42	2	5
42-44	4	9
44-46	5	14
46-48	14	28
48-50	4	32
50-52	3	35

Since N = 35 and n/2 = 17.5 hence median class = Less than 46-48

Here; l = 46, cf = 14, f = 14 and h = 2

Median can be calculated as follows:

This value of median verifies the median shown in ogive.

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (in kg)	50-55	55-60	60-65	65-70	70-75	75-80
Number of farms	2	8	12	24	38	16

Change this distribution to a more than type distribution, and draw its ogive.

Solution:

Production yield	Cumulative frequency
More than 50	100
More than 55	98
More than 60	90
More than 65	78
More than 70	54
More than 75	16

---

GO BACK TO CLASS X MATHS ALL CHAPTERS SOLUTION