NCERT Solutions Class 10^th Maths Chapter 14: Statistics

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.

Which method did you use for finding the mean, and why?

SOLUTION

Class Intervals = Number of Plants	Frequency (f_i) = Number of houses	Mid-point (x_i) = (U.L + L.L)/2	Product of f_i and 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
	∑ f_i = 20		∑ f_ix_i = 162

The values of f_i and x_i are numerically small, so we will use the direct method to calculate the mean.

Mean = x? =∑ f_ix_i/∑ f_i

= 162/20 = 8.1

Mean number of plant per house = 8.1

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

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

SOLUTION

Class Intervals = Daily wages	Frequency (f_i) = Number of workers	Mid-point (x_i) = (U.L + L.L)/2	u_i = (x_i - 150)/20	Product of f_i and u_i
100 - 120	12	110	-2	-24
120 - 140	14	130	-1	-14
140 - 160	8	150	0	0
160 - 180	6	170	1	6
180 - 200	10	190	2	20
	∑ f_i = 50			∑ f_iu_i = -12

Let us assume the value of mean to be A = 150.

Class interval = h = U.L - L.L = 20

True Mean = x? = A + h∑ f_iu_i/∑ f_i

= 150 + 20(-12)/50

= 145.2

Mean daily wage of the workers = Rs 145.2

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.

SOLUTION

Class Intervals = Daily pocket allowance (in Rs)	Frequency (f_i) = Number of children	Mid-point (x_i) = (U.L + L.L)/2	Product of f_i and x_i
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
	∑ f_i = f + 44		∑ f_ix_i = 752 + 20f

Mean = x? = ∑f_ix_i/∑f_i = (752 + 20f)/(f + 44)

We know that x? = 18. Therefore

18 = (752 + 20f)/(f + 44)

18(f + 44) = 752 + 20f

18f + 792 = 752 + 20f

40 = 2f

f = 20

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

SOLUTION

Class Intervals = Number of heartbeats per minutes	Frequency (f_i) = Number of women	Mid-point (x_i) = (U.L + L.L)/2	u_i = (x_i - 75.5)/3	Product of f_i and u_i
65 - 68	2	66.5	-3	-6
68 - 71	4	69.5	-2	-8
71 - 74	3	72.5	-1	-3
74 - 77	8	75.5	0	0
77 - 80	7	78.5	1	7
80 - 83	4	81.5	2	8
83 - 86	2	84.5	3	6
	∑ f_i = 30			∑ f_iu_i = 4

Let us assume the mean to be A = 75.5

Class height = U.L - L.L = 3

True Mean = x? = A + h∑ f_iu_i/∑ f_i

= 75.5 + 3(4)/10

= 75.9

Mean heart beats per minute of the women is 75.9

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.

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

SOLUTION

The given data does not have continuous class intervals. Therefore, we will add 0.5 to all the Upper Limit values and subtract 0.5 from all the Lower Limit values.

The given values for f_i are numerically big, so we will use the step-deviation method.

Class Intervals = Number of mangoes	Frequency (f_i) = Number of boxes	Mid-point (x_i) = (U.L + L.L)/2	d_i = x_i - A	Product of f_i and d_i
49.5 - 52.5	15	51	-6	90
52.5 - 55.5	110	54	-3	-330
55.5 - 58.5	135	57	0	0
58.5 - 61.5	115	60	3	345
61.5 - 64.5	25	63	6	150
	∑ f_i = 400			∑ f_id_i = 75

Let us assume the mean to be A = 57

Class height = h = U.L - L.L = 3

True Mean = x? = A + h∑ f_id_i/∑ f_i

= 57 + 0.1875

= 57.19

Mean number of mangoes kept in one box is 57.19

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

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

SOLUTION

Class Intervals = Daily expenditure	Frequency (f_i) = Number of households	Mid-point (x_i) = (U.L + L.L)/2	u_i = (x_i - 150)/50	Product of f_i and u_i
100 - 150	4	125	-2	-8
150 - 200	5	175	-1	-5
200 - 250	12	225	0	0
250 - 300	2	275	1	2
300 - 350	2	325	2	4
	∑ f_i = 25			∑ f_iu_i = -7

Let us assume the mean to be A = 225

Class height = h = U.L - L.L = 50

True Mean = x? = A + h∑ f_iu_i/∑ f_i

= 225 + 50(-7)/25

= 211

Mean daily expenditure on food is Rs 211.

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:

Find the mean concentration of SO₂ in the air.

SOLUTION

Class Intervals = Concentration of SO₂ in ppm	Frequency (f_i)	Mid-point (x_i) = (U.L + L.L)/2	Product of f_i and x_i
0.00 - 0.04	4	0.02	0.08
0.04 - 0.08	9	0.06	0.54
0.08 - 0.12	9	0.1	0.9
0.12 - 0.16	2	0.14	0.28
0.16 - 0.20	4	0.18	0.72
0.2 - 0.24	2	0.22	0.44
	∑ f_i = 30		∑ f_ix_i = 2.96

Mean = x? = ∑ f_ix_i/∑ f_i

= 2.96/30

= 0.0986

Mean concentration of SO₂ in the air is 0.0986 ppm.

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.

SOLUTION

Class Intervals = Number of Days	Frequency (f_i) = Number of students	Mid-point (x_i) = (U.L + L.L)/2	Product of f_i and x_i
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
	∑ f_i = 40		∑ f_ix_i = 499

Mean = x? = ∑ f_ix_i/∑ f_i

= 499/40

= 12.48

Mean number of days a student was absent is 12.48

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

SOLUTION

Class Intervals = Literacy rate (in %)	Frequency (f_i) = Number of cities	Mid-point (x_i) = (U.L + L.L)/2	u_i = (x_i - 70)/10	Product of f_i and u_i
45 - 55	3	50	-2	-6
55 - 65	10	60	-1	-10
65 - 75	11	70	0	0
75 - 85	8	80	1	8
85 - 95	3	90	2	6
	∑ f_i = 35			∑ f_iu_i = -2

Let us assume the mean to be A = 70

Class height = U.L - L.L = 10

True Mean = x? = A + h∑ f_iu_i/∑ f_i

= 70 + 10(-2)/35

= 69.428

Mean literacy rate of a city is 69.428 %

Exercise 14.2

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

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

SOLUTION

The highest frequency in the given data is 23. Therefore, the modal class will be 35 - 45.

l = 35

f₁ = 23

f₂ = 14

f₀ = 21

Class width = h = 10

Mode = 36.81 years

Class Intervals = Age (in years)	Frequency (f_i) = Number of patients	Mid-point (x_i) = (U.L + L.L)/2	Product of f_i and x_i
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
	∑ f_i = 80		∑ f_ix_i = 2830

Mean = x? = ∑ f_ix_i/∑ f_i

= 2830/80

= 35.375 years

Mean age of a patient is 35.375 years

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

Determine the modal lifetimes of the components.

SOLUTION

The highest frequency in the given data is 61. Therefore, the modal class will be 60 - 80.

l = 60

f₁ = 61

f₂ = 38

f₀ = 52

Class width = 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 :

SOLUTION

The highest frequency in the given data is 40. Therefore, the modal class will be 1500 - 2000.

l = 1500

f₁ = 40

f₂ = 33

f₀ = 24

Class width = h = 500

Class Intervals = Expenditure	Frequency (f_i) = Number of families	Mid-point (x_i) = (U.L + L.L)/2	u_i = (x_i - 2750)/500	Product of f_i and u_i
1000 - 1500	24	1250	-3	-72
1500 - 2000	40	1750	-2	-80
2000 - 2500	33	2250	-1	-33
2500 - 3000	28	2750	0	0
3000 - 3500	30	3250	1	30
3500 - 4000	22	3750	2	44
4000 - 4500	16	4250	3	48
4500 - 5000	7	4750	4	28
	∑ f_i = 200			∑ f_iu_i = -35

Let us assume the mean to be A = 2750

True mean = x? = A + h∑ f_iu_i/∑ f_i

= 2750 + 500(-35)/200

= 2662.5

Mean monthly expenditure of the families = Rs 2662.5

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.

SOLUTION

The highest frequency in the given data is 10. Therefore, the modal class will be 30 - 35.

l = 30

f₁ = 10

f₂ = 3

f₀ = 9

Class width = h = 5

Class Intervals = Age (in years)	Frequency (f_i) = Number of patients	Mid-point (x_i) = (U.L + L.L)/2	Product of f_i and x_i
15 - 20	3	17.5	52.5
20 - 25	8	22.5	180
25 - 30	9	27.5	247.5
30 - 35	10	32.5	325
35 - 40	3	37.5	112.5
40 - 45	0	42.5	0
45 - 50	0	47.5	0
50 - 55	2	52.7	105.5
	∑ f_i = 35		∑ f_ix_i = 1022.5

Mean = x? = ∑f_ix_i/∑f_i

= 1022.5/35

= 29.214

Mean student-teacher ratio of the States/U.T.s = 29.214

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

Find the mode of the data.

SOLUTION

The highest frequency in the given data is 18. Therefore, the modal class will be 4000 - 5000.

l = 4000

f₁ = 18

f₂ = 9

f₀ = 4

Class width = 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 summarised it in the table given below. Find the mode of the data :

SOLUTION

The highest frequency in the given data is 20. Therefore, the modal class will be 40 - 50.

l = 40

f₁ = 20

f₂ = 11

f₀ = 12

Class width = h = 10

Exercise 14.3

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.

SOLUTION

Class Interval = Monthly consumption	Frequency (f_i) = Number of consumers	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

n = 68

n/2 = 34

Therefore, the median class will be 125 - 145 and cumulative frequency is 42

l = 125

f = 20

cf = 22

Class height = h = 20

Median monthly consumption = 137 units

The highest frequency in the given data is 20. Therefore, the modal class will be 125 - 145.

l = 125

f₁ = 20

f₂ = 14

f₀ = 13

Class width = h = 20

Mode = 103.823

Class Intervals = Monthly Consumption	Frequency (f_i) = Number of consumers	Mid-point (x_i) = (U.L + L.L)/2	u_i = (x_i - 135)/20	Product of f_i and u_i
65 - 85	4	75	-3	-12
85 - 105	5	95	-2	-10
105 - 125	13	115	-1	-13
125 - 145	20	135	0	0
145 - 165	14	155	1	14
165 - 185	8	175	2	16
185 - 205	4	195	3	12
	∑ f_i = 68			∑ f_iu_i = 7

Let us assume the mean to be A = 135

True Mean = x? = A + h∑ f_iu_i/∑ f_i

= 135 + 20(7)/68

= 137.06

Mean monthly consumption is 137.06 units.

On comparison of mean, median and mode of the data, we can conclude that they are approximately equal to each other.

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

SOLUTION

Class Interval	Frequency (f_i)	Cumulative frequency
0 - 10	5	5
10 - 20	x	5 + x
20 - 30	20	25 + x
30 - 40	15	40 + x
40 - 50	y	40 + x + y
50 - 60	5	45 + x + y
	n = 60

n = 60

n/2 = 30

Therefore, the median class will be 20 - 30 with cumulative frequency 25 + x

l = 20

f = 20

cf = 5 + x

Class height = h = 10

∑ f_i = 45 + x + y = n

45 + 8 + y = 60

53 + y = 60

y = 7

Hence, the values of x and y are 8 and 7 respectively.

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 year.

SOLUTION

Class Interval	Frequency (f_i)	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
	n = 100

n = 100

n/2 = 50

Therefore, the median class will be 35 - 40 with cumulative frequency = 78

l = 35

f = 33

cf = 45

Class height = h = 5

Median age = 35.75 years

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

Find the median length of the leaves.

(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

SOLUTION

Since the data isn't continuous, we will add 0.5 to all the Upper Limit values and subtract 0.5 from all the Lower Limit values.

Class Interval	Frequency (f_i)	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
	n = 40

n = 40

n/2 = 20

Therefore, the median class will be 144.5 - 153.5 with cumulative frequency = 29

l = 144.5

f = 12

cf = 17

Class height = h = 9

Median length of the leaves = 146.25 mm.

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

Find the median life time of a lamp.

SOLUTION

Class Interval	Frequency (f_i)	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
	n = 400

n = 400

n/2 = 200

Therefore, the median class will be 3000 - 3500 with cumulative frequency = 216

l = 3000

f = 86

cf = 130

Class height = h = 500

Median life time of a lamp = 3406.976 hours

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:

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

Class Interval	Frequency (f_i)	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
	n = 100

n = 100

n/2 = 50

Therefore, the median class will be 7 - 10 with cumulative frequency = 76

l = 7

f = 40

cf = 36

Class height = h = 3

Median number of letters = 8.05

Class Intervals = Number of letters	Frequency (f_i) = Number of surnames	Mid-point (x_i) = (U.L + L.L)/2	Product of f_i and x_i
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
	∑ f_i = 100		∑ f_ix_i = 825

Mean = x? = ∑ f_ix_i/∑ f_i

= 825/100 = 8.25

Mean number of letters = 8.25

The highest frequency in the given data is 40. Therefore, the modal class will be 7 - 10.

l = 7

f₁ = 40

f₂ = 16

f₀ = 30

Class width = h = 3

Modal number of letters = 7.882

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

SOLUTION

Class Interval	Frequency (f_i)	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
	n = 30

n = 30

n/2 = 15

Therefore, the median class will be 55 - 60 with cumulative frequency = 19

l = 55

f = 6

cf = 13

Class height = h = 5

Median weight of the students = 56.667 kg

Exercise 14.4

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

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

SOLUTION

We need to convert the data into less than type distribution and find cumulative frequencies.

Daily Income	Frequency (f_i) = Number of workers	Cumulative frequency
Less than 120	12	12
Less than 140	14	26
Less than 160	8	34
Less than 180	6	40
Less than 200	10	50
	n = 50

Using the upper limits and the corresponding cumulative frequencies, we can obtain the points (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50). We will plot a graph with these coordinates and join the points to obtain the less than type ogive curve.

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

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

Using the upper limits and the corresponding cumulative frequencies, we can obtain the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). We will plot a graph with these coordinates and join the points to obtain the less than type ogive curve.

Sum of frequencies is 35. Therefore, we will locate the point 35/2 = 17.5 on the y-axis or the cumulative frequency axis.

Produce a line parallel to x-axis through the point (0, 17.5) until it meets the ogive curve.

Draw the perpendicular at the point at which this line intersects the ogive curve.

The coordinates of this point are found out to be (46.5, 17.5). Therefore, 46.5 is the median weight of the students from the given data.

Verification :

We need to find the actual frequencies.

Weight	Frequency (f_i) = Number of students	Cumulative frequency
Less than 38	0	0
Less than 40	3 - 0 = 3	3
Less than 42	5 - 3 = 2	5
Less than 44	9 - 5 = 4	9
Less than 46	14 - 9 = 5	14
Less than 48	28 - 14 = 14	28
Less than 50	32 - 28 = 4	32
Less than 52	35 - 32 = 3	35
	n = 35

n = 35

n/2 = 17.5

Therefore, the median class will be 46 - 48 with cumulative frequency = 28

l = 46

f = 14

cf = 14

Class height = h = 2

Median weight of the students = 46.5 kg

Hence, the median weight of students has been verified.

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

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

SOLUTION

We need to convert the given data to more than type distribution and find out the cumulative frequencies.

Production yeild	Frequency (f_i) = Number of farms	Cumulative frequency
More than 50	2	100
More than 55	8	100 - 2 = 98
More than 60	12	98 - 8 = 90
More than 65	24	90 - 12 = 78
More than 70	38	78 - 24 = 54
More than 75	16	54 - 38 = 16
	n = 50

Using the upper limits and the corresponding cumulative frequencies, we can obtain the points (50, 100), (55, 90), (60, 90), (65, 78), (70, 54) and (75, 16). We will plot a graph with these coordinates and join the points to obtain the more than type ogive curve.