NCERT Solutions for Class 10 Mathematics chapter 14 Statistics

Q:

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 (in ₹) Number of families

1000 – 1500
1500– 2000
2000– 2500
2500–3000
3000–3500
3500–4000
4000–4500
4500–5000 24
40
33
28
30
22
16
7

A:

\begin{array}{l} We find class mark for each interval by using the following relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 2750 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}

Expenditure (in ₹)	Number of families (f_i)	x_i	d_i = x_i - 2750	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 2750}{h} \end{array}$	f_iu_i
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	0	0	0
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
Total	200				-35

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 2750 + \frac{- 35}{200} \times 500 \\ = 2662.5 \\ Therefore, mean monthly expenditure is ₹ 2662.5. \\ From the given data, we have \\ l = 1500, f_{1} = 40, f_{0} = 24, f_{2} = 33, h = 500 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 1500 + (\frac{40 - 24}{2 \times 40 - 24 - 33}) \times 500 \\ = 1847.826 \approx 1847.83 \\ So, modal monthly expenditure is ₹ 1847.83. \end{array}$

Q:

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

Production yield (in kg/ha) 50-55 55-60 60-65 65-70 70-75 75-80

Number of farms 2 8 12 24 38 16

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

A:

Following is the cumulative frequency distributions of more than type of the given data. .

$\begin{array}{l} Production yield (in kg/ha) \end{array}$	$\begin{array}{l} Number of farms \\ (Cumulative frequency) \end{array}$
More than or equal to 50	100
More than or equal to 55	100 - 2 = 98
More than or equal to 60	98 - 8 = 90
More than or equal to 65	90 - 12 = 78
More than or equal to 70	78 - 24 = 54
More than or equal to 75	54 - 38 = 16

$\begin{array}{l} \end{array}$

Now, draw the ogive by plotting the points : (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).

Q:

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

A:

\begin{array}{l} We find the cumulative frequency (c f) as below in the \\ following table on the basis of the given information . \end{array}

Weight (in kg)	Number of students (f_i)	Cumulative frequency (cf )
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
Total (n)	30

$\begin{array}{l} We have, n = 30. \\ ∴ \frac{n}{2} = \frac{30}{2} = 15 \\ Cumulative frequency (c f) just greater than \frac{n}{2} = 15 is 19 \\ which belongs to interval 55 - 60. \\ ∴ Median class = 55 - 60 \\ Now, we have \\ l = 55, \\ Cumulative frequency (c f) of the class preceding the \\ median class = 13, \\ Frequency of median class (f) = 6, \\ Class size (h) = 5 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h \\ ∴ Median = 55 + (\frac{\frac{30}{2} - 13}{6}) \times 5 \\ or Median = 56.67 \\ Hence, median weight of students is 56.67 kg . \end{array}$

Q:

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

Life time (in hours) Number of lamps

1500– 2000
2000– 2500
2500–3000
3000–3500
3500–4000
4000–4500
4500–5000 14
56
60
86
74
62
48

Find the median life time of a lamp.

A:

\begin{array}{l} We write the cumulative frequency (c f) as below in the \\ following table on the basis of the given information . \end{array}

Life time (in hours)	Number of lamps (f_i)	Cumulative frequency (cf)
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
Total (n)	400

$\begin{array}{l} We have, n = 400. \\ ∴ \frac{n}{2} = \frac{400}{2} = 200 \\ Cumulative frequency (c f) just greater than \frac{n}{2} = 200 is 216 \\ which belongs to interval 3000 - 3500. \\ ∴ Median class = 3000 - 3500 \\ Now, we have \\ l = 3000, \\ Cumulative frequency (c f) of the class preceding the \\ median class = 130, \\ Frequency of median class (f) = 86, \\ Class size (h) = 500 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h \\ ∴ Median = 3000 + (\frac{\frac{400}{2} - 130}{86}) \times 500 \\ or Median = 3406.98 \\ Hence, median life time of a lamp is 3406.98 hours . \end{array}$

Q:

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.

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

A:

\begin{array}{l} From the given data, we have \\ l = 40, f_{1} = 20, f_{0} = 12, f_{2} = 11, h = 10 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 40 + (\frac{20 - 12}{2 \times 20 - 12 - 11}) \times 10 \\ = 44.7 \\ ∴ Mode of the given data is 44.7. \end{array}

Q:

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

4000– 5000

5000– 6000

6000–7000

7000–8000

8000–9000

9000–10000

10000–11000 4

18

9

7

6

3

1

1

Find the mode of the data.

A:

\begin{array}{l} From the given data, we have \\ l = 4000, f_{1} = 18, f_{0} = 4, f_{2} = 9, h = 1000 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 4000 + (\frac{18 - 4}{2 \times 18 - 4 - 9}) \times 1000 \\ = 4608.695 \\ Mode of the given data is 4608.7 (approx.) \end{array}

Q:

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

Lifetimes (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.

A:

\begin{array}{l} From the given data, we have \\ l = 60, f_{1} = 61, f_{0} = 52, f_{2} = 38, h = 20 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 60 + (\frac{61 - 52}{2 \times 61 - 52 - 38}) \times 20 \\ = 65.625 \\ So, modal lifetime of electrical components is 65 .625 hours . \end{array}

Q:

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?

A:

$\begin{array}{l} Here, we observe that class marks and frequencies are small \\ quantities . \\ So, we use direct method to compute the mean and proceed \\ as below . \end{array}$

Number of plants	Number of houses (f_i)	x_i	f_ix_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
Total	20		162

$\begin{array}{l} Mean = \bar{x} = \frac{\sum f_{i} x_{_{i}}}{\sum f_{i}} = \frac{162}{20} = 8.1 \\ Therefore, mean number of plants per house is 8 .1 . \end{array}$

Q:

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

Weight (in kg) Number of students

Less than 38

Less than 40

Less than 42

Less than 44

Less than 46

Less than 48

Less than 50

Less than 52 0

3

5

9

14

28

32

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.

A:

Following is the given cumulative frequency distributions of less than type

$\begin{array}{l} Weight (in kg) \\ upper class limits \end{array}$	$\begin{array}{l} Number of students \\ (Cumulative frequency) \end{array}$
Less than 38	0
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

$Taking upper class limits on x-axis and cumulative frequencies on y-axis, plot the points
(38, 0); (40, 3); (42, 5); (44, 9); (46, 14); (48, 28); (50, 32) and (52, 35). Now join these points
to obtain the ogive as below.$

$Now Now$

Now locate 17.5 (half of 35) on y-axis and draw a line parallel to x-axis to cut the graph at a point.
Draw a perpendicular from this point to x-axis. The point where this perpendicular meets the x-axis
determines the median of the data. From the graph we see the median is 46.5.

$\begin{array}{l} Now, we mark the point A(17 .5, 46 .5) . So median of the given data is 46 .5 . \\ We find class intervals with their respective frequencies as the following . \end{array}$

Weight (in kg)	Frequency (f)	Cumulative Frequency
Less than 38	0	0
38 - 40	3 - 0 = 3	3
40 - 42	5 - 3 = 2	5
42 - 44	9 - 5 = 4	9
44 - 46	14 - 9 = 5	14
46 - 48	28 - 14 = 14	28
48 - 50	32 - 28 = 4	32
50 - 52	35 - 32 = 3	35
Total (n)	35

$\begin{array}{l} \frac{n}{2} = \frac{35}{2} = 17.5 \\ Cumulative frequency just greater that 17 .5 is 28 which belongs \\ to class interval 46 - 48. \\ ∴ Median class = 46 - 48 \\ Now, we have \\ l = 46, \\ Cumulative frequency (c f) of the class preceding the \\ median class = 14, \\ Frequency of median class (f) = 14, \\ Class size (h) = 2 \end{array}$

$\begin{array}{l} We know that \\ Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h \\ ∴ Median = 46 + (\frac{\frac{35}{2} - 14}{14}) \times 2 \\ or Median = 46.5 \\ So median of the given data is 46.5. \\ Hence, value of the median is verified . \end{array}$

Q:

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

Daily income (in ₹) 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.

A:

We find less than type cumulative frequency table as below .

Daily income (in ₹)	Cumulative Frequency
Less than 120	12
Less than 140	12 + 14 = 26
Less than 160	26 + 8 = 34
Less than 180	34 + 6 = 40
Less than 200	40 + 10 = 50

$\begin{array}{l} Now, taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we draw its \\ ogive as following . \end{array}$

Q:

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:

Length (in mm) Number of leaves

118 – 126

127– 135

136– 144

145–153

154–162

163–171

172–180 3

5

9

12

5

4

2

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)

A:

\begin{array}{l} The given class intervals are not continuous . We convert it to continuous class intervals by subtracting 0 .5 from each \\ of lower boundary and adding 0 .5 in each of upper boundary . \\ We write the class interval, number of leaves (f_{i}), and cumulative frequency (c f) as below in the following table \\ on the basis of the given information . \end{array}

Class interval	Number of leaves (f_i)	Cumulative frequency (cf)
117.5 - 126.5	3	3
126.5 - 135.5	5	8
13.5 - 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
Total	40

$\begin{array}{l} We have, n = 40. \\ ∴ \frac{n}{2} = \frac{40}{2} = 20 \\ Cumulative frequency (c f) just greater than \frac{n}{2} = 20 is 29 \\ which belongs to interval 144.5 - 153.5. \\ ∴ Median class = 144.5 - 153.5 \\ Now, we have \\ l = 144.5, \\ Cumulative frequency (c f) of the class preceding the \\ median class = 17, \\ Frequency of median class (f) = 12, \\ Class size (h) = 9 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h \\ ∴ Median = 144.5 + (\frac{\frac{40}{2} - 17}{12}) \times 9 \\ or Median = 146.75 \\ Hence, median length of leaves is 146.75 mm . \end{array}$

Q:

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.

Class interval Frequency

Below 20

Below 25

Below 30

Below 35

Below 40

Below 45

Below 50

Below 55

Below 60 2

6

24

45

78

89

92

98

100

A:

\begin{array}{l} We write the class interval, number of policy holders (f_{i}), and cumulative frequency (c f) as below in the following table \\ on the basis of the given information . \end{array}

Class interval	Number of policy holders (f_i)	Cumulative frequency (cf)
18 - 20	2	2
20 - 25	6 - 2 = 4	6
25 - 30	24 - 6 = 18	24
30 - 35	45 - 24 = 21	45
35 - 40	78 - 45 = 33	78
40 - 45	89 - 78 = 11	89
45 - 50	92 - 89 = 3	92
50 - 55	98 - 92 = 6	98
55 - 60	100 - 98 = 2	100
Total (n)	100

$\begin{array}{l} We have, n = 100. \\ ∴ \frac{n}{2} = \frac{100}{2} = 50 \\ Cumulative frequency (c f) just greater than \frac{n}{2} = 50 is 78 \\ which belongs to interval 35 - 40. \\ ∴ Median class = 35 - 40 \\ Now, we have \\ l = 35, \\ Cumulative frequency (c f) of the class preceding the \\ median class = 45, \\ Frequency of median class (f) = 33, \end{array}$

$\begin{array}{l} Class size (h) = 5 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h \\ ∴ Median = 35 + (\frac{\frac{100}{2} - 45}{33}) \times 5 \\ or x = 35.76 \\ Hence, median age is 35.76 years . \end{array}$

Q:

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

Class interval Frequency

0 – 10

10– 20

20– 30

30–40

40–50

50–60 5

x

20

15

y

5

Total 60

A:

We find cumulative frequency of the given data as below.

Class interval

Frequency

Cumulative frequency

0 – 10

10– 20

20– 30

30–40

40–50

50–60

5

x

20

15

y

5

5+x

25+x

40+x

40+x+y

45+x+y

Total

60

Here,

n = 60

or 45+x+y = 60

or x+y =15 ...(1)

Median of data is given as 28.5 that lies in the interval 20–30.

So, median class = 20–30

\begin{array}{l} From the given data, we have \\ l = 20, \\ Cumulative frequency (c f) of the class preceding the \\ median class = 5 + x, \\ Frequency of median class (f) = 20, \\ Class size (h) = 10 \\ Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h \\ or 28.5 = 20 + (\frac{\frac{60}{2} - 5 - x}{20}) \times 10 \\ or x = 8 \\ On putting this value of x in equation (1), we get \\ x + y = 15 \\ or         8 + y = 15 \\ or y = 15 - 8 = 7 \\ Hence, values of x and y are 8 and 7 respectively . \end{array}

Q:

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 / U .T.

15– 20

20– 25

25–30

30–35

35–40

40–45

45–50

50–55 3

8

9

10

3

0

0

2

A:

\begin{array}{l} From the given data, we have \\ l = 30, f_{1} = 10, f_{0} = 9, f_{2} = 3, h = 5 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 30 + (\frac{10 - 9}{2 \times 10 - 9 - 3}) \times 5 \\ = 30.625 \approx 30.6 \\ Interpretation: \\ Most of the states/UT have a teacher student ratio as 30 .6 . \\ Now, we find class mark for each interval by using the \\ following relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 32.5 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}

Number of students per teacher	Number of states/UT (f_i)	x_i	d_i = x_i - 32.5	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 32.5}{h} \end{array}$ b	f_iu_i
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	0	0	0
35 - 40	3	37.5	5	1	3
40 - 45	0	42.5	10	2	0
45 - 50	0	47.5	15	3	0
50 - 55	2	52.5	20	4	8
Total	35				-23

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 32.5 + \frac{- 23}{35} \times 5 \\ = 29.2 \\ Interpretation: \\ Average teacher student ratio of the states/UT is 29.2. \end{array}$

Q:

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.

A:

\begin{array}{l} We find class mark for each interval by using the following relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 30 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}

Age (in years)	Number of patients (f_i)	Class mark (x_i)	d_i = x_i - 30	f_id_i
5 -15	6	10	-20	-120
15 - 25	11	20	-10	-110
25 - 35	21	30	0	0
35 - 45	23	40	10	230
45 - 55	14	50	20	280
55 - 65	5	60	30	150

$\begin{array}{l} Total = 80 Total = 430 \\ Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ = 30 + \frac{430}{80} \\ = 35.375 \approx 35.38 \\ Therefore, mean of the given data is 35.38 years . \\ Modal class is 35 - 45. \\ l = 35, f_{1} = 23, f_{0} = 21, f_{2} = 14, h = 10 \end{array}$

$\begin{array}{l} Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 35 + (\frac{23 - 21}{2 \times 23 - 21 - 14}) \times 10 \\ = 36.8 \\ Mode is 36 .8 which represents that maximum number of patients admitted in hospital are of 36 .8 years . \end{array}$

$\begin{array}{l} While on average the age of a patient admitted to the hospital is 35.38 years \end{array}$

$\begin{array}{l} \end{array}$

Q:

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 85-95

Number of cities 3 10 11 8 3

A:

\begin{array}{l} We find class mark for each interval by using the following relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 10 \\ Here, we take assumed mean(a) = 70 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}

Literacy rate (in %)	Number of cities f_i	x_i	d_i = x_i - 70	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 70}{h} \end{array}$	f_iu_i
45 - 55	3	50	-20	-2	-6
55 - 65	10	60	-10	-1	-10
65 - 75	11	70	0	0	0
75 - 85	8	80	10	1	8
85 - 95	3	90	20	2	6
Total	35				-2

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 70 + (\frac{- 2}{35}) \times 10 = 69.43 \\ Therefore, mean literacy rate is 69.43. \end{array}$

Q:

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

A:

\begin{array}{l} We find class mark for each interval by using the following \\ relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 17 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}

Number of days	Number of students f_i	x_i	d_i = x_i - 17	f_id_i
0 - 6	11	3	-14	-154
6 - 10	10	8	-9	-90
10 - 14	7	12	-5	-35
14 - 20	4	17	0	0
20 - 28	4	24	7	28
28 - 38	3	33	16	48
38 - 40	1	39	22	22
Total	40			-181

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ = 17 + \frac{- 181}{40} \\ = 12.475 \approx 12.48 \\ Therefore, mean number of days for which a student was absent is 12.48. \end{array}$

Q:

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

0.04– 0.08

0.08 – 0.12

0.12–0.16

0.16–0.20

0.20–0.24 4

9

9

2

4

2

Find the mean concentration of SO₂ in the air.

A:

\begin{array}{l} We find class mark for each interval by using the following \\ relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 0.04 \\ Here, we take assumed mean(a) = 0.14 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}

Concentration of SO₂ (in ppm)	Frequency (f_i)	x_i	d_i = x_i - 0.14	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 0.14}{h} \end{array}$ b	f_iu_i
0.00 - 0.04	4	0.02	-0.12	-3	-12
0.04 - 0.08	9	0.06	-0.08	-2	-18
0.08 - 0.12	9	0.10	-0.04	-1	-9
0.12 - 0.16	2	0.14	0	0	0
0.16 - 0.20	4	0.18	0.04	1	4
0.20 - 0.24	2	0.22	0.08	2	4
Total	30				-31

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 0.14 + (\frac{- 31}{30}) \times 0.04 \\ \approx 0.099 ppm \\ {Therefore, mean concentration of SO}_{2} in the air is 0.099 ppm . \end{array}$

Q:

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

Daily Expenditure (in ₹) 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.

A:

\begin{array}{l} We find class mark for each interval by using the following relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 50 \\ Here, we take assumed mean(a) = 225 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}

Class interval	f_i	x_i	d_i = x_i - 225	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 57}{h} \end{array}$	f_iu_i
100 - 150	4	125	-100	-2	-8
150 - 200	5	175	-50	-1	-5
200 - 250	12	225	0	0	0
250 - 300	2	275	50	1	2
300 - 350	2	325	100	2	4
Total					-7

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 225 + (\frac{- 7}{25}) \times 50 = 211 \\ Therefore, mean daily expenditure is ₹ 211. \end{array}$

Q:

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 59-61 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?

A:

\begin{array}{l} The given class intervals are not continuous . So, we add 0 .5 to upper class limit and subtract 0 .5 to lower class limit \\ of each interval . We find class mark for each interval by using the following \\ relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 3 \\ Here, we take assumed mean(a) = 57 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}

Class interval	f_i	x_i	d_i = x_i - 57	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 57}{h} \end{array}$	$f_{i} u_{i}$
49.5 - 52.5	15	51	-6	-2	-30
52.5 - 55.5	110	54	-3	-1	-110
55.5 - 58.5	135	57	0	0	0
58.5 - 61.5	115	60	3	1	115
61.5 - 64.5	25	63	6	2	50
Total	400				25

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 57 + (\frac{25}{400}) \times 3 = 57.1875 \approx 57.19 \\ Therefore, mean number of mangoes is 57.19. \\ We have chosen step deviation method as values of f_{i}, d_{i} are big \\ and there is a common multiple between all d_{i} . \end{array}$

Q:

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.

Number of heartbeats per minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86

Number of women 2 4 3 8 7 4 2

A:

\begin{array}{l} We find class mark for each interval by using the following \\ relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 3 \\ Here, we take assumed mean(a) = 75.5 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}

Number of heartbeats per minute	Number of women (f_i)	x_i	d_i = x_i - 75.5	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 75.5}{h} \end{array}$	f_iu_i
65-68	2	66.5	-9	-3	-6
68-71	4	69.5	-6	-2	-8
71-74	3	72.5	-3	-1	-3
74-77	8	75.5	0	0	0
77-80	7	78.5	3	1	7
80-83	4	81.5	6	2	8
83-86	2	84.5	9	3	6
Total	30				4

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 75.5 + (\frac{4}{30}) \times 3 = 75.9 \\ Therefore, mean heartbeats per minute is 75.9. \end{array}$

Q:

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

Daily pocket allowance (in ₹) 11-13 13-15 15-17 17-19 18-21 21-23 23-25

Number of children 7 6 9 13 f 5 4

A:

\begin{array}{l} We find class mark for each interval by using the following \\ relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 2 \\ Given mean(a) = 18 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}

Daily pocket allowance (in ₹)	Number of children (f_i)	x_i	d_i = x_i - 18	f_id_i
11 - 13	7	12	-6	-42
13 - 15	6	14	-4	-24
15 - 17	9	16	-2	-18
17 - 19	13	18	0	0
19 - 21	f	20	2	2f
21 - 23	5	22	4	20
23 - 25	4	24	6	24
Total	$\sum f_{i} = 44 + f$			2f - 40

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ or 18 = 18 + \frac{2 f - 40}{44 + f} \\ or f = 20 \\ Therefore, missing frequency f is 20. \end{array}$

Q:

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

Daily wages (in ₹) 500-120 520-140 540-160 560-180 580-200

Number of workers 12 14 8 6 10

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

A:

\begin{array}{l} We find class mark for each interval by using the following \\ relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 20 \\ Here, we take assumed mean(a) = 550 and proceed as below . \end{array}

Daily wages (in ₹)	Number of workers (f_i)	x_i	d_i = x_i - 550	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 550}{h} \end{array}$	f_iu_i
500 - 520	12	510	-40	-2	-24
520 - 540	14	530	-20	-1	-14
540 - 560	8	550	0	0	0
560 - 580	6	570	20	1	6
580 - 600	10	590	40	2	20
Total	50				-12

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 550 + (\frac{- 12}{50}) \times 20 = 545.2 \end{array}$

$\begin{array}{l} Therefore, mean daily wages of the workers is ₹ 545.20 \end{array}$

Q:

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.

A:

\begin{array}{l} We find the cumulative frequency (c f) as below in the \\ following table on the basis of the given information . \end{array}

Number of letters	Number of surnames (f_i)	Cumulative frequency (cf)
1 - 4	6	6
4 - 7	30	36
7 - 10	40	76
10 - 13	16	92
13 - 16	4	96
16 - 19	4	100

$\begin{array}{l} Total(n) = 100 \\ We have, n = 100. \\ ∴ \frac{n}{2} = \frac{100}{2} = 50 \\ Cumulative frequency (c f) just greater than \frac{n}{2} = 50 is 76 \\ which belongs to interval 7 - 10. \\ ∴ Median class = 7 - 10 \end{array}$

$\begin{array}{l} Now, we have \\ l = 7, \\ Cumulative frequency (c f) of the class preceding the \\ median class = 36, \\ Frequency of median class (f) = 40, \\ Class size (h) = 3 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h \\ ∴ Median = 7 + (\frac{\frac{100}{2} - 36}{40}) \times 3 \\ or Median = 8.05 \\ ∴ Median number of letters in the surnames is 8 .05 . \\ Now, we find class mark for each interval by using the \\ following relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 11.5 \end{array}$

We proceed as below to find d_i, u_i, f_iu_i .

Number of letters	Number of surnames (f_i)	x_i	d_i = x_i - a	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - a}{h} \end{array}$	f_iu_i
1 - 4	6	2.5	-9	-3	-18
4 - 7	30	5.5	-6	-2	-60
7 - 10	40	8.5	-3	-1	-40
10 - 13	16	11.5	0	0	0
13 - 16	4	14.5	3	1	4
16 - 19	4	17.5	6	2	8
Total (n)	100				-106

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 11.5 + \frac{- 106}{100} \times 3 \\ = 8.32 \\ We know that \\ 3 Median = Mode + 2 Mean \\ 3 \times 8.05 = Mode + 2 \times 8.32 \\ or Mode = 7.51 \\ Hence, mean number of letters in the surnames is 8.32 \\ and modal size of the surnames is 7.51. \end{array}$

Q:

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 consumers

65 – 85

85– 105

105– 125

125–145

145–165

165–185

185–205 4

5

13

20

14

8

4

A:

\begin{array}{l} We find class mark for each interval by using the \\ following relation . \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 135 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}

Monthly consumption (in units)	Number of consumers (f_i)	x_i	d_i = x_i - 135	$\begin{array}{l} u_{i} \\ = \frac{x_{i} - 135}{h} \end{array}$	f_iu_i
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	0	0	0
145 - 165	14	155	20	1	14
165 - 185	8	175	40	2	16
185 - 205	4	195	60	3	12
Total	68				7

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 135 + \frac{7}{68} \times 20 \\ = 137.058 \\ Modal class = 125 - 145 \\ From the given data, we have \\ l = 125, f_{1} = 20, f_{0} = 13, f_{2} = 14, h = 20 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 125 + (\frac{20 - 13}{2 \times 20 - 13 - 14}) \times 20 \\ = 135.76 \\ We know that \\ 3 Median = Mode + 2 Mean \\ or Median = \frac{135.76 + 2 \times 137.058}{3} = 136.625 \end{array}$

We observe all these three measures are approximately same.

NCERT Solutions For Class 10 Mathematics Chapter 14 Statistics

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Sample Paper

Exam Weightage

Expenditure (in ₹)	Number of families
1000 – 1500 1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000	24 40 33 28 30 22 16 7

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

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

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

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

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

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

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

Weight (in kg)	Number of students
Less than 38 Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52	0 3 5 9 14 28 32 35

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

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

Class interval	Frequency
Below 20 Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60	2 6 24 45 78 89 92 98 100

Class interval	Frequency
0 – 10 10– 20 20– 30 30–40 40–50 50–60	5 x 20 15 y 5
Total	60

Number of students per teacher	Number of states / U .T.
15– 20 20– 25 25–30 30–35 35–40 40–45 45–50 50–55	3 8 9 10 3 0 0 2

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

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

Concentration of SO₂ (in ppm)	Frequency
0.00 – 0.04 0.04– 0.08 0.08 – 0.12 0.12–0.16 0.16–0.20 0.20–0.24	4 9 9 2 4 2

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

Number of mangoes	50-52	53-55	56-58	59-61	62-64
Number of boxes	15	110	135	115	25

Daily pocket allowance (in ₹)	11-13	13-15	15-17	17-19	18-21	21-23	23-25
Number of children	7	6	9	13	f	5	4

NCERT Solutions For Class 10 Mathematics Chapter 14 Statistics

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The following table gives production yield per hectare of wheat of 100 farms of a village. Production yield (in kg/ha) 50-55 55-60 60-65 65-70 70-75 75-80 Number of farms 2 8 12 24 38 16 Change the distribution to a more than type distribution, and draw its ogive.

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

The following table gives the distribution of the life time of 400 neon lamps: Life time (in hours) Number of lamps 1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000 14 56 60 86 74 62 48 Find the median life time of a lamp.

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. 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

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

The following distribution gives the daily income of 50 workers of a factory. Daily income (in ₹) 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.

If the median of the distribution given below is 28.5, find the values of x and y. Class interval Frequency 0 – 10 10– 20 20– 30 30–40 40–50 50–60 5 x 20 15 y 5 Total 60

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.

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 85-95 Number of cities 3 10 11 8 3

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

The table below shows the daily expenditure on food of 25 households in a locality. Daily Expenditure (in ₹) 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.

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f. Daily pocket allowance (in ₹) 11-13 13-15 15-17 17-19 18-21 21-23 23-25 Number of children 7 6 9 13 f 5 4

Consider the following distribution of daily wages of 50 workers of a factory. Daily wages (in ₹) 500-120 520-140 540-160 560-180 580-200 Number of workers 12 14 8 6 10 Find the mean daily wages of the workers of the factory by using an appropriate method.