NCERT Solutions for Class 10 Maths Chapter 14 - Statistics

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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	Number of housesn
0 – 2	1
2 – 4	2
4 – 6	1
6 – 8	5
8 – 10	6
10 – 12	2
12 – 14	3

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

Sol. Let us find mean of the data by direct method because the figures are small.

Class (Number of plants)	Frequency f_i (Number of houses)	Class marks x_i	f_i × 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
Total	N=20		162

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

$$\text{We have, N =}\space\sum_{i=1}^{7}f_i=20\\\text{and\space}\sum_{i=1}^{7}f_ix_i=162$$

Then, mean of the data of

$$\bar x=\frac{1}{\text{N}}×\sum_{i=1}^{7}f_ix_i\\=\frac{1}{20}×162=8.1$$

Hence, the required mean of the data is 8.1 plants.

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

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

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

Sol. Here, we use step deviation method for finding the mean of the data.

Daily wages (in ₹)	Number of workers f_i	Class mark x_i	d_i = x_i – 150	$$u_i=\frac{d_i}{h}=\frac{x_i-150}{20}$$	f_i u_i
100-120	12	110	– 40	– 2	– 24
120-140	14	130	-20	-1	-14
140-160	8	150=a	0	0	0
160-180	6	170	20	1	6
180-200	10	190	40	2	20
Total	N = 50				– 12

Here, a (Assumed mean) = 150

h (Class width) = 20

N = 50

and Σf_iu_i = – 12

By step deviation method, we have
$$\text{Mean =}a+h×\frac{1}{h}\sum f_iu_i\\=\begin{Bmatrix}150+20×\frac{1}{50}×(-12)\end{Bmatrix}\\\begin{Bmatrix}150-\frac{24}{5}\end{Bmatrix}$$

= {150 – 4.80} = ₹ 145.20

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

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

Sol.

Daily pocket allowance (in ₹)	Number of children f_i	Class mark x_i	d_i = x_i – 18	f_i×d_i
11 – 13	7	12	-6	-42
13 – 15	6	14	-4	-24
15 – 17	9	16	-2	-18
17 – 19	13	18=a	0	0
19 – 21	f	20	2	2f
21 – 23	5	22	4	20
23 – 25	4	24	6	24
Total	N = 44 + f			2f – 40

Here,

$$\bar x=₹\space18$$

From the table, we have

a (assumed mean) = 18, N = 44 + f

$$\text{Now,\space mean}=a+\frac{1}{\text{N}}×\Sigma f_id_i$$

Then, substituting the value as given above, we have

$$18=18+\frac{1}{44-f}×(2f-40)\\\Rarr\space 0=\frac{2f-40}{44+f}$$

⇒ 2f = 40

⇒ f = 20

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

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

Sol.

Number of heart beats per minute	Number of women f_i	Class mark x_i	$$u_i=\frac{x_i-a}{h}=\frac{x_i-75.5}{3}$$	f_i u_i
65 – 68	2	66.5	$$u_i=\frac{66.5-75.5}{3}=-3$$	– 6
68 – 71	4	69.5	$$u_2=\frac{69.5-75.5}{3}=-2$$	-8
71 – 74	3	72.5	$$u_3=\frac{72.5-75.5}{3}=-3$$	-3
74 – 77	8	75.5 = a	$$u_4=\frac{75.5-75.5}{3}=0$$	0
77 – 80	7	78.5	$$u_5=\frac{78.5-75.5}{3}=1$$	7
80-83	4	81.5	$$u_6=\frac{81.5-75.5}{3}=2$$	8
83 – 86	2	84.5	$$u_7=\frac{84.5-75.5}{3}=3$$	6
Total	N = 30			4

Here, a (Assumed mean) = 75.5

h (Class width) = 3

N = 30

By step deviation method,

$$\text{Mean}=a+h×\frac{1}{h}\Sigma f_iu_i\\=75.5+3×\frac{1}{30}×4$$

= 75.5 + 0.4 = 75.9 heart beats

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

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

Sol. Since, given data is not continuous, so we subtract and add 0.5 respectively in the lower and upper limit of each class.

Number of mangoes	Number of boxes (f_i)	Class mark (x_i)	$$u_i=\frac{x_i-a}{h}=\frac{x_i-57}{3}$$	f_i u_i
49.5 – 52.5	15	51	-2	-30
52.5 – 55.5	110	54	-1	-110
55.5 – 58.5	135	a = 57	0	0
58.5 – 61.5	115	60	1	115
61.5 – 64.5	25	63	2	50
Total	N = 400			25

Here, a = 57

h = 3

N = 400

$$\text{Now,\space \text{Mean}}=a+h×\frac{1}{h}×\Sigma f_iu_i\\=57+3×\frac{1}{400}×25\\=57+\frac{75}{400}$$

= 57 + 0.1875 = 57 + 0.19 = 57.19

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

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

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

Sol.

Daily expenditure (in ₹)	Number of households (f_i)	Class mark (x_i)	$$u_i=\frac{x_i-a}{h}=\frac{x_i-225}{50}$$	f_i u_i
100 – 150	4	125	-2	-8
150 – 200	5	175	-1	-5
200 – 250	12	a = 225	0	0
250 – 300	2	275	1	2
300 – 350	2	325	2	4
TOTAL	N = 25			– 7

Here,

a = 225

h = 50

N = 25

$$\text{Now,\space Mean =a+h}\space×\frac{1}{h}×\Sigma f_iu_i\\=225+50×\frac{1}{25}×(\normalsize-7)$$

= 225 – 14 = ₹ 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.

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.

Sol.

Concentration of SO₂ (in ppm)	f_i	Class mark (x_i)	$$u_i=\frac{x_i-a}{h}=\frac{x_i-0.10}{0.04}$$	f_i u_i
0.00 – 0.04	4	0.02	-2	-8
0.04 – 0.08	9	0.06	-1	-9
0.04 – 0.08	9	0.06	-1	-9
0.08 – 0.12	9	0.10 = a	0	0
0.12 – 0.16	2	0.14	1	2
0.16 – 0.20	4	0.18	2	8
0.20 – 0.24	2	0.22	3	6
Total	N=30			-1

Here, a = 0.10

h = 0.04

N = 30

$$\text{Now,\space Mean}=a+h×\frac{1}{\text{N}}×\Sigma f_iu_i\\=0.10+0.04×\frac{1}{30}×(\normalsize-1)\\=0.10=\frac{0.04}{30}$$

= 0.10 – 0.001 = 0.099 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.

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

Sol. Here, the class size varies but we will apply assumed mean method.

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

Here, a = 17

N = 40

By assumed mean method,

$$\text{Mean}=a+\frac{1}{\text{N}}×\Sigma f_id_i\\=17+\frac{1}{40}×(-181)\\=17-\frac{18.1}{4}$$

= 12.48 days

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

Literacy rate (in %)	Number of cities
45 – 55	3
55-65	10
65-75	11
75 – 85	8
85 – 95	3

Sol.

Literacy rate (in %)	Number of cities (f_i)	Class marks (x_i)	$$u_i=\frac{x_i-a}{h}=\frac{x_i-70}{10}$$	f_i u_i
45 – 55	3	50	-2	-6
55-65	10	60	-1	-10
65-75	11	70=a	0	0
75 – 85	8	80	1	8
85 – 95	3	90	2	6
Total	N=35			-2

Here, a = 70

h = 10

N = 25

By step deviation method,

$$\text{Mean}=a+h×\frac{1}{h}×\Sigma f_iu_i\\=70+10×\frac{1}{35}×(\normalsize-2)\\=70-\frac{4}{7} = 70 – 0.57 = 69.43%$$

Exercise 14.2

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

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

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

Sol.

Age (in years)	Number of patients (f_i)	Class mark (x_i)	$$u_i=\frac{x_i-a}{h}=\frac{x_i-30}{10}$$	f_i u_i
5 – 15	6	10	-2	-12
15 – 25	11	20	-1	-11
25 – 35	21	30=a	0	0
35 – 45	23	40	1	23
45 – 55	14	50	2	28
55 – 65	5	60	3	15
Total	N=80			43

Here a = 30, h = 10, N = 80

Mode :

For the given data, we have the modal class 35 – 43, with maximum frequency 23.

$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h$$

Here, l = 35, f_m = 23, f₁ = 21, f₂ = 14, h = 10

$$=35+\begin{Bmatrix*}[r]\frac{23-21}{46-21-14}\end{Bmatrix*}×10\\=35+\frac{20}{11}$$

= 35 + 1.8 = 36.8 yr

Mean :

Now, by step deviation method,

$$\text{Mean = a+h}×\frac{1}{\text{N}}×\Sigma f_iu_i\\=30+10×\frac{1}{80}×43$$

= 30 + 5.37 = 35.37 yr

Thus, mode = 36.8 yr and mean = 35.57 yr. So, we conclude that the maximum number of patients admitted in the hospital are of age 36.8 yr (approx.), whereas on an average age of a patient admitted to the hospital is 35.37 yr.

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

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

Determine the modal lifetimes of the components.

Sol. Modal class of the given data is 60–80. Because it has largest frequency among the given classes of the data.

Here, (Lower limit of modal class) l = 60, f_m = 61, f₁ = 52, f₂ = 38 and (Class width) h = 20

$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=60+\begin{Bmatrix*}[r]\frac{61-52}{122-52-38}\end{Bmatrix*}×20\\=60+\frac{9×20}{32}\\=60+\frac{45}{8}$$

= 60 + 5.625 = 65.625 h

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 mean monthly expenditure.

Expenditure (in ₹)	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

Sol. For mode : Modal class of the data is (1500-2000) because it has maximum frequency among the given classes of the data.

(Lower limit of modal class) l = 1500, f_m = 40, f₁ = 24, f₂ = 33, (Class width) h = 500

$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=1500+\begin{Bmatrix*}[r]\frac{40-24}{80-24-33}\end{Bmatrix*}×500\\=1500+\frac{16×500}{23}\\=1500+\frac{8000}{23}$$

= 1500 + 347.83

= ₹ 1847.83

For mean :

Expenditure (in ₹)	Number of families (f_i)	Class mark (x_i)	$$u_i=\frac{x_i-A}{h}$$	f_iu_i
1000 – 1500	24	1250	-3	-72
1500 – 2000	40	1750	-2	-80
2000 – 2500	33	2250	-1	-33
2500 – 3000	28	2750=a	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
Total	200			-35

So, a = 2750, h = 500, N = 200 and Σf_iu_i = – 35

Now, by step deviation method,

$$\text{Mean = a+h}×\frac{1}{h}×\Sigma f_iu_i\\=2750+500×\frac{1}{200}×(-35)\\=2750-\frac{5×35}{2}$$

= 2750 – 87.50 = ₹ 2662.50

Thus, the modal monthly expenditure is ₹ 1847.83 and the mean monthly expenditure is ₹ 2662.50.

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	0
45 – 50	0
50 – 55	2

Sol.

Number of students per teacher	Number of states/UT(f_i)	Class mark(x_i)	$$u_i=\frac{x_i-a}{h}=\frac{x_i-32.5}{5}$$	f_i × u_i
15 – 20	3	17.5	-3	-9
20 – 25	8	22.5	-2	-16
25 – 30	9	27.5	-1	-9
30 – 35	10	32.5=a	0	0
35 – 40	3	37.5	1	3
40 – 45	0	42.5	2	0
45 – 50	0	47.5	2	0
50 – 55	2	52.5	4	8
Total	N=35			-23

Mode : Since, largest frequency is fm = 10, so its modal class is (30–35)

∴ (Lower limit of modal class) l = 30, fm = 10, f1 = 9, f2 = 3 (Class width) h = 5

$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=30+\begin{Bmatrix*}[r]\frac{10-9}{20-9-3}\end{Bmatrix*}×5\\=30+\frac{5}{8}$$

= 30 + 0.625

= 30.625 = 30.6

Mean :

Here, a = 32.5, h = 5, N = 35 and Σf_iu_i = – 23

Now, by step deviation method,

$$\text{Mean = a+h}×\frac{1}{h}×\Sigma f_iu_i\\=32.5+5×\frac{1}{35}×(-23)\\=32.5-\frac{23}{7}$$

= 32.5 – 3.3 = 29.2

Hence, mode = 30.6 and mean = 29.2. We conclude that most states/UT have a student teacher ratio of 30.6 and on an average ratio is 29.2.

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 – 10000	1
10000 – 11000	1

Find the mode of the data.

Sol. Since, maximum frequency is fm = 18, so its modal class is 4000-5000.

∴ (Lower limit of modal class) l = 4000, f_m = 18, f₁ = 4, f₂ = 9, (Class width) h = 1000

$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=4000+\begin{Bmatrix*}[r]\frac{18-4}{36-4-9}\end{Bmatrix*}×1000\\=4000+\frac{14000}{23}$$

= 4000 + 608.7

= 4608.7 runs

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 given below.Find the mode of the data

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

Sol. Since, maximum frequency is f_m = 20, so its modal class is (40-50).

∴ (Lower limit of modal class) l = 40, f_m = 20, f₁ = 12, f₂ = 11, (Class width) h = 10

$$\text{Mode = l}\space+\space\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=40+\begin{Bmatrix*}[r]\frac{20-12}{40-12-11}\end{Bmatrix*}×10\\=40+\frac{80}{17}$$

= 40 + 4.71

= 44.71 cars.

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.

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

Sol.

Monthly consumption (in units)	Number of consumers (f_i)	Cumulative frequency (cf)
65 – 85	4	4
85 – 105	5	9
105 – 125	13	cf=22
125 – 145	20=f	42
145 – 165	14	56
165 – 185	8	64
185-205	4	68
Total	N=68

$$\textbf{(i) For median} :\\\text{Here, N = 68 gives}\space\frac{N}{2}=34$$

So, we have the median class (125 – 145).

(Lower median class) l = 125, N = 68, f = 20, cf = 22, (Class width) h = 20

$$\text{Median}=l+\begin{Bmatrix*}[r]\frac{\frac{N}{2}-cf}{f}\end{Bmatrix*}×h\\=125+\begin{Bmatrix*}[r]\frac{34-22}{20}\end{Bmatrix*}×20\\=125+\frac{12}{20}×20$$

= 125 + 12 = 137 units

(ii) For mode : Since, maximum frequency is f_m= 20 so its modal class is 125–145. Then f₁ = 13, f₂ = 14,
f_m = 20, l = 125 and h = 20

$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=125+\begin{Bmatrix*}[r]\frac{20-13}{40-13-14}\end{Bmatrix*}×20\\=125+\frac{7×20}{13}\\=125+\frac{140}{13}$$

= 125 + 10.76 = 135.76 units

(iii) For mean

Monthly consumption (in units)	Number of consumers (fi)	Class mark (x_i)	$$u_i=\frac{x_i-a}{h}=\frac{x_i-135}{20}$$	f_i u_i
65 – 85	4	75	-3	-12
85 – 105	5	95	-2	-10
105 – 125	13	115	-1	-13
125 – 145	20	a=135	0	0
145 – 165	14	155	1	14
165 – 185	8	175	2	16
185 – 205	4	195	3	12
Total	N = 68			7

Here, N = 68, a = 135, h = 20 and Σf_iu_i = 7

Now, by step deviation method

$$\text{Mean = a + h}×\frac{1}{N}×\Sigma f_iu_i\\=135+20×\frac{1}{68}×7\\=135+\frac{35}{17}$$

= 135 + 2.05 = 137.05 units

Hence, the three measures are approximately the same in this cases.

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

Sol.

Class interval	Frequency (f)	Cumulative frequency
0 – 10	5	5
10 – 20	x	5 + x = cf
20 – 30	20 = f	25 + x
30 – 40	15	40 + x
40 – 50	y	40 + x + y
50 – 60	5	45 + x + y
Total	60

Given, median = 28.5 lies in the class interval (20–30).

Thus, median class is (20–30).

Here, l = 20, f = 20, cf = 5 + x (Class width) h = 10, (Total observations) N = 60

$$\text{Median = l}+\begin{Bmatrix*}[r]\frac{\frac{N}{2}-cf}{f}\end{Bmatrix*}×h\\\therefore\space 28.5=20 + \begin{Bmatrix*}[r]\frac{\frac{60}{2}-(5+x)}{20}\end{Bmatrix*}×10\\\Rarr\space 8.5=\frac{25-x}{2}$$

⇒ 17 = 25 – x

⇒ x = 8

From the given table, we have

i.e., x + y + 45 = 60

⇒ x + y = 15

⇒ y = 15 – x

= 15 – 8 = 7

⇒ y = 7

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

3. A life insurance agent found the following data for distribution of ages of 100 policy holders.

Calculate the median age, if the policies are only given only to persons having age 18 year onwards but less than 60 year.

Age (in years)	Number of Policy holders
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

Sol.

Age (in years)	Number of Policy holders	Cumulative frequency
Below 20	2=2	2
20 – 25	(6 – 2) = 4	6
25 – 30	(24 – 6) = 18	24
30 – 35	(45 – 24) = 21	45 = cf
(Median class) 35 – 40	(78 – 45) = 33 = f	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

$$\because\space\frac{N}{2}=\frac{100}{2}=50$$

Here, l = 35, f = 33, cf = 45, h = 5, N = 100

$$\text{Median = l + }\begin{Bmatrix*}[r]\frac{\frac{N}{2}-cf}{f}\end{Bmatrix*}×h\\=35+\begin{Bmatrix*}[r]\frac{50-45}{33}\end{Bmatrix*}×5\\=35+\frac{25}{33}$$

= 35 + 0.76 = 35.76 age

Hence,the median age is 35.76 years.

4. The length 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.

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

Sol. Since, given classes is not continuous, so it needs to be converted into continuous classes for finding the median. The classes, then change to 117.5–126.5, 126.5–135.5, ... 171.5–180.5.

Length (in mm)	Number of leaves f_i	Cumulative frequency
117.5 – 126.5	3	3
126.5 – 135.5	5	(3 + 5) = 8
135.5 – 144.5	9	(8 + 9) = 17 = cf
144.5 – 153.5	12 = f	(17 + 12) = 29 (Median class)
153.5 – 162.5	5	(29 + 5) = 34
162.5 – 171.5	4	(34 + 4) = 38
171.5 – 180.5	2	(38 + 2) = 40
Total	N = 40

$$\because\space\frac{N}{2}=\frac{40}{2}=20$$

Since, cumulative frequency 20 lies in the interval 144.5 – 153.5

Here, l = 144.5, f = 12, cf = 17, h = 9, N = 40

$$\text{Median = l}\space+\space\begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix} ×h\\=144.5+\begin{Bmatrix}\frac{20-17}{12}\end{Bmatrix}×9\\=144.5+\frac{3×3}{4}\\=144.5+\frac{9}{4}$$

= 144.5 + 2.25

= 146.75 mm

Hence, the median length of the leaves in 146.75 mm.

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

Life time (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.

Sol.

Life time (in hours) Number of lamps f_i Cumulative frequency cf 1500 – 2000 14 14 2000 – 2500 56 (14 + 56) = 70 2500 – 3000 60 (70 + 60) = 130 = cf 3000 – 3500 86 = f (130 + 86) = 216 (Median class) 3500 – 4000 74 (216 + 74) = 290 4000 – 4500 62 (290 + 62) = 352 4500 – 5000 48 (352 + 48) = 400 Total N = 400

$$\because\space\frac{N}{2}=\frac{400}{2}=200$$

Since, cumulative frequency 20 lies in the interval 3000–3500.

Here, l = 3000, f = 86, cf = 130, h = 500, N = 400

$$\text{Median = l}\space+\space\begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix}×h\\=3000+\frac{\begin{Bmatrix}200-130\end{Bmatrix}}{86}×500\\=3000+\frac{70×500}{86}\\=3000+\frac{35000}{86}$$

= 3000 + 406.98 = 3406.98 h

Hence, median life time of a lamp is 3406.98 h.

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

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

Sol. (i)

Number of letters	Number of surnames	Cumulative frequency
1 – 4	6	6
4 – 7	30	6 + 30 = 36 = cf
07 – 10	40 = f	36 + 40 = 76 (median class)
10 – 13	16	76 + 16 = 92
13 – 16	4	92 + 4 = 96
16 – 19	4	96 + 4 = 100
Total	N = 100

$$\because\space\frac{N}{2}=\frac{100}{2}=50$$

Since, cumulative frequency 50 lies in the interval 7 – 10.

Here, l = 7, f = 40, cf = 36, h = 3, N = 100

$$\text{Median = l }+\begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix}×h\\=7+\begin{Bmatrix}\frac{50-36}{40}\end{Bmatrix}×3\\=7+\frac{14×3}{40}\\=7+\frac{21}{20}=7+1.05$$

= 8.05

Hence, the median number of letters in the surnames is 8.05.

(ii) Modal class is (7 – 10), because it has maximum frequency as f_m = 40.

Here, l = 7, f_m = 40, f₁ = 30, f₂ = 16, h = 3

$$\text{Mode = l}\space+\space\begin{Bmatrix}\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix}×h\\=7+\begin{Bmatrix}\frac{40-30}{80-30-16}\end{Bmatrix}×3\\=7+\frac{30}{34}$$

= 7 + 0.88

= 7.88

Hence, the modal size of the surnames is 7.88.

(iii)

Number of letters	f_i	Class mark (x_i)	$$u_i=\frac{x_i-a}{h}=\frac{x_i-8.5}{3}$$	f_i × u_i
1 – 4	6	2.5	-2	-12
4 – 7	30	5.5	-1	-30
7 – 10	40	8.5 = a	0	0
10 – 13	16	11.5	1	16
13 – 16	4	14.5	2	8
16 – 19	4	17.5	3	12
Total	N = 100			-6

Here, N = 100, a = 8.5, h = 3 and Σf_iu_i = – 6

Now, by step deviation method,

$$\text{Mean = a+h×}\frac{1}{N}×\Sigma f_iu_i\\=8.5+3×\frac{1}{100}×(-6)\\=8.5-\frac{18}{100}$$

= 8.5 – 0.18

= 8.32

Hence, the mean number of letters in the surname is 8.32.

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

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

Sol.

Weight (in kg)	Number of students (f_i)	Cumulative frequency
40 – 45	2	2
45 – 50	3	2+3=5
50 – 55	8	5 + 8 = 13 = cf
55 – 60	6 = f	13 + 6 = 19 (median class)
60 – 65	6	19 + 6 = 25
65 – 70	3	25 + 3 = 28
70 – 75	2	28 + 2 = 30
Total	N=30

$$\because\space\frac{N}{2}=\frac{30}{2}=15$$

Since, cumulative frequency 15 lies in the interval 55–60.

Here, l = 55, f = 6, cf = 13, h = 5, N = 30

$$\text{Median = l} + \begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix}×h\\55+\begin{Bmatrix}\frac{15-13}{6}\end{Bmatrix}×5\\=55+\frac{5}{3}$$

= 55 + 1.67

= 56.67 kg

Hence, the median weight of the students is 56.67 kg.

Exercise 14.4

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

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

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

Sol.

Daily income (in ₹)	Number of workers (Frequency)	Cumulative frequency less than type
100 – 120	12	Less than 120	12=12
120 – 140	14	Less than 140	(12 + 14) = 26
140 – 160	8	Less than 160	(26 + 8) = 34
160 – 180	6	Less than 180	(34 + 6) = 40
180 – 200	10	Less than 200	(40 + 10) = 50
Total	N=50

$$\because\space\text{N = 50, gives}\space\frac{n}{2}=25$$

On the graph, we will plot the points (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50).

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

Weight (in kg)	Number of students
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

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

Sol.

Weight (in kg)	Number of students (Frequency) f_i	Cumulative frequency less than type
Less than 38	0 = 0	Less than 38	0
38 – 40	(3 – 0) = 3	Less than 40	3
40 – 42	(5 – 3) = 2	Less than 42	5
42 – 44	(9 – 5) = 4	Less than 44	9
44 – 46	(14 – 9) = 5	Less than 46	14 = cf
46 – 48	(28 – 14) = 14 = f	Less than 48	28 (median class)
48 – 50	(32 – 28) = 4	Less than 50	32
50 – 52	(35 – 32) = 3	Less than 52	35
Total	N = 35

$$\because\space\frac{N}{2}=\frac{35}{2}=17.5$$

To draw the 'less than' type ogive, we plot the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35) on the graph.

Take point 17.5 taking on y-axis draw a line parallel to x-axis meet at a point L′ and draw a perpendicular line from M′ to the x-axis, the intersection point of X-axis is the median.

Median from the graph = 46.5 kg

Median class is (46-48)

We have, l = 46, f = 14, cf = 14, h = 2, N = 35

From the table :

$$\text{Median = l} + \begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix}×h\\=46+\begin{Bmatrix}\frac{\frac{35}{2}-14}{14}\end{Bmatrix}×2\\=46+\frac{1}{2}$$

= 46 + 0.5

= 46.5 kg

Hence, the median is same as we have noticed from the graph.

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

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

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

Sol.

Production yield (in kg/ha)	Number of farms (Frequency)	Cumulative frequency more than type
50 – 55	2	More than 50	98 + 2 = 100
55 – 60	8	More than 55	8 + 90 = 98
60 – 65	12	More than 60	12 + 78 = 90
65 – 70	24	More than 65	24 + 54 = 78
70 – 75	38	More than 70	38 + 16 = 54
75 – 80	16	More than 75	=16
Total	N = 100