Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 2

Question 14. Find the mean of the following frequency distribution:

Class interval:	25 – 29	30 – 34	35 – 39	40 – 44	45 – 49	50 – 54	55 – 59
Frequency:	14	22	16	6	5	3	4

Solution:

Let’s consider the assumed mean (A) = 42

Class interval	Mid – value xi	di = xi – 42	ui = (xi – 42)/5	fi	fiui
25 – 29	27	-15	-3	14	-42
30 – 34	32	-10	-2	22	-44
35 – 39	37	-5	-1	16	-16
40 – 44	42	0	0	6	0
45 – 49	47	5	1	5	5
50 – 54	52	10	2	3	6
55 – 59	57	15	3	4	12
				N = 70	Σ fiui = -79

From the table it’s seen that,

A = 42 and h = 5

Mean = A + h x (Σf_i u_i/N)

= 42 + 5 x (-79/70)

= 42 – 79/14

= 42 – 5.643

= 36.357

Question 15. For the following distribution, calculate mean using all suitable methods:

Size of item:	1 – 4	4 – 9	9 – 16	16 – 20
Frequency:	6	12	26	20

Solution:

By direct method

Class interval	Mid value xi	Frequency fi	fixi
1 – 4	2.5	6	15
4 – 9	6.5	12	18
9 – 16	12.5	26	325
16 – 27	21.5	20	430
		N = 64	Sum = 848

Mean = (sum/N) + A

= 848/64

= 13.25

By assuming mean method

Let the assumed mean (A) = 65

Class interval	Mid value xi	ui = (xi – A) = xi – 65	Frequency fi	fiui
1 – 4	2.5	-4	6	-25
4 – 9	6.5	0	12	0
9 – 16	12.5	6	26	196
16 – 27	21.5	15	20	300
			N = 64	Sum = 432

Mean = A + sum/N

= 6.5 + 6.75

= 13.25

Question 16. The weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost if living index.

Cost of living index	Number of students	Cost of living index	Number of students
1400 – 1500	5	1700 – 1800	9
1500 – 1600	10	1800 – 1900	6
1600 – 1700	20	1900 – 2000	2

Solution:

Let the assumed mean (A) = 1650

Class interval	Mid value xi	di = xi – A = xi – 1650		Frequency fi	fiui
1400 – 1500	1450	-200	-2	5	-10
1500  – 1600	1550	-100	-1	10	-10
1600 – 1700	1650	0	0	20	0
1700 – 1800	1750	100	1	9	9
1800 – 1900	1850	200	2	6	12
1900 – 2000	1950	300	3	2	6
				N = 52	Sum = 7

We have

A = 16, h = 100

Mean = A + h (sum/N)

= 1650 + (175/13)

= 21625/13

= 1663.46

Question 17. The following table shows the marks scored by 140 students in an examination of a certain paper:

Marks:	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Number of students:	20	24	40	36	20

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

Solution:

(i) Direct method:

Class interval	Mid value xi	Frequency fi	fixi
0 – 10	5	20	100
10 – 20	15	24	360
20 – 30	25	40	1000
30 – 40	35	36	1260
40 – 50	45	20	900
		N = 140	Sum = 3620

Mean = sum/N

= 3620/140

= 25.857

(ii) Assumed mean method:

Let the assumed mean = 25

Mean = A + (sum/N)

Class interval	Mid value xi	ui = (xi – A)	Frequency fi	fiui
0 – 10	5	-20	20	-400
10 – 20	15	-10	24	-240
20 – 30	25	0	40	0
30 – 40	35	10	36	360
40 – 50	45	20	20	400
			N = 140	Sum = 120

Mean = A + (sum/N)

= 25 + (120/140)

= 25 + 0.857

= 25.857

(iii) Step deviation method:

Let the assumed mean (A) = 25

Class interval	Mid value xi	di = xi – A = xi – 25		Frequency fi	fiui
0 – 10	5	-20	-2	20	-40
10 – 20	15	-10	-1	24	-24
20 – 30	25	0	0	40	0
30 – 40	35	10	1	36	36
40 – 50	45	20	2	20	40
				N = 140	Sum = 12

Mean = A + h(sum/N)

= 25 + 10(12/140)

= 25 + 0.857

= 25.857

Question 18. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the miss frequency f₁ and f₂. 

Class:	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100	100 – 120
Frequency:	5	f1	10	f2	7	8

Solution:

Class interval	Mid value xi	Frequency fi	fixi
0 – 20	10	5	50
20 – 40	30	fi	30fi
40 – 60	50	10	500
60 – 80	70	f2	70f2
80 – 100	90	7	630
100 – 120	110	8	880
		N = 50	Sum = 30f1 + 70f2 + 2060

Given,

Sum of frequency = 50

5 + f₁ + 10 + f₂ + 7 + 8 = 50

f₁ + f₂ = 20

3f₁ + 3f₂ = 60        —(1) [Multiply both side by 3]

and mean = 62.8

Sum/N = 62.8

(30f₁ + 70f₂ + 2060)/50 = 62.8

30f₁ + 70f₂ = 3140 – 2060

30f₁ + 70f₂ = 1080

3f₁ + 7f₂ = 108      —(2) [divide it by 10]

Subtract equation (1) from equation (2)

3f₁ + 7f₂ – 3f₁ – 3f₂ = 108 – 60

4f₂ = 48

f₂ = 12

Put value of f₂ in equation (1)

3f₁ + 3(12) = 60

f₁ = 24/3 = 8

f₁ = 8, f₂ = 12

Question 19. The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs 18.00. Find out the missing frequency.

Class interval:	11 – 13	13 – 15	15 – 17	17 – 19	19 – 21	21 – 23	23 – 25
Frequency:	7	6	9	13	–	5	4

Solution:

Given mean = 18,

Let the missing frequency be v

Class interval	Mid interval xi	Frequency fi	fixi
11 – 13	12	7	84
13 – 15	14	6	88
15 – 17	16	9	144
17 – 19	18	13	234
19 – 21	20	x	20x
21 – 23	22	5	110
23 – 25	14	4	56
		N = 44 + x	Sum = 752 + 20x

Mean = sum/N

18 = 752 + 20×44 + x 

792 + 18x = 752 + 20x

2x = 40

x = 20

Question 20. If the mean of the following distribution is 27. Find the value of p.

Class:	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Frequency:	8	P	12	13	10

Solution:

Class interval	Mid value xi	Frequency fi	fixi
0 – 10	5	8	40
10 – 20	15	P	152
20 – 30	25	12	300
30 – 40	35	13	455
40 – 50	45	16	450
		N = 43 + P	Sum = 1245 + 15p

Given mean = 27

Mean = sum/N

1245 + 15p43 + p = 27

1245 + 15p = 1161 + 27p

12p = 84 

P =7

Question 21. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain 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 packing box. Which method of finding the mean did you choose?

Solution:

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

We may observe that class internals are not continuous

There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (x_i) may be obtained by using the relation

x_i = upperlimit + lowerclasslimit2

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculated d_i, u_i, f_iu_i as follows

Class interval	Frequency fi	Mid values xi	di = xi – A = xi – 25		fiui
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	N = 400				Sum = 25

Now we have N

Sum = 25

Mean = A +h (sum/N)

= 57 + 3 (45/400)

= 57 + 3/16

= 57 + 0.1875

= 57.19

Clearly mean number of mangoes kept in packing box is 57.19

Question 22. 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:

We may calculate class mark (x_i) for each interval by using the relation

x_i = upperlimit + lowerclasslimit2

Class size = 50

Now, Taking 225 as assumed mean (x_i) we may calculate d_i, u_i, f_iu_i as follows

Daily expenditure	Frequency f1	Mid value xi	di = xi – 225		fiui
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
	N = 25				Sum = -7

Now we may observe that

N = 25

Sum = -7

225 + 50 (-7/25)

225 – 14 = 211

So, mean daily expenditure on food is Rs 211

Question 23. To find out the concentration of SO₂ in the air (in parts per million i.e ppm) the data was collected for localities 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:

We may find class marks for each interval by using the relation

x = upperlimit + lowerclasslimit2x = 

Class size of this data = 0.04

Now taking 0.04 assumed mean (x_i) we may calculate d_i, u_i, f_iu_i as follows

Concentration of SO2	Frequency f1	Class interval xi`	di = xi – 0.14	ui	fiui
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	N = 30				Sum = -31

From the table we may observe that 

N = 30

Sum = -31

= 0.14 + (0.04)(-31/30)

= 0.099 ppm

So mean concentration of SO₂ in the air is 0.099 ppm.

Question 24. 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:

We may find class mark of each interval by using the relation

x =  upperlimit + lowerclasslimit2x = 

Now, taking 16 as assumed mean (a) we may

Calculate d_i and f_id_i as follows

Number of days	Number of students fi	Xi	d = xi + 10	fidi
0 – 6	11	3	-13	-143
6 – 10	10	8	-8	-280
10 – 14	7	12	-4	-28
14 – 20	7	16	0	0
20 – 28	8	24	8	32
28 – 36	3	33	17	51
30 – 40	1	39	23	23
Total	N = 40			Sum = -145

Now we may observe that

N = 40

Sum = -145

= 16 + (-145/40)

= 16 – 3.625 

= 12.38

So mean number of days is 12.38 days, for which student was absent.

Question 25. 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

Solution:

We may find class marks by using the relation

x = upperlimit + lowerclasslimit2x = 

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) wrong

Calculate d_i, u_i, f_iu_i as follows

Literacy rate (in %)	Number of cities (fi)	Mid value xi	di = xi – 70	ui = di – 50	fiui
45 – 55	3	50	-20	-20	-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	N = 35				Sum = -2

Now we may observe that

N = 35

Sum = -2 

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%

Question 21. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain 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 packing box. Which method of finding the mean did you choose?

Solution:

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

We may observe that class internals are not continuous

There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (x_i) may be obtained by using the relation

x_i = upperlimit + lowerclasslimit2

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculated d_i, u_i, f_iu_i as follows

Class interval	Frequency fi	Mid values xi	di = xi – A = xi – 25		fiui
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	N = 400				Sum = 25

Now we have N

Sum = 25

Mean = A +h (sum/N)

= 57 + 3 (45/400)

= 57 + 3/16

= 57 + 0.1875

= 57.19

Clearly mean number of mangoes kept in packing box is 57.19

Question 22. 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:

We may calculate class mark (x_i) for each interval by using the relation

x_i = upperlimit + lowerclasslimit2

Class size = 50

Now, Taking 225 as assumed mean (x_i) we may calculate d_i, u_i, f_iu_i as follows

Daily expenditure	Frequency f1	Mid value xi	di = xi – 225		fiui
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
	N = 25				Sum = -7

Now we may observe that

N = 25

Sum = -7

225 + 50 (-7/25)

225 – 14 = 211

So, mean daily expenditure on food is Rs 211

Question 23. To find out the concentration of SO₂ in the air (in parts per million i.e ppm) the data was collected for localities 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:

We may find class marks for each interval by using the relation

x = upperlimit + lowerclasslimit2x = 

Class size of this data = 0.04

Now taking 0.04 assumed mean (x_i) we may calculate d_i, u_i, f_iu_i as follows

Concentration of SO2	Frequency f1	Class interval xi`	di = xi – 0.14	ui	fiui
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	N = 30				Sum = -31

From the table we may observe that 

N = 30

Sum = -31

= 0.14 + (0.04)(-31/30)

= 0.099 ppm

So mean concentration of SO₂ in the air is 0.099 ppm.

Question 24. 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:

We may find class mark of each interval by using the relation

x =  upperlimit + lowerclasslimit2x = 

Now, taking 16 as assumed mean (a) we may

Calculate d_i and f_id_i as follows

Number of days	Number of students fi	Xi	d = xi + 10	fidi
0 – 6	11	3	-13	-143
6 – 10	10	8	-8	-280
10 – 14	7	12	-4	-28
14 – 20	7	16	0	0
20 – 28	8	24	8	32
28 – 36	3	33	17	51
30 – 40	1	39	23	23
Total	N = 40			Sum = -145

Now we may observe that

N = 40

Sum = -145

= 16 + (-145/40)

= 16 – 3.625 

= 12.38

So mean number of days is 12.38 days, for which student was absent.

Question 25. 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

Solution:

We may find class marks by using the relation

x = upperlimit + lowerclasslimit2x = 

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) wrong

Calculate d_i, u_i, f_iu_i as follows

Literacy rate (in %)	Number of cities (fi)	Mid value xi	di = xi – 70	ui = di – 50	fiui
45 – 55	3	50	-20	-20	-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	N = 35				Sum = -2

Now we may observe that

N = 35

Sum = -2 

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%