NCERT Solutions Class 10th Maths Chapter – 13 Statistics Exercise – 13.1

June 5, 2022

NCERT Solutions Class 10th Maths Chapter – 13 Statistics

Textbook	NCERT
class	10^th
Subject	Mathematics
Chapter	13^th
Chapter Name	Statistics
Category	Class 10th Mathematics
Medium	English
Source	last doubt

NCERT Solutions Class 10th Maths Chapter – 13 Statistics

Chapter – 13

Statistics

Exercise – 13.1

1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of Houses	1	2	1	5	6	2	3

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

Solution: In order to find the mean value, we will use the direct method because the numerical value of f_i and x_i are small.
Find the midpoint of the given interval using the formula.
Midpoint (x_i) = (upper limit + lower limit)/2

No. of plants(Class interval)	No. of housesFrequency (f_i)	Mid-point (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
	Sum f_i= 20		Sum f_ix_i = 162

The formula to find the mean is:
Mean = x̄ = ∑f_ix_i /∑f_i
= 162/20
= 8.1
Therefore, the mean number of plants per house is 8.1

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

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

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

Solution: Find the midpoint of the given interval using the formula.
Midpoint (x_i) = (upper limit + lower limit)/2
In this case, the value of the mid-point (x_i) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.
So, u_i= (x_i – A)/h = u_i= (x_i – 150)/20
Substitute and find the values as follows:

Daily wages(Class interval)	Number of workersfrequency (f_i)	Mid-point (x_i)	u_i= (x_i – 150)/20	f_iu_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
Total	Sum f_i= 50			Sum f_iu_i = -12

So, the formula to find out the mean is:
Mean = x̄ = A + h∑f_iu_i /∑f_i=150 + (20 × -12/50) = 150 – 4.8 = 145.20
Thus, mean daily wage of the workers = Rs. 145.20

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

Daily Pocket Allowance(in c)	11-13	13-15	15-17	17-19	19-21	21-23	23-35
Number of children	7	6	9	13	f	5	4

Solution: To find out the missing frequency, use the mean formula.
Here, the value of mid-point (x_i) mean x̄ = 18

Class interval	Number of children (f_i)	Mid-point (x_i)	f_ix_i
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18 = A	234
19-21	f	20	20f
21-23	5	22	110
23-25	4	24	96
Total	f_i = 44+f		Sum f_ix_i = 752+20f

The mean formula is Mean = x̄ = ∑f_ix_i /∑f_i= (752+20f)/(44+f)
Now substitute the values and equate to find the missing frequency (f)
⇒ 18 = (752+20f)/(44+f)
⇒ 18(44+f) = (752+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 – 752 = 20f – 18f
⇒ 40 = 2f
⇒ f = 20
So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarized 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

Solution: From the given data, let us assume the mean as A = 75.5

x_i= (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the u_iand f_iu_i as follows:

Class Interval	Number of women (f_i)	Mid-point (x_i)	u_i = (x_i – 75.5)/h	f_iu_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	3	8
83-86	2	84.5	3	6
	Sum f_i= 30			Sum f_iu_i= 4

Mean = x̄ = A + h∑f_iu_i /∑f_i
= 75.5 + 3×(4/30)
= 75.5 + 4/10
= 75.5 + 0.4
= 75.9

Therefore, the mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained a 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?

Solution: Since the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between the two intervals is 1
Here, assumed mean (A) = 57
Class size (h) = 3
Here, the step deviation is used because the frequency values are big.

Class Interval	Number of boxes (f_i)	Mid-point (x_i)	d_i = x_i – A	f_id_i
49.5-52.5	15	51	-6	90
52.5-55.5	110	54	-3	-330
55.5-58.5	135	57 = A	0	0
58.5-61.5	115	60	3	345
61.5-64.5	25	63	6	150
	Sum f_i = 400			Sum f_id_i = 75

The formula to find out the Mean is: Mean = x̄ = A +h ∑f_id_i /∑f_i

= 57 + 3(75/400)
= 57 + 0.1875
= 57.19
Therefore, the mean number of mangoes kept in a packing box is 57.19

6. The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

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

Solution: Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
Let is assume the mean (A) = 225
Class size (h) = 50

Class Interval	Number of households (f_i)	Mid-point (x_i)	d_i = x_i – A	u_i= d_i/50	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
	Sum f_i = 25				Sum f_iu_i = -7

Mean = x̄ = A +h∑f_iu_i /∑f_i

_{= 225+50(-7/25)

= 225-14

= 211

Therefore, the mean daily expenditure on food is 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:

The 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: To find out the mean, first find the midpoint of the given frequencies as follows:

The concentration of SO₂(in ppm)	Frequency (f_i)	Mid-point (x_i)	f_ix_i
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.10	0.90
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.20	0.40
Total	Sum f_i = 30		Sum (f_ix_i) = 2.96

The formula to find out the mean is Mean = x̄ = ∑f_ix_i /∑f_i

= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO2 in the air is 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	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: Find the midpoint of the given interval using the formula.

Midpoint (x_i) = (upper limit + lower limit)/2

Class interval	Frequency (f_i)	Mid-point (x_i)	f_ix_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
	Sum f_i = 40		Sum f_ix_i = 499

The mean formula is, Mean = x̄ = ∑f_ix_i /∑f_i

= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.

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

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

Solution: Find the midpoint of the given interval using the formula.

Midpoint (x_i) = (upper limit + lower limit)/2

In this case, the value of the mid-point (x_i) is very large, so let us assume the mean value, A = 70, and the class interval is h = 10.

So, u_i= (x_i-A)/h = u_i= (x_i-70)/10

Substitute and find the values as follows:

Class Interval	Frequency (f_i)	(x_i)	d_i = x_i – a	u_i = d_i/h	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
	Sum f_i = 35				Sum f_iu_i = -2

So, Mean = x̄ = A+(∑f_iu_i /∑f_i)×h

= 70+(-2/35)×10 = 69.42

Therefore, the mean literacy part = 69.42

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