NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14

Q1 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.

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

Answer:

Number of plants	Number of houses $f_{i}$	Classmark $x_{i}$	$f_{i} x_{i}$
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
	$\sum f_{i}$ =20		$\sum f_{i} x_{i}$ =162

Mean,

$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$= \frac{162}{20} = 8.1$

We used the direct method in this as the values of $x_{i} a n d f_{i}$ are small.

Q2 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.

Answer:

Let the assumed mean be a = 550

Daily Wages	Number of workers $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$f_{i} d_{i}$
500-520	12	510	-40	-480
520-540	14	530	-20	-280
540-560	8	550	0	0
560-580	6	570	20	120
580-600	10	590	40	400
	$\sum f_{i}$ = 50			$\sum f_{i} x_{i}$ = -240

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$

$= 550 + \frac{- 240}{50} = 550 - 4.8 = 545.20$

Therefore, the mean daily wages of the workers of the factory is Rs. 545.20

Q3 Following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Answer:

Daily pocket allowance	Number of children $f_{i}$	Classmark $x_{i}$	$f_{i} x_{i}$
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
	$\sum f_{i}$ =44+f		$\sum f_{i} x_{i}$ =752+20f

Mean,

$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$⟹ 18 = \frac{752 + 20 f}{44 + f}$

$⟹ 18 (44 + f) = (752 + 20 f) ⟹ 2 f = 40 ⟹ f = 20$

Therefore the missing f = 20

Q4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

1745918959444

Answer:

Let the assumed mean be a = 75.5

No. of heartbeats per minute	Number of women $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$f_{i} d_{i}$
65-68	2	66.5	-9	-18
68-71	4	69.5	-6	-24
71-74	3	72.5	-3	-9
74-77	8	75.5	0	0
77-80	7	78.5	3	21
80-83	4	81.5	6	24
83-86	2	84.5	9	18
	$\sum f_{i}$ =30			$\sum f_{i} x_{i}$ =12

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$

$= 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$

Therefore, the mean heartbeats per minute of these women are 75.9

Q5 In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers 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?

Answer:

Let the assumed mean be a = 57

Number of mangoes	Number of boxes $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$f_{i} d_{i}$
50-52	15	51	-6	-90
53-55	110	54	-3	-330
56-58	135	57	0	0
59-61	115	60	3	345
62-64	25	63	6	150
	$\sum f_{i}$ =400			$\sum f_{i} x_{i}$ =75

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$

$= 57 + \frac{75}{400} = 57 + 0.1875 = 57.1875 \approx 57.19$

Therefore, the mean number of mangoes kept in a packing box is approx 57.19

Q6 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 rupees	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Answer:

Let the assumed mean be a = 225 and h = 50

Daily Expenditure	Number of households $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$u_{i} = \frac{d_{i}}{h}$	$f_{i} u_{i}$
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				$\sum f_{i} x_{i}$ = -7

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h$

$= 225 + \frac{- 7}{25} \times 50 = 225 - 14 = 211$

Therefore, the mean daily expenditure on food is Rs. 211

Q7 To find out the concentration of $S O_{2}$ 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 $S O_{2}$ in the air.

Answer:

Class Interval	Frequency $f_{i}$	Classmark $x_{i}$	$f_{i} x_{i}$
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.10	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	0.22	0.44
	$\sum f_{i}$ =30		$\sum f_{i} x_{i}$ =2.96

Mean,

$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$= \frac{2.96}{30} = 0.099$

Therefore, the mean concentration of $S O_{2}$ in the air is 0.099 ppm

Q8 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.

1636093576677

Answer:

Number of days	Number of Students $f_{i}$	Classmark $x_{i}$	$f_{i} x_{i}$
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

	$\sum f_{i}$ =40		$\sum f_{i} x_{i}$ =499

Mean,

$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$= \frac{499}{40} = 12.475$ $= \frac{499}{40} = 12.475 \approx 12.48$

Therefore, the mean number of days a student was absent is 12.48 days.

Q9 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer:

Let the assumed mean be a = 75 and h = 10

Literacy rates	Number of cities $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$u_{i} = \frac{d_{i}}{h}$	$f_{i} u_{i}$
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				$\sum f_{i} x_{i}$ = -2

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h$

$= 70 + \frac{- 2}{35} \times 10 = 70 - 0.57 = 69.43$

Therefore, the mean mean literacy rate is 69.43%

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics

NCERT Solutions Class 10 Maths Chapter 13: Exercise 13.1

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Q4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Answer:

Answer:

Answer:

Answer:

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Answer:

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