NCERT solutions for class 10 maths Chapter 14 Statistics 2021

NCERT solutions for class 10 maths Chapter 14 Statistics

Edited By Ramraj Saini | Updated on Sep 09, 2023 12:11 PM IST | #CBSE Class 10th

NCERT Solutions For Statistics Class 10 Chapter 14

NCERT Solutions for Class 10 Maths Chapter 14 Statistics are provided here. These NCERT Solutions are created by expert team at Careers360 keeping in mind latest syllabus and pattern of CBSE 2023-24. Students preparing for the Class 10 board exams must go through the Statistics Class 10 Maths chapter 14 NCERT solutions. This is one of the most important chapters in NCERT books for Class 10 Maths. The NCERT solutions for Class 10 Maths Chapter 14 Statistics covers each and every question of the exercise comprehensively and provide step by step solutions.

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NCERT Solutions For Statistics Class 10 Chapter 14
NCERT Solutions for Class 10 Maths Chapter 14 Statistics PDF Free Download
NCERT Solutions For Statistics Class 10 Chapter 14 - Important Formulae
NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Intext Questions and Exercise)
Statistics Class 10 Maths - Important Topics
NCERT solutions for Class 10 maths - Chapter Wise
Key Features of NCERT Statistics Class 10 Maths Solutions Chapter 14
NCERT Exemplar solutions - Subject Wise
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NCERT solutions for class 10 maths Chapter 14 Statistics

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics PDF Free Download

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NCERT Solutions For Statistics Class 10 Chapter 14 - Important Formulae

Measures of Central Tendency - Mean, Median, and Mode:

>> Mean:

Direct Method: The mean, represented as X, can be calculated directly using the formula:

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X = ∑(f_ix_i) / ∑f_i

Where 'f_i' denotes the frequency of the value 'x_i'.

Assumed Mean Method: Alternatively, the mean can be calculated using the Assumed Mean Method:

X = a + ∑(f_id_i) / ∑f_i

Where 'a' is an assumed mean and 'd_i' is the deviation of each value 'x_i' from the assumed mean.

Step Deviation Method: Another approach is the Step Deviation Method:

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X = a + [∑(f_iu_i) / ∑f_i]h

Where 'u_i' represents the step deviations and 'h' is the class interval.

>> Median:

The median is the central value in a set of observations, and its calculation depends on the number of observations.
For an odd number of observations,

Median = [value of the (n+1)/2]th term in the ordered set

In the case of an even number of observations

Median = average of the middle two values

>> Mode:

The mode represents the value that appears most frequently in a dataset.
The formula to calculate mode is:

Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] h

Where,

l = lower limit of the modal class

f₁ = frequency of the modal class

f₀ = frequency of the class before the modal class

f₂ = frequency of the class after the modal class

H = Class interval

Free download NCERT Solutions for Class 10 Maths Chapter 14 Statistics PDF for CBSE Exam.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Intext Questions and Exercise)

NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.1

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 $x_i$	$f_ix_i$
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

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{162}{20} = 8.1$
We used the direct method in this as the values of $x_i\ and\ f_i$ 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 $x_i$	$d_i = x_i -a$	$f_id_i$
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			$\sum f_ix_i$ =-240

Mean,

$\overline x =a + \frac{\sum f_id_i}{\sum f_i}$
$= 550 + \frac{-240}{50} = 550-4.8 = 545.20$
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 $x_i$	$f_ix_i$
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		$\sum f_ix_i$ =752+20f

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$\implies 18 = \frac{752+20f}{44+f}$

$\\ \implies 18(44+f) =( 752+20f) \\ \implies 2f = 40 \\ \implies f = 20$
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.

1636093324773

Answer:

Let the assumed mean be a = 75.5

No. of heartbeats per minute	Number of women $f_i$	Classmark $x_i$	$d_i = x_i -a$	$f_id_i$
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			$\sum f_ix_i$ =12

Mean,

$\overline x =a + \frac{\sum f_id_i}{\sum f_i}$
$= 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$
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 $x_i$	$d_i = x_i -a$	$f_id_i$
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			$\sum f_ix_i$ =75

Mean,

$\overline x =a + \frac{\sum f_id_i}{\sum f_i}$
$= 57+ \frac{75}{400} = 57+0.1875 = 57.1875 \approx 57.19$
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 $x_i$	$d_i = x_i -a$	$u_i = \frac{d_i}{h}$	$f_iu_i$
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				$\sum f_ix_i$ = -7

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 225 + \frac{-7}{25}\times50 = 225 -14 = 211$
Therefore, the mean daily expenditure on food is Rs. 211

Q7 To find out the concentration of $SO_2$ 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 $SO_2$ in the air.

Answer:

Class Interval	Frequency $f_i$	Classmark $x_i$	$f_ix_i$
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	0.22	0.44
	$\sum f_i$ =30		$\sum f_ix_i$ =2.96

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{2.96}{30} = 0.099$
Therefore, the mean concentration of $SO_2$ 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 $x_i$	$f_ix_i$
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

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{499}{40} = 12.475$ $= \frac{499}{40} = 12.475\approx 12.48$
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 $x_i$	$d_i = x_i -a$	$u_i = \frac{d_i}{h}$	$f_iu_i$
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				$\sum f_ix_i$ = -2

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 70 + \frac{-2}{35}\times10 = 70 -0.57 = 69.43$
Therefore, the mean mean literacy rate is 69.43%

NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.2

Q1 The following table shows the ages of the patients admitted in a hospital during a year:

1640759472805

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

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 23 and hence the modal class = 35-45

Lower limit (l) of modal class = 35, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 23, frequency ( $f_0$ ) of class preceding the modal class = 21, frequency ( $f_2$ ) of class succeeding the modal class = 14.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 35 + \left(\frac{23-21}{2(23)-21-14} \right).10 \\ \\ = 35 + \frac{2}{11}.10$

$= 36.8$

Now,

Age	Number of patients $f_i$	Classmark $x_i$	$f_ix_i$
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
	$\sum f_i$ =80		$\sum f_ix_i$ =2830

Mean,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$= \frac{2830}{80} = 35.37$
The maximum number of patients are in the age group of 36.8, whereas the average age of all the patients is 35.37.

Q2 The following data gives information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 61 and hence the modal class = 60-80

Lower limit (l) of modal class = 60, class size (h) = 20

Frequency ( $f_1$ ) of the modal class = 61 frequency ( $f_0$ ) of class preceding the modal class = 52, frequency ( $f_2$ ) of class succeeding the modal class = 38.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 60 + \left(\frac{61-52}{2(61)-52-38} \right).20 \\ \\ = 60 + \frac{9}{32}.20$

$= 65.62$

Thus, the modal lifetime of 225 electrical components is 65.62 hours

Q3 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 :

1640759588002

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 40 and hence the modal class = 1500-2000

Lower limit (l) of modal class = 1500, class size (h) = 500

Frequency ( $f_1$ ) of the modal class = 40 frequency ( $f_0$ ) of class preceding the modal class = 24, frequency ( $f_2$ ) of class succeeding the modal class = 33.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 1500 + \left(\frac{40-24}{2(40)-24-33} \right).500 \\ \\ = 1500 + \frac{16}{23}.500$

$= 1847.82$

Thus, the Mode of the data is Rs. 1847.82

Now,

Let the assumed mean be a = 2750 and h = 500

Expenditure	Number of families $f_i$	Classmark $x_i$	$d_i = x_i -a$	$u_i = \frac{d_i}{h}$	$f_iu_i$
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
	$\sum f_i$ =200				$\sum f_ix_i$ = -35

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 2750 + \frac{-35}{200}\times500 = 2750 -87.5 = 2662.50$

Thus, the Mean monthly expenditure is Rs. 2662.50

Q4 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.

1640759627719

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 10 and hence the modal class = 30-35

Lower limit (l) of modal class = 30, class size (h) = 5

Frequency ( $f_1$ ) of the modal class = 10 frequency ( $f_0$ ) of class preceding the modal class = 9, frequency ( $f_2$ ) of class succeeding the modal class = 3

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 30 + \left(\frac{10-9}{2(10)-9-3} \right).5 \\ \\ = 30 + \frac{1}{8}.5$

$= 30.625$

Thus, Mode of the data is 30.625

Now,

Let the assumed mean be a = 32.5 and h = 5

Class	Number of states $f_i$	Classmark $x_i$	$d_i = x_i -a$	$u_i = \frac{d_i}{h}$	$f_iu_i$
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
	$\sum f_i$ =35				$\sum f_ix_i$ = -23

Mean,

$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 32.5 + \frac{-23}{35}\times5= 29.22$

Thus, the Mean of the data is 29.22

Q5 The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 18 and hence the modal class = 4000-5000

Lower limit (l) of modal class = 4000, class size (h) = 1000

Frequency ( $f_1$ ) of the modal class = 18 frequency ( $f_0$ ) of class preceding the modal class = 4, frequency ( $f_2$ ) of class succeeding the modal class = 9

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ =4000 + \left(\frac{18-4}{2(18)-4-9} \right).1000 \\ \\ = 4000 + \frac{14}{23}.1000$

$= 4608.70$

Thus, Mode of the data is 4608.70

Q6 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 :

1640759687964

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 20 and hence the modal class = 40-50

Lower limit (l) of modal class = 40, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 20 frequency ( $f_0$ ) of class preceding the modal class = 12, frequency ( $f_2$ ) of class succeeding the modal class = 11

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ =40+ \left(\frac{20-12}{2(20)-12-11} \right).10 \\ \\ = 40 + \frac{8}{17}.10$

$= 44.70$

Thus, Mode of the data is 44.70

Statistics Class 10 Maths Chapter 14 Solution Excercise: 14.3

Q1 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.

1640759732201

Answer:

Let the assumed mean be a = 130 and h = 20

Class	Number of consumers $f_i$	Cumulative Frequency	Classmark $x_i$	$d_i = x_i -a$	$u_i = \frac{d_i}{h}$	$f_iu_i$
65-85	4	4	70	-60	-3	-12
85-105	5	9	90	-40	-2	-10
105-125	13	22	110	-20	-1	-13
125-145	20	42	130	0	0	0
145-165	14	56	150	20	1	14
165-185	8	64	170	40	2	16
185-205	4	68	190	60	3	12
		$\sum f_i = N$ = 68				$\sum f_ix_i$ = 7

MEDIAN:
$N= 68 \implies \frac{N}{2} = 34$
$\therefore$ Median class = 125-145; Cumulative Frequency = 42; Lower limit, l = 125;

c.f. = 22; f = 20; h = 20
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 125 + \left (\frac{34-22}{20} \right ).20 \\ \\ = 125 + 12$

$= 137$

Thus, the median of the data is 137

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125-145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency ( $f_1$ ) of the modal class = 20; frequency ( $f_0$ ) of class preceding the modal class = 13, frequency ( $f_2$ ) of class succeeding the modal class = 14.

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 125 + \left(\frac{20-13}{2(20)-13-14} \right).20 \\ \\ = 125 + \frac{7}{13}.20$

$= 135.76$

Thus, Mode of the data is 135.76

MEAN:

Mean,
$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
$= 130 + \frac{7}{68}\times20 = 137.05$

Thus, the Mean of the data is 137.05

Q2 If the median of the distribution given below is 28.5, find the values of x and y.

1640759760891

Answer:

Class	Number of consumers $f_i$	Cumulative Frequency
0-10	5	5
10-20	x	5+x
20-30	20	25+x
30-40	15	40+x
40-50	y	40+x+y
50-60	5	45+x+y
	$\sum f_i = N$ = 60

$N= 60 \implies \frac{N}{2} = 30$

Now,
Given median = 28.5 which lies in the class 20-30

Therefore, Median class = 20-30
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ \implies28.5= 20 + \left (\frac{30-5-x}{20} \right ).10 \\ \\ \implies8.5=\frac{25-x}{2} \\ \implies 25-x = 8.5(2) \\ \implies x = 25 - 17 = 8$

Also,

$\\ 60 = 45 + x+y \\ \implies x+y = 60-45 = 15 \\ \implies y = 15-x = 15-8 \ \ \ (\because x =8) \\ \implies y = 7$

Therefore, the required values are: x=8 and y=7

Q3 A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

1640759806014

Answer:

Class	Frequency $f_i$	Cumulative Frequency
15-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

$N= 100 \implies \frac{N}{2} = 50$
Therefore, Median class = 35-45
Frequency corresponding to median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 35 + \left (\frac{50-45}{33} \right ).5 \\ \\$

$= 35.75$

Thus, the median age is 35.75 years.

Q4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table :

1640759831563

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 - 180.5.)

Answer:

The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.

Class	Frequency $f_i$	Cumulative Frequency
117.5-126.5	3	3
126.5-135.5	5	8
135.5-144.5	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

$\dpi{100} N= 40 \implies \frac{N}{2} = 20$
Therefore, Median class = 144.5-153.5

Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 144.5 + \left (\frac{20-17}{12} \right ).9 \\ \\$

$= 146.75$

Thus, the median length of the leaves is 146.75 mm

Q5 The following table gives the distribution of the lifetime of 400 neon lamps :

1640759970074 Find the median lifetime of a lamp.

Answer:

Class	Frequency $f_i$	Cumulative Frequency
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

$\dpi{100} N= 400 \implies \frac{N}{2} = 200$
Therefore, Median class = 3000-3500

Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 3000 + \left (\frac{200-130}{86} \right ).500 \\ \\ = 3000+406.97$

= 3406.97

Thus, the median lifetime of a lamp is 3406.97 hours

$= 146.75$

Thus, the median length of the leaves is 146.75 mm

Q6 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:

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.

Answer:

Class	Number of surnames $f_i$	Cumulative Frequency	Classmark $x_i$	$f_ix_i$
1-4	6	6	2.5	15
4-7	30	36	5.5	165
7-10	40	76	8.5	340
10-13	16	92	11.5	184
13-16	4	96	14.5	51
16-19	4	100	17.5	70
		$\sum f_i = N$ = 100		$\sum f_ix_i$ = 825

MEDIAN:
$N= 100 \implies \frac{N}{2} = 50$ $\therefore$ Median class = 7-10; Lower limit, l = 7;

Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 7+ \left (\frac{50-36}{40} \right ).3 \\ \\$

$= 8.05$

Thus, the median of the data is 8.05

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 7-10
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency ( $f_1$ ) of the modal class = 40; frequency ( $f_0$ ) of class preceding the modal class = 30, frequency ( $f_2$ ) of class succeeding the modal class = 16

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

$\\ = 7 + \left(\frac{40-30}{2(40)-30-16} \right).3 \\ \\ = 125 + \frac{10}{34}.3$

$= 7.88$

Thus, Mode of the data is 7.88

MEAN:

Mean,
$\overline x =\frac{\sum f_ix_i}{\sum f_i}$
$= \frac{825}{100} = 8.25$

Thus, the Mean of the data is 8.25

Q7 The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

1640760021634

Answer:

Class	Number of students $f_i$	Cumulative Frequency
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

MEDIAN:
$N= 30 \implies \frac{N}{2} = 15$
$\therefore$ Median class = 55-60; Lower limit, l = 55;

Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 55+ \left (\frac{15-13}{6} \right ).5 \\ \\ = 55+\frac{2}{6}.5$

$= 56.67$

Thus, the median weight of the student is 56.67 kg

Q1.The following distribution gives the daily income of 50 workers of a factory.

Answer:

Daily Income (Upper-Class Limit)	Cumulative Frequency
Less than 120	12
Less than 140	26
Less than 160	34
Less than 180	40
Less than 200	50

Now,

Taking upper-class interval on the x-axis and their respective frequencies on the y-axis,

1640760151886

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.

Answer:

Taking upper-class interval on the x-axis and their respective frequencies on the y-axis,

1640760282843

$N= 35 \implies \frac{N}{2} = 17.5$

Marking a point on the curve whose ordinate is 17.5 gives an x-ordinate= 46.5.

Hence, the Median of the data is 46.5

Now,

Weight (Class)	Frequency	Cumulative Frequency
>38	0	0
38-40	3	3
40-42	2	5
42-44	4	9
44-46	5	14
46-48	14	28
48-50	4	32
50-52	3	35

$N= 35 \implies \frac{N}{2} = 17.5$
$\therefore$ Median class = 46-48; Lower limit, l = 46;

Cumulative frequency of preceding class, c.f. = 14; f = 14; h = 2
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
$\\ = 46+ \left (\frac{17.5-14}{14} \right ).2 \\ \\ = 46+\frac{7}{14}$

$= 46.5$

Thus, the median using formula is 46.5 which verifies the result.

Q3 The following table gives production yield per hectare of wheat of 100 farms of a village. 1640760343873 Change the distribution to a more than type distribution, and draw its ogive.

Answer:

Production yield (Upper-Class Limit)	Cumulative Frequency
More than or equal to 50	100
More than or equal to 55	98
More than or equal to 60	90
More than or equal to 65	78
More than or equal to 70	54
More than or equal to 75	16

Now,

Taking lower class limit on the x-axis and their respective frequencies on the y-axis,

1640760369574

Statistics Class 10 Maths - Important Topics

A direct method of finding mean
Assumed mean method of finding mean
Step deviation method of finding mean
Median
Mode

NCERT Solutions for Class 10 subject wise

NCERT solutions for Class 10 maths - Chapter Wise

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Key Features of NCERT Statistics Class 10 Maths Solutions Chapter 14

Valuable Student Resources: These materials offer invaluable guidance to students studying statistics in Class 10.

Enhanced Academic Performance: By utilizing class 10 math chapter 14 resources, students can significantly improve their performance on questions related to chapter 14 class 10 maths, ensuring they achieve high marks.

Versatile Study Aid: Whether for exam preparation or chapter review, this resource proves to be a versatile tool for students.

Expertly Crafted Solutions: Professional subject specialists have meticulously answered the statistics questions, providing accurate and clear explanations.

CBSE Aligned: The content within these materials is in complete alignment with the CBSE Syllabus (2022-23) and follows the guidelines set forth by the board.

Also, check - NCERT Solutions for Class 10

How to use NCERT solutions for class 10 maths chapter 14 Statistics?

First of all, go through NCERT Textbook to have an understanding of Statistics Class 10 chapter.
Learn the techniques to calculate the mean, median, mode.
Once you learn the techniques and formulae, then move to the practice exercises available in the Statistics Class 10 chapter.
While practising, if you get stuck anywhere then you can take the help of NCERT Solutions for chapter 14 class 10 maths PDF Download, which cab be used offline.

NCERT Exemplar solutions - Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. What is the weightage of the chapter statistics for CBSE board exam?

Two chapters - statistics and probability combined have 11 marks weightage in the CBSE class 10 board exam. Refer to NCERT exemplar for more problems based on Class 10 Mathematics NCERT syllabus. Also practice the CBSE Class 10 Mathematics previous year papers for good score in the exam. Class 10 Maths chapter 14 examples are discussed above in this article, go and check to become familiar with concepts.

2. Will the NCERT Solutions for Class 10 Maths Chapter 14 aid in passing the CBSE exams?

Yes, Statistics class 10 solutions is a crucial chapter for Class 10 Maths and includes various math tricks and shortcuts for efficient calculations. Statistics math can aid in understanding and passing the math subject in the CBSE exams. Students should practice problems from the statistics class 10 pdf to do well in the exam.

3. How do the Chapter 14 statistics class 10 ncert solutions assist in understanding the areas and volumes of geometrical shapes?

The NCERT Solutions for class 10 maths statistics are designed in an interactive format to make it easy for students to grasp the concepts. The PDF of solutions can be used as a primary study material to improve problem-solving skills and accuracy. Interested students can use NCERT solutions class 10 maths chapter 14 pdf download from the Careers360 website and study both offline and online mode.

4. How many questions are present in NCERT Solutions for Class 10 Maths Chapter 14?

The chapter 14 statistics 10th class has a total of 25 questions across 4 exercises. The statistics class 10 exercise 14.1 has 9 questions, the second has 6, the third has 7, and the fourth has 3 questions. Practice the class 10 maths chapter 14 solutions pdf download to score well in the exam.

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hindi 10th class mp board sample paper

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https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://school.careers360.com/download/sample-papers/mp-board-10th-hindi-model-paper&ved=2ahUKEwjO3YvJu5KEAxWAR2wGHSLpAiQQFnoECBMQAQ&usg=AOvVaw2qFFjVeuiZZJsx0b35oL1x .

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Thanking you

Read Complete Answer

Shivanshu 04 February,2024

CBSE class 10th date sheet 11/1/2024 2024

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

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Sajal Trivedi 12 January,2024

iam in 7 th plz send 7th previous paper

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The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

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Sajal Trivedi 25 October,2023

my son's dob is 13/02/2009 is hi eligible for 10th board exam in 2024,,from cbse. olz reply me

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

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can Hindi marks be replace with IT 402(additional subject) in CBSE class 10th board

That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.

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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 the molecular mass of the substance.

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

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT solutions for class 10 maths Chapter 14 Statistics

NCERT Solutions For Statistics Class 10 Chapter 14

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Intext Questions and Exercise)

Answer:

Answer:

Answer:

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: