NCERT Solutions Class 10th Maths Chapter – 13 Statistics Exercise 13.2

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 Maths solution
Medium	English
Source	last doubt

NCERT Solution Class 10th Maths Chapter – 13 Statistics Exercise 13.2 – In This Chapter We will read about Statistics, What do you mean by statistics?, What are the three types of statistics?, Why is it called statistics?, Who is the father of statistics?, What is scope of statistics?, What is statistics and its types?, What are the 5 importance of statistics?, What is statistics and its formula? etc.

NCERT Solutions Class 10th Maths Chapter – 13 Statistics

Chapter – 13

Statistics

Exercise – 13.2

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

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

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

Solution: To find out the modal class, let us the consider the class interval with high frequency
Here, the greatest frequency = 23, so the modal class = 35 – 45,
l = 35,
class width (h) = 10,
f_m = 23,
f₁ = 21 and f₂ = 14
The formula to find the mode is
Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h
Substitute the values in the formula, we get
Mode = 35+[(23-21)/(46-21-14)]×10
Mode = 35+(20/11) = 35+1.8
Mode = 36.8 year
So the mode of the given data = 36.8 year
Calculation of Mean:
First find the midpoint using the formula, x_i= (upper limit +lower limit)/2

Class Interval	Frequency (f_i)	Mid-point (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

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

= 2830/80
= 35.37 years
Therefore, the mean of the given data = 35.37 years

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

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

Determine the modal lifetimes of the components.

Solution: From the given data the modal class is 60–80.
l = 60,
The frequencies are:
f_m = 61, f₁ = 52, f₂ = 38 and h = 20
The formula to find the mode is
Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h
Substitute the values in the formula, we get
Mode =60+[(61-52)/(122-52-38)]×20
Mode = 60+((9 x 20)/32)
Mode = 60+(45/8) = 60+ 5.625
Therefore, a modal lifetime of the components = 65.625 hours.

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 the
mean monthly expenditure:

Expenditure	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

Solution: Given data:

Modal class = 1500-2000,
l = 1500,
Frequencies:
f_m = 40 f₁ = 24, f₂ = 33 and
h = 500
Mode formula:
Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h
Substitute the values in the formula, we get
Mode =1500+[(40-24)/(80-24-33)]×500
Mode = 1500+((16×500)/23)
Mode = 1500+(8000/23) = 1500 + 347.83
Therefore, the modal monthly expenditure of the families = Rupees 1847.83
The calculation for mean:
First find the midpoint using the formula, x_i=(upper limit +lower limit)/2
Let us assume a mean, A be 2750

Class Interval	f_i	x_i	d_i = x_i– a	u_i= 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
	f_i = 200				f_iu_i = -35

The formula to calculate the mean, Mean = x̄ = a +(∑f_iu_i /∑f_i)×h

Substitute the values in the given formula
= 2750+(-35/200)×500
= 2750-87.50
= 2662.50
So, the mean monthly expenditure of the families = Rupees 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

No of Students per teacher	Number of states / U.T
15-20	3
20-25	8
25-30	9
30-35	10
35-40	3
40-45	0
45-50	0
50-55	2

Solution: Given data:
Modal class = 30 – 35,
l = 30,
Class width (h) = 5,
f_m = 10, f₁ = 9 and f₂ = 3
Mode Formula:
Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h
Substitute the values in the given formula
Mode = 30+((10-9)/(20-9-3))×5
Mode = 30+(5/8) = 30+0.625
Mode = 30.625
Therefore, the mode of the given data = 30.625
Calculation of mean:
Find the midpoint using the formula, x_i=(upper limit +lower limit)/2

Class Interval	Frequency (f_i)	Mid-point (x_i)	f_ix_i
15-20	3	17.5	52.5
20-25	8	22.5	180.0
25-30	9	27.5	247.5
30-35	10	32.5	325.0
35-40	3	37.5	112.5
40-45	0	42.5	0
45-50	0	47.5	0
50-55	2	52.5	105.5
	Sumf_i = 35		Sum f_ix_i = 1022.5

Mean = x̄ = ∑f_ix_i /∑f_i

= 1022.5/35
= 29.2
Therefore, mean = 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.

Run Scored	Number of Batsman
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.

Solution: Given data:
Modal class = 4000 – 5000,
l = 4000,
class width (h) = 1000,
f_m = 18, f₁ = 4 and f₂ = 9
Mode Formula:
Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h
Substitute the values
Mode = 4000+((18-4)/(36-4-9))×1000
Mode = 4000+(14000/23) = 4000+608.695
Mode = 4608.695
Mode = 4608.7 (approximately)
Thus, the mode of the given data is 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 summarized it in the table 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

Solution: Given Data:
Modal class = 40 – 50, l = 40,
Class width (h) = 10, f_m = 20, f₁ = 12 and f₂ = 11
Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h
Substitute the values
Mode = 40+((20-12)/(40-12-11))×10
Mode = 40 + (80/17) = 40 + 4.7 = 44.7
Thus, the mode of the given data is 44.7 cars

NCERT Solutions Class 10th Maths All Chapter

You Can Join Our Social Account

Youtube	Click here
Facebook	Click here
Instagram	Click here
Twitter	Click here
Linkedin	Click here
Telegram	Click here
Website	Click here