NCERT Solutions Class 10th Maths Chapter – 13 Statistics Exercise 13.3

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 13.3 – 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 13.3

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers in a locality. Find the median, mean, and mode of the data and compare them.

Monthly consumption(in units)	No. of customers
65-85	4
85-105	5
105-125	13
125-145	20
145-165	14
165-185	8
185-205	4

Solution: Find the cumulative frequency of the given data as follows:

Class Interval	Frequency	Cumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	8	64
185-205	4	68
	N=68

From the table, it is observed that, n = 68 and hence n/2=34
Hence, the median class is 125-145 with cumulative frequency = 42
Where, l = 125, n = 68, C_f= 22, f = 20, h = 20
Median is calculated as follows:

=125+((34−22)/20) × 20
=125+12 = 137
Therefore, median = 137
To calculate the mode:
Modal class = 125-145,
f₁=20, f₀=13, f₂=14 & h = 20
Mode formula:
Mode = l+ [(f₁-f₀)/(2f₁-f₀-f₂)]×h
Mode = 125 + ((20-13)/(40-13-14))×20
=125+(140/13)
=125+10.77
=135.77
Therefore, mode = 135.77
Calculate the Mean:

Class Interval	f_i	x_i	d_i=x_i-a	u_i=d_i/h	f_iu_i
65-85	4	75	-60	-3	-12
85-105	5	95	-40	-2	-10
105-125	13	115	-20	-1	-13
125-145	20	135	0	0	0
145-165	14	155	20	1	14
165-185	8	175	40	2	16
185-205	4	195	60	3	12
	Sum f_i= 68				Sum f_iu_i= 7

x̄ =a+h ∑f_iu_i/∑f_i =135+20(7/68)

Mean=137.05

In this case, mean, median, and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5 then, find the value of x & y.

Class Interval	Frequency
0-10	5
10-20	x
20-30	20
30-40	15
40-50	y
50-60	5
Total	60

Solution: Given data, n = 60
Median of the given data = 28.5
Where, n/2 = 30
Median class is 20 – 30 with a cumulative frequency = 25+x
Lower limit of median class, l = 20,
C_f = 5+x,
f = 20 & h = 10

Substitute the values
28.5=20+((30−5−x)/20) × 10
8.5 = (25 – x)/2
17 = 25-x
Therefore, x =8
Now, from cumulative frequency, we can identify the value of x + y as follows:
Since,
60=5+20+15+5+x+y
Now, substitute the value of x, to find y
60 = 5+20+15+5+8+y
y = 60-53
y = 7
Therefore, the value of x = 8 and y = 7.

3. The 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 the persons whose age is 18 years onwards but less than 60 years.

Age (in years)	Number of policyholders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Solution:

Class interval	Frequency	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

Given data: n = 100 and n/2 = 50
Median class = 35-45
Then, l = 35, c_f = 45, f = 33 & h = 5

Median = 35+((50-45)/33) × 5
= 35 + (5/33)5
= 35.75
Therefore, the median age = 35.75 years.

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

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	5
163-171	4
172-180	2

Find the median length of leaves.

Solution: Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

Class Interval	Frequency	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	3840
171.5-180.5	2

So, the data obtained are:
n = 40 and n/2 = 20
Median class = 144.5-153.5
then, l = 144.5,
c_f = 17, f = 12 & h = 9

Median = 144.5+((20-17)/12)×9
= 144.5+(9/4)
= 146.75 mm
Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a lifetime of 400 neon lamps.

Lifetime (in hours)	Number of lamps
1500-2000	14
2000-2500	56
2500-3000	60
3000-3500	86
3500-4000	74
4000-4500	62
4500-5000	48

Find the median lifetime of a lamp.

Solution:

Class Interval	Frequency	Cumulative
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

Data:
n = 400 &n/2 = 200
Median class = 3000 – 3500
Therefore, l = 3000, C_f= 130,
f = 86 & h = 500

Median = 3000 + ((200-130)/86) × 500
= 3000 + (35000/86)
= 3000 + 406.97
= 3406.97
Therefore, the median lifetime of the lamps = 3406.97 hours

6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of the modal in the surnames.

Solution: To calculate the median:

Class Interval	Frequency	Cumulative Frequency
1-4	6	6
4-7	30	36
7-10	40	76
10-13	16	92
13-16	4	96
16-19	4	100

Given:
n = 100 &n/2 = 50
Median class = 7-10
Therefore, l = 7, C_f = 36, f = 40 & h = 3

Median = 7+((50-36)/40) × 3
Median = 7+42/40
Median=8.05
Calculate the Mode:
Modal class = 7-10,
Where, l = 7, f₁ = 40, f₀ = 30, f₂ = 16 & h = 3

Mode = 7+((40-30)/(2×40-30-16)) × 3
= 7+(30/34)
= 7.88
Therefore mode = 7.88
Calculate the Mean:

Class Interval	f_i	x_i	f_ix_i
1-4	6	2.5	15
4-7	30	5.5	165
7-10	40	8.5	340
10-13	16	11.5	184
13-16	4	14.5	51
16-19	4	17.5	70
	Sum f_i = 100		Sum f_ix_i = 825

Mean = x̄ = ∑f_ix_i /∑f_i

Mean = 825/100 = 8.25
Therefore, mean = 8.25

7. The distributions below give a weight of 30 students in a class. Find the median weight of a student.

Weight(in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75
Number of students	2	3	8	6	6	3	2

Solution:

Class Interval	Frequency	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

Given: n = 30 and n/2= 15
Median class = 55-60
l = 55, C_f = 13, f = 6 & h = 5

Median = 55+((15-13)/6)×5
Median=55 + (10/6) = 55+1.666
Median =56.67
Therefore, the median weight of the students = 56.67

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