NCERT Solutions for class 10 Maths Chapter 14- Statistics Exercise 14.3 have been provided here, prepared by our subject experts. These solutions are designed in accordance with NCERT syllabus and guidelines prescribed by CBSE.
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Exercise 14.1 Solutions 6 Question ( 6 long)
Exercise 14.2 Solutions 7 Question ( 7 long)
Exercise 14.4 Solutions 3 Question ( 3 long)
Access Answers of Maths NCERT class 10 Chapter 14 â€“ Statistics Exercise 14.3
1. The following frequency distribution gives the monthly consumption of an 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 henceÂ 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+((34âˆ’22)/20) Ã— 20
=125+12 = 137
Therefore, median = 137
To calculate the mode:
Modal class = 125-145,Â
f_{1}=20, f_{0}=13, f_{2}=14Â & h = 20
Mode formula:
Mode = l+ [(f_{1}-f_{0})/(2f_{1}-f_{0}-f_{2})]Ã—h
Mode = 125 + ((20-13)/(40-13-14))Ã—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_{i}u_{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 |
SumÂ f_{i}= 68 |
SumÂ f_{i}u_{i}= 7 |
xÌ„ =a+h âˆ‘f_{i}u_{i}/âˆ‘f_{i}Â =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:Â
Given data, n = 60
Median of the given data = 28.5
Where, n/2Â = 30
Median class is 20 â€“ 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+((30âˆ’5âˆ’x)/20) Ã— 10
8.5 = (25 â€“ 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 policy holders. Calculate the median age, if policies are given only to theÂ persons whose age is 18 years onwards but less than the 60 years.
Age (in years) |
Number of policy holder |
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 andÂ n/2Â = 50
Median class = 35-45
Then, l = 35, c_{f} = 45, f = 33 & h = 5
Median = 35+((50-45)/33) Ã— 5
= 35 + (5/33)5Â
= 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 as 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.Â Â Â Â Â Â Â Â Â Â Â Â Â
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 |
38 |
171.5-180.5 |
2 |
40 |
So, the data obtained are:
n = 40 andÂ n/2Â = 20
Median class = 144.5-153.5
then, l = 144.5,
cf = 17, f = 12 & h = 9
Median = 144.5+((20-17)/12)Ã—9
= 144.5+(9/4)
= 146.75Â mm
Therefore, the median length of the leaves = 146.75 mm.
5. The following table gives the distribution of a life time 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/2Â = 200
Median class = 3000 â€“ 3500
Therefore, l = 3000, C_{f }= 130,
f = 86 & h = 500
Median = 3000 + ((200-130)/86) Ã— 500
= 3000 + (35000/86)
= 3000 + 406.97
= 3406.97
Therefore, the median life time 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 modal in the surnames.
Solution:Â
To calculate 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/2Â = 50
Median class = 7-10
Therefore, l = 7, C_{f} = 36, f = 40 & h = 3
Median = 7+((50-36)/40) Ã— 3
Median = 7+42/40
Median=8.05
Calculate the Mode:
Modal class = 7-10,
Where, l = 7, f_{1} = 40, f_{0} = 30, f_{2} = 16 & h = 3
Mode = 7+((40-30)/(2Ã—40-30-16)) Ã— 3
= 7+(30/34)
= 7.88
Therefore mode = 7.88
Calculate the Mean:
Class Interval |
f_{i} |
x_{i} |
f_{i}x_{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}Â = 100 |
Sum f_{i}x_{i}Â = 825 |
Mean = xÌ„ = âˆ‘f_{i }x_{i}Â /âˆ‘f_{iÂ }
Mean = 825/100 = 8.25
Therefore, mean = 8.25
Â 7. The distributions of below give a weight of 30 students of 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 andÂ n/2= 15
Median class = 55-60
l = 55, C_{f} = 13, f = 6 & h = 5
Median = 55+((15-13)/6)Ã—5
Median=55 + (10/6) = 55+1.666
Median =56.67
Therefore, the median weight of the students = 56.67
In this exercise, students will solve questions based on finding the median of a grouped data, which is a measure of central tendency which gives the value of the middle-most observation in the data.
A group of collection of data will be provided to the students to find the median for cases having odd and even number of observations. To solve complete problems for class 10 Maths chapter 14 exercise-wise, students can visit us any time. Also get few more study materials such as notes, books, question papers, etc. to practice well.
The problems in Ex.14.3 is solved as per the guidance and formulas, explained in the example questions before the exercise. NCERT solutions serve as the best materials for students who are looking for solutions for all the subjects class-wise and chapter-wise.
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