Exercise 14.3 Page: 287
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 hence n/2=34
Hence, the median class is 125-145 with cumulative frequency = 42
Where, l = 125, n = 68, Cf = 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 | fi | xi | di=xi-a | ui=di/h | fiui |
| 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 fi= 68 | Sum fiui= 7 |
x̄ =a+h ∑fiui/∑fi =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,
Cf = 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 policy holders. Calculate the median age, if policies are given only to the 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 and n/2 = 50
Median class = 35-45
Then, l = 35, cf = 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 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.
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 and n/2 = 20
Median class = 144.5-153.5
then, l = 144.5,
cf = 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 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/2 = 200
Median class = 3000 – 3500
Therefore, l = 3000, Cf = 130,
f = 86 & h = 500
Median = 3000 + ((200-130)/86) × 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:
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/2 = 50
Median class = 7-10
Therefore, l = 7, Cf = 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 | fi | xi | fixi |
| 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 fi = 100 | Sum fixi = 825 |
Mean = x̄ = ∑fi xi /∑fi
Mean = 825/100 = 8.25
Therefore, mean = 8.25
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 and n/2= 15
Median class = 55-60
l = 55, Cf = 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
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Class 10 Maths Chapter 14, Statistics, is one of the most important of all the chapter present in the textbook. The weightage of this chapter in the CBSE Term II exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4( it could vary as per question).
Topics covered in Chapter 14, Statistics are;
- Mean of Grouped Data
- Mode of Grouped Data
- Median of Grouped Data
- Graphical Representation of Cumulative Frequency Distribution
List of Exercises in class 10 Maths Chapter 14 :
Exercise 14.1 Solutions 9 Question ( 9 long)
Exercise 14.2 Solutions 6 Question ( 6 long)
Exercise 14.3 Solutions 7 Question ( 7 long)
Exercise 14.4 Solutions 3 Question ( 3 long)
NCERT solutions for Class 10 Maths Chapter 14- Statistics are made available for students who want to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions, such that, students find it easy to understand the fundamentals quickly.
The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.
You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form or in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the CBSE Term II exam.
Key Features of NCERT Solutions for Class 10 Maths Chapter 14- Statistics
- The solutions for the statistics chapter works as a reference for the students.
- It will help students to score marks against the questions asked from the statistics chapter.
- Students can prepare and do the revision for chapter 14 with this source.
- The questions of statistics are solved by subject experts.
- The content of the material is as per the term-wise CBSE Syllabus (2021-22) and guidelines.
Statistics can also be understood in a much better way by using the other solutions which are provided at RAJ EDUCATION HUB. The solutions are prepared to help students perform well in the CBSE exams.