NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.2 will help students understand an important concept of central tendency in statistics, the mode of a dataset. The exercise is mainly focused on explaining how to find the mode of a certain dataset either as a grouped or ungrouped frequency distribution. The concepts form a solid base for understanding the essential aspects of complex statistical mathematics at advanced levels. So, let’s get a deeper insight into this crucial topic of statistics by exploring the methods of solving questions of exercise.
In statistics, mode refers to that value which occurs most often in a data set. In other words, it is the observation that has the highest frequency in the dataset. Just like the mean, the mode is also divided into two types based on the type of data being processed, which is grouped and ungrouped data. For ungrouped data, it is easy to determine the mode because it is just the value that occurs most frequently. However, in the case of grouped data, which is collected in class intervals, it is not easy to find the Mode.
For grouped data, we would rather find the modal class, the class interval having the greatest frequency, and approximate the mode in this class. This mode gives us a sense of the most common value or range in the data. While a dataset can be multimodal (having more than one mode), this exercise focuses on datasets with a single mode.
The exercise mainly focuses on calculating the mode of grouped data, which is an important part of central tendencies. Let’s explore the formulas and key concepts of this important topic:
In exercise 12.2, we calculate the mode of grouped data from the frequency distribution of different class intervals. We will identify the modal class, the class with the largest frequency, to determine the mode. It is not possible to determine the exact mode in a grouped dataset, so we apply a formula to estimate the mode in the modal class. The formula to find the mode of grouped data can be expressed as:
Here,
Mode is the most frequent value in a dataset, while the mean is the average value of that dataset; in other words, it is the value close to all the values of data. In some cases, the Mode can be lower than the Mean, and in others, it can be equal to or higher than the Mean. Both of these entities of the statistics are different and used for different situations. For example, in a class, the marks scored by all the students in an exam may vary. The marks scored by most of the students are the mode of that class, while the average of these marks is the mean.
To master your statistics skills, especially the concept of the mode, immerse yourself in the NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.2 from today.
1. The following table shows the ages of the patients admitted in a hospital during a year :
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Sol.
From the given data, we have the modal class 35-45.
{∵ It has largest frequency among the given classes of the data}
So, ℓ=35,fm=23,f1=21,f2=14
and h=10
Mode =ℓ+{2fm−f1−f2fm−f1}×h
=35+{46−21−1423−21}×10=35+1120
=36.8 years
Now, let us find the mean of the data :
a=30, h=10, N=80 and ∑fiui=43
Mean =a+h× N1×Σfiui=30+10×801×43
=30+5.37=35.37 years
Thus, mode =36.8 years and mean =35.37 years.
So, we conclude that the maximum number of patients admitted in the hospital are of the age 36.8 years (approx), whereas on an average the age of a patient admitted to the hospital is 35.37 years.
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Determine the modal lifetimes of the components.
Sol.
The modal class of the given data is 60-80.
Here, ℓ=60,f1=61,f0=52,f2=38 and h=
20.
Mode =ℓ+{2f1−f0−f2f1−f0}×h
=60+{122−52−3861−52}×20
=60+329×20=60+845
=60+5.625
=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:
Sol.
Mean =∑fi∑fiXi=200532500=2662.5
Modal class =1500−2000
Mode =ℓ+{2f1−f0−f2f1−f0}×h
=1500+{2×40−24−3340−24}×500
=1500+80−5716×500=1847.83.
4. The following distribution gives the statewise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
Sol.
Modal class is (30-35) and its frequency is 10.
So, ℓ=30,f1=10,f0=9,f2=3, h=5.
Mode =ℓ+{2f1−f0−f2f1−f0}×h
=30+{20−9−310−9}×5=30+85=30.6
a=32.5, h=5, N=35 and Σfiui=−23.
By step-deviation method,
Mean =a+h× N1×Σfiui
=32.5+5×351×(−23)
=32.5−723=32.5−3.3=29.2
Hence, Mode =30.6 and Mean =29.2 We conclude that most states/U.T. have a student teacher ratio of 30.6 and on an average, the ratio is 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 :
Find the mode of the data.
Sol.
Modal class =4000−5000
Mode =ℓ+{2f1−f0−f2f1−f0}×h
=4000+{2×18−4−918−4}×1000
=4000+{2314}×1000
=4608.69
6. 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.
Sol.
Modal class =40−50
Mode =40+{2×20−12−1120−12}×10
=40+{40−238}×10
=40+4.706
=44.706
(Session 2025 - 26)