Mode questions with solutions are given here for students to help them with their exam preparations. It is one of three measures of central tendency – Mean, mode and median.
Mode represents the data whose frequency is the highest. For example, given a dataset 11, 16, 12, 23, 16, 14, 12, 13, 11, 14, 15, 16, 17, 16, for this dataset we have frequency distribution table:
|
Number |
Frequency |
|
11 |
2 |
|
12 |
2 |
|
13 |
1 |
|
14 |
2 |
|
15 |
1 |
|
16 |
4 |
|
17 |
1 |
|
23 |
1 |
We see that 16 occurs the most. Thus, the mode or the modal value of the given data is 16.
Types of the dataset according to multiple modes:
- Bimodal – two modal values for the given dataset
- Trimodal – three modal values for the given dataset
- Multimodal – more than three modal values for the given dataset.
Learn more about Mode.
Following are some important facts about mode:
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Mode Questions with Solutions
Now, let us solve a few questions based on mode.
Question 1:
If the ratio of the mode to the median is 2 : 3 for a dataset. Find the ratio of mode to mean.
Solution:
The empirical relationship between mean mode and median is
Mode = 3 Median – 2 Mean.
Let the modal value be 2x, and the median be 3x, then
2x = 3.3x – 2 Mean
⇒ 2 Mean = 9x – 2x
⇒ Mean = 7x/2
∴ Mode : Mean = 2x/(7x/2) = 4/7 = 4 : 7.
Question 2:
Find the mode of the following data:
|
Value of x |
5 |
10 |
15 |
20 |
25 |
30 |
35 |
|
Frequency |
3 |
6 |
10 |
6 |
10 |
5 |
9 |
Solution:
From the given frequency distribution table there are two modes, 15 and 25. Hence, the dataset is bimodal.
Question 3:
For any given data, the mean is 45.5, and the median is 43. Find the modal value.
Solution:
We know that,
Mode = 3 Median – 2 Mean
∴ Mode = 3 × 43 – 2 × 45.5
= 129 – 91 = 38.
Mode = 38.
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Mode for a grouped data distribution: To calculate the mode of a grouped data distribution, we need first to determine the modal class. Modal Class: The class interval which has the maximum frequency. Let the kth class interval be the modal class. Then mode is given by the following formula:
\(\begin{array}{l}Mode = l +\left [ \frac{f_{k}-f_{k-1}}{2f_{k}-f_{k-1}-f_{k+1}} \right ]\times h\end{array} \)
where, l = lower limit of the modal class fk = frequency of the modal class fk – 1 = frequency of the class before the modal class fk – 1 = frequency of the class after the modal class h = Width of each class interval. |
Question 4:
Find the mode of the following data distribution:
|
Classes |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |
|
Frequency |
12 |
14 |
10 |
13 |
14 |
18 |
10 |
Solution:
The modal class is 60-70. Now,
l = 60
fk = 18
fk – 1 = 14
fk + 1 = 10
h = 70 – 60 = 10
= 60 + [(18 – 14)/(2 × 18 – 14 – 10)] × 10
= 60 + [4/12] × 10
= 60 + 10/3
= 63.333 (approx).
Question 5:
Find the mode of the following data distribution:
|
Class-Interval |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
|
frequency |
12 |
35 |
45 |
25 |
13 |
Solution:
Clearly, the modal class is 30-40, then
l = 30
fk = 45
fk – 1 = 35
fk + 1 = 25
h = 40 – 30 = 10
= 30 + [(45 – 35)/(2 × 45 – 35 – 25)] × 10
= 30 + [10/30] × 10
= 30 + 10/3
= 100/3 = 33.333.
Also Read:
Question 6:
The mode of the following data is 36. Find the missing frequency in it.
|
Class Interval |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
Frequency |
4 |
6 |
… |
9 |
5 |
Solution:
Let the missing value be x, since the modal value is 36, therefore the modal class is 30-40.
l = 30
fk = 9
fk – 1 = x
fk + 1 = 5
h = 40 – 30 = 10
⇒ 30 + [(9 – x)/(2 × 9 – x – 5)] × 10 = 36
⇒ 30 + [(9 – x)/(13 – x)] × 10 = 36
⇒ [(9 – x)/(13 – x)] = (36 – 30)/10
⇒ [(9 – x)/(13 – x)] = 6/10
⇒ 10 × (9 – x) = 6(13 – x)
⇒ 90 – 10x = 78 – 6x
⇒ 4x = 12
⇒ x = 3
∴ the missing frequency is 3
Question 7:
Calculate the mode of the following data distribution:
|
Age (in years) |
0-5 |
5-10 |
10-15 |
15-20 |
20-25 |
25-30 |
30-35 |
|
Number of patients |
6 |
11 |
18 |
24 |
17 |
13 |
5 |
Solution:
Clearly, the modal class is 15-20, then
l = 15
fk = 24
fk – 1 = 18
fk + 1 = 17
h = 20 – 15 = 5
= 15 + [(24 – 18)/(2 × 24 – 18 – 17)] × 5
= 15 + [6/13] × 5
= 15 + 30/13
= 17.3 years (approx.)
Question 8:
Calculate the mode of the following data distribution:
|
Marks in % |
40-50 |
50-60 |
60-70 |
70-80 |
80-90 |
90-100 |
|
Number of students |
5 |
4 |
7 |
8 |
6 |
5 |
Solution:
Clearly, the modal class is 70-80, then
l = 70
fk = 8
fk – 1 = 7
fk + 1 = 6
h = 80 – 70 = 10
= 70 + [(8 – 7)/(2 × 8 – 7 – 6)] × 10
= 70 + [1/3] × 10
= 70 + 10/3
= 220/3 = 73.34 (approx.)
Question 9:
Calculate the mode of weekly wages from the following frequency distribution:
|
Wages (in ₹) |
Number of workers |
|
30-40 |
10 |
|
40-50 |
20 |
|
50-60 |
40 |
|
60-70 |
16 |
|
70-80 |
8 |
|
80-90 |
6 |
Solution:
Clearly, the modal class is 50-60, then
l = 50
fk = 40
fk – 1 = 20
fk + 1 = 16
h = 60 – 50 = 10
= 50 + [(40 – 20)/(2 × 40 – 20 – 16)] × 10
= 50 + [20/44] × 10
= 50 + 50/11
= 600/3 = 54.55 (approx.)
Question 10:
Calculate the mode of the following data distribution:
|
Class Interval |
10-14 |
14-18 |
18-22 |
22-26 |
26-30 |
30-34 |
34-38 |
38-42 |
|
Frequency |
8 |
6 |
11 |
20 |
25 |
22 |
10 |
4 |
Solution:
Clearly, the modal class is 26-30, then
l = 26
fk = 25
fk – 1 = 20
fk + 1 = 22
h = 30 – 26= 4
= 26 + [(25 – 20)/(2 × 25 – 20 – 22)] × 4
= 26 + [⅝] × 4
= 26 + 5/2
= 57/2= 28.5.
Related Articles |
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Practice Questions on Mode
1. Mode of the scores 4, 4, 6, 6, 2, 3, 3, 3, 3, 3, 4, 5, 6, 5, 6, 6 is _______.
2. If the mode of the given data set 3, 4, 3, 5, 4, 6, 6 and x is 4. Find the value of x.
3. Find the mode of the following data distribution:
|
Class-Interval |
Frequency |
|
0-5 |
2 |
|
5-10 |
10 |
|
10-15 |
8 |
|
15-20 |
6 |
|
20-25 |
4 |
4. Find the mode of the discrete data:
|
Variable |
Frequency |
Variable |
Frequency |
|
1 |
5 |
5 |
11 |
|
2 |
4 |
6 |
13 |
|
3 |
6 |
7 |
9 |
|
4 |
8 |
8 |
10 |
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