Mean, Median and Mode are three measures of central tendency, and these are the important topics in statistics. In this article, you will get mean, median, and mode questions and answers, along with some practice questions. These questions and answers will help you to practise for the board exams. Also, these questions cover various types of problems asked on mean, median and mode for Classes 9 and 10.
What is the mean formula?
As we know, the mean is the average of the given data set. This can be calculated using the formula given below.
Mean for ungrouped data = Sum of observations/Number of observations
Click here to get more information about mean.
What is the formula for the median?
The median is the middlemost data value of an ordered data set. This can be estimated using the formulas given below.
When n is odd:
Median = (n + 1)/2th observation
When n is even:
Median = [(n/2)th observation + {(n/2)+1}th observation]/2
Learn more about median here.
What is the mode?
Mode is the most frequently occurring value in the data set.
Get more information about mode here.
Mean Median Mode Questions and Answers
1. Find the mean of the first 10 odd integers.
Solution:
First 10 odd integers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Mean = Sum of the first 10 odd integers/Number of such integers
= (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/10
= 100/10
= 10
Therefore, the mean of the first 10 odd integers is 10.
2. What is the median of the following data set?
32, 6, 21, 10, 8, 11, 12, 36, 17, 16, 15, 18, 40, 24, 21, 23, 24, 24, 29, 16, 32, 31, 10, 30, 35, 32, 18, 39, 12, 20
Solution:
The ascending order of the given data set is:
6, 8, 10, 10, 11, 12, 12, 15, 16, 16, 17, 18, 18, 20, 21, 21, 23, 24, 24, 24, 29, 30, 31, 32, 32, 32, 35, 36, 39, 40
Number of values in the data set = n = 30
n/2 = 30/2 = 15
15th data value = 21
(n/2) +1 = 16
16th data value = 21
Median = [(n/2)th observation + {(n/2)+1}th observation]/2
= (15th data value + 16th data value)/2
= (21 + 21)/2
= 21
3. Identify the mode for the following data set:
21, 19, 62, 21, 66, 28, 66, 48, 79, 59, 28, 62, 63, 63, 48, 66, 59, 66, 94, 79, 19 94
Solution:
Let us write the given data set in ascending order as follows:
19, 19, 21, 21, 28, 28, 48, 48, 59, 59, 62, 62, 63, 63, 66, 66, 66, 66, 79, 79, 94, 94
Here, we can observe that the number 66 occurred the maximum number of times.
Thus, the mode of the given data set is 66.
| Mean formula for grouped data
Direct method: \(\begin{array}{l}Mean,\ \overline{x}=\frac{\sum_{i=1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}\end{array} \)
Here, ∑fi = Sum of all frequencies Assumed mean method: \(\begin{array}{l}Mean =\ (\overline{x})=a+\frac{\sum f_id_i}{\sum f_i}\end{array} \)
Here, a = Assumed mean di = xi – a ∑fi = Sum of all frequencies Step-deviation method: \(\begin{array}{l}Mean\ (\overline{x})=a+h\frac{\sum f_iu_i}{\sum f_i}\end{array} \)
Here, a = Assumed mean ui = (xi – a)/h h = Class size ∑fi = Sum of all frequencies |
4. Consider the following frequency distribution. Calculate the mean weight of students.
| Weight (in kg) | 31-35 | 36 – 40 | 41 – 45 | 46 – 50 | 51 – 55 | 56 – 60 | 61 – 65 | 66 – 70 | 71 – 75 |
| Number of Students | 9 | 6 | 15 | 3 | 1 | 2 | 2 | 1 | 1 |
Solution:
The given distribution has discontinuous class intervals, so we need to make them continuous.
| Class intervals | Number of students (fi) | Class mark (xi) | di = xi – a | fidi |
| 30.5 – 35.5 | 9 | 33 | -10 | -90 |
| 35.5 – 40.5 | 6 | 38 | -5 | -25 |
| 40.5 – 45.5 | 15 | 43 = a | 0 | 0 |
| 45.5 – 50.5 | 3 | 48 | 5 | 15 |
| 50.5 – 55.5 | 1 | 53 | 10 | 10 |
| 55.5 – 60.5 | 2 | 58 | 15 | 30 |
| 60.5 – 65.5 | 2 | 63 | 20 | 40 |
| 65.5 – 70.5 | 1 | 68 | 25 | 25 |
| 70.5 – 75.5 | 1 | 73 | 30 | 30 |
| Total | ∑fi = 40 | ∑fidi = 35 |
Here, ∑fi = 40 and ∑fidi = 35
By Assumed mean method,
Mean = a + (∑fidi/∑fi)
= 43 + (35/40)
= 43 + 0.875
= 43.875
Therefore, the mean weight of the students is 43.875 kg.
5. Find the mean for the following distribution.
| xi | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
Solution:
| xi | fi | fixi |
| 15 | 3 | 45 |
| 21 | 5 | 105 |
| 27 | 6 | 162 |
| 30 | 7 | 210 |
| 35 | 8 | 280 |
| Total | ∑fi = 29 | ∑fixi = 802 |
Mean = ∑fixi/∑fi
= 802/29
= 27.655
| Median formula for grouped data
\(\begin{array}{l}Median = l + \left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\end{array} \)
Here, l = Lower limit of the median class f = Frequency of the median class cf = Cumulative frequency of the class preceding the median class N = Sum of frequencies h = Class size |
6. Calculate the median marks of students from the following distribution.
| Marks | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 |
| Number of Students | 7 | 10 | 10 | 20 | 20 | 15 | 8 |
Solution:
| Class interval | Number of students (frequency) | Cumulative frequency |
| 10 – 20 | 7 | 7 |
| 20 – 30 | 10 | 17 |
| 30 – 40 | 10 | 27 = cf |
| 40 – 50 | 20 = f | 47 |
| 50 – 60 | 20 | 67 |
| 60 – 70 | 15 | 82 |
| 70 – 80 | 8 | 90 |
N/2 = 90/2 = 45
Cumulative frequency greater and nearer to 45 is 47, which lies in the interval 40 – 50
Median class is 40 – 50.
Lower limit of the median class = l = 40
Class size = h = 10
Frequency of the median class = f = 20
Cumulative frequency of the class preceding the median class = cf = 27
As we know,
Median = 40 + [(45 – 27)/20] × 10
= 40 + (18/2)
= 40 + 9
= 49
Hence, the median marks of the students = 49.
| Mode formula for grouped data
\(\begin{array}{l}Mode = l + \left ( \frac{f_1-f_0}{2f_1-f_0-f_2} \right )\times h\end{array} \)
Here, l = Lower limit of the modal class h = Class size f1 = Frequency of the modal class f0 = Frequency of the class preceding the modal class f2 = Frequency of the class succeeding the modal class |
7. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
| Run Scored | Number of Batsman |
| 3000-4000 | 4 |
| 4000-5000 | 18 |
| 5000-6000 | 9 |
| 6000-7000 | 7 |
| 7000-8000 | 6 |
| 8000-9000 | 3 |
| 9000-10000 | 1 |
| 10000-11000 | 1 |
Find the mode of the above distribution.
Solution:
From the given,
Modal class = 4000 – 5000
Lower limit of the modal class = l = 4000
Class width (h) = 1000
Frequency of the modal class = f1 = 18,
Frequency of the class preceding the modal class = f0 = 4
Frequency of the class succeeding the modal class = f2 = 9
Mode = l + [(f1 – f0)/ (2f1 – f0 – f2)] × h
Mode = 4000 + [(18 – 4)/(36 – 4 – 9)] × 1000
= 4000 + (14000/23)
= 4000 + 608.695
= 4608.695
= 4608.7 (approximately)
Thus, the mode of the given data is 4608.7 runs
8. If the median of a distribution given below is 28.5, then find the value of x and y.
| CI | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | Total |
| Frequency | 5 | x | 20 | 15 | y | 5 | 60 |
Solution:
From the given,
N/2 = 60/2 = 30
Median of the given data = 28.5
Median class is 20 – 30 with a cumulative frequency = 25 + x.
Lower limit of median class = l = 20
Frequency of the median class = f = 20
Cumulative frequency of the class preceding the median class = cf = 5 + x
Class size = h = 10
28.5 = 20 + [(30 − 5 − x)/ 20] × 10
28.5 – 20 = (25 – x)/2
8.5 × 2 = 25 – x
17 = 25 – x
x = 25 – 17
Therefore, x = 8
Also,
60 = 5 + 20 + 15 + 5 + x + y
60 = 5 + 20 + 15 + 5 + 8 + y
y = 60 – 53
y = 7
Therefore, the value of x = 8 and y = 7.
| Empirical Relation between Mean Median and Mode
Mode = 3 Median – 2 Mean (or) Mean – Mode = 3 (Mean – Median) |
9. For a moderately skewed distribution, the mean and median are respectively 26.8 and 27.9. What is the mode of the distribution?
Solution:
Given,
Mean = 26.8
Median = 27.9
Using the relationship between mean, median and mode,
Mode = 3 Median – 2 Mean
= 3 × 27.9 – 2 × 26.8
= 83.7 – 53.6
= 30.1
Therefore, the mode of the distribution is 30.1.
10. If the mean of the given frequency distribution is 35, then find the missing frequency y. Also, calculate the median and mode for the distribution.
| Class | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Frequency | 2 | 4 | 7 | y | 1 |
Solution:
| Class | Frequency (fi) | Class mark (xi) | fixi |
| 10 – 20 | 2 | 15 | 30 |
| 20 – 30 | 4 | 25 | 100 |
| 30 – 40 | 7 | 35 | 245 |
| 40 – 50 | y | 45 | 45y |
| 50 – 60 | 1 | 55 | 55 |
| Total | ∑fi = 14 + y | ∑fixi = 430 + 45y |
As we know,
Mean = ∑fixi/∑fi
Given that the mean of the distribution is 35.
So, (430 + 45y)/(14 + y) = 35
430 + 45y = (14 + y)35
430 + 45y = 490 + 35y
45y – 35y = 490 – 430
10y = 60
y = 6
Thus, the missing frequency is 6.
Now, we can calculate the mode as follows:
| Class | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Frequency | 2 | 4 = f0 | 7 = f1 | 6 = f2 | 1 |
Here,
l = 30
h = 10
Mode = l + [(f1 – f0)/ (2f1 – f0 – f2)] × h
Mode = 30 + [(7 – 4)/(14 – 4 – 6)] × 10
= 30 + (30/4)
= 30 + 7.5
= 37.5
Now, using the formula Mode = 3Median – 2 Mean, we can get the value of median.
37.5 = 3 Median – 2 (35)
3 Median = 37.5 + 70
Median = 107.7/3
= 35.9
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Practice Questions on Mean Median Mode
1. Find the mean, median and mode of the following data:
23, 18, 24, 23, 31, 37, 28, 30, 25, 40, 35, 35, 27, 25
2. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer, and details are tabulated as given below:
| Mileage (km/l) | 10 – 12 | 12 – 14 | 14 – 16 | 16 – 18 |
| Number of cars | 7 | 12 | 18 | 13 |
Find the mean mileage.
The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?
3. Find the median of the following frequency distribution.
| xi | 15 | 20 | 22 | 17 | 30 | 25 |
| fi | 6 | 9 | 4 | 4 | 10 | 6 |
4. Calculate the mode for the following distribution.
| Cl | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |
| Frequency | 4 | 4 | 13 | 5 | 6 | 5 | 2 | 1 |
5. Suppose the mean and median of distribution are 10.14 and 8, respectively. Find the mode of the distribution using the empirical relationship between mean, median and mode.
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