Relationship among mean, median and mode

1. NADEEM UDDIN
ASSOCIATE PROFESSOR
OF STATISTICS
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2. Relationship Among Mean, Median and Mode:
i) Median lies between mean and mode.
ii) Median closer to mean than mode.
In moderately skewed distribution, the following approximate
relation holds good.
Mean – Mode = 3 (Mean – Median)
OR
Mode = 3 Median – 2 Mean

3. Interpretation of Results:
i) If Mean = Median = Mode then the distribution is said to be
symmetrical.

4. ii) If Mean > Median > Mode, then the distribution is said to
be positively skewed.

5. iii) If Mean < Median < Mode, then the distribution is said to
be negatively skewed.

6. Classes 2.0 – 2.2 2.3 – 2.5 2.6 – 2.8 2.9 – 3.1 3.2 – 3.4
Frequency 10 20 30 20 10
Example-1
Compute Mean, Median and Mode for the following distribution
and comment on the result.
Solution:
Classes F x fx C.B C.f
2.0 – 2.2
2.3 – 2.5
2.6 – 2.8
2.9 – 3.1
3.2 – 3.4
10
20
30
20
10
2.1
2.4
2.7
3.0
3.3
21
48
81
60
33
1.95 – 2.25
2.25 – 2.55
2.55 – 2.85
2.85 – 3.15
3.15 – 3.45
10
30
60
80
90
Sum 𝑓 = 90 243𝑓𝑥 =

7. 1st, we find the mean:
Mean =
243
90
2nd, we find the median:
f
2
th term =
90
2
th term = 45thterm
45th term lies in the class 2.6 – 2.8.
Therefore 2.55 – 2.85 is the median class.
l = 2.55, h = 2.85 − 2.55 = 0.3, f = 30,
f
2
= 45, c. f = 30
Mean = 2.7
Mean =
fx
f

8. Median = 2.55 +
0.3
30
45 − 30
Median = 2.55 +
0.3
30
15
Median = 2.55 + 0.15
Median = 2.7
Median = l +
h
f
f
2
− c. f
Substitute the values, we get:

9. 3rd, we find the mode:
Since the highest frequency occur in the class 2.6 – 2.8.
Therefore 2.55 – 2.85 is the modal class.
l = 2.55, h = 2.85 – 2.55 = 0.3, fm = 30, f1 = 20, f2 = 20
Substitute the values, we get:
Mode = l + h
fm − f1
2fm − f1 − f2

10. Mode = 2.55 + 0.3
30 − 20
2(30) − 20 − 20
Mode = 2.55 + 0.3
10
60 − 40
Mode = 2.55 + 0.3
10
20
Mode = 2.55 + 0.15
Mode = 2.7

11. Comment:
Since Mean = Median = Mode = 2.7
So that the given distribution is symmetrical.