Quartile is that value which divides the total distribution into four equal parts. So there are three quartiles, i.e. Q₁, Q₂ and Q₃. Q₁, Q₂ and Q₃ are termed as first quartile, second quartile and third quartile or lower quartile, middle quartile and upper quartile respectively. Quartiles, deciles and percentiles all are positional measures with following difference.
Quartiles divide the series into four equal parts, deciles divide the series into 10 equal parts, and percentiles divide the series into 100 equal parts. Method of their estimation is also similar.

The relationship between mean, median and mode depends upon the nature of the distribution. A distribution may be symmetrical or asymmetrical.
In asymmetrical distribution the mean, median and mode are equal
i.e. Mean(AM) = Median(M)
= Mode(Mo)
In a highly asymmetrical distribution it is not possible to find a relationship among the averages. But in a moderately asymmetric distribution the difference between the mean and mode is three times the difference between the mean and median.
i.e. Mean-Mode = 3(Mean-Median)
In other words, In a normal Distribution,
Mean = Median = Mode
But in a moderately skewed distribution,
Mode = 3Median - 2Mean
This formula can also be used to find mode in case of bimodal series. If this formula is used to find mode, such a mode is called empirical mode. Karl Pearson expressed this relationship as:
Mode = mean - 3 [mean - median]
Mode = 3 median - 2 mean and Median = mode + ()
Knowing any two values, the third can be computed.
Example: Given median = 20.6, mode = 26 Find mean.
Mode = 3 median - 2 mean
\(\therefore\) Mean = () [3 median - mode)
\(\therefore\) Mean = () [3(20.6) - (26)]
\(\therefore\) Mean = () [35.8]
\(\therefore\) Mean = 17.9
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Mode is the most frequent item in the series.
Merits:
(a) It is easy to calculate and simple to understand.
(b) It is not affected by the extreme values.
(c) The value of mode can be can be determined graphically.
(d) Its value can be determined in case of open-end class interval.
(e) The mode is the most representative of the distribution.
Demerits:
(a) It is not suitable for further mathematical treatments.
(b) The value of mode cannot always be determined.
(c) The value of mode is not based on each and every items of the series.
(d) The mode is strictly defined.
(e) It is difficult to calculate when one of the observations is zero or the sum of the observations is zero.

Calculation of Arithmetic mean by direct method

\(\bar x=\frac{\sum_{i=1}^nf_im_i}{\sum_{i=1}^nf_i}=\frac{1560}{60}=26\)
Calculation of Median

\(Median=size\ of\frac{N}{2}\ Item\)
Median = size of \(\frac{60}{20}\) observation 30th observation lies in 20-30
We can find median by using the formula equal to
\(Median=l_1+\frac{(\frac{N}{2}-C)}{f}(i)\)
Where l₁ = 20; f = 20;\(\frac{N}{2}=30\);C = 20; i = 10
\(Median=20+\frac{(30-20)}{20}(10)=25\)
Mode = 3Median - 2Mean
Mode = 3 (25) - 2 (26)
Mode = 23

\(Q_1=\frac{n}{4}th\ observation\frac{30}{4}=7.5th\ observation\)
Q₁ class is 15-25
\(Q_1=l_1+\frac{(\frac{N}{4}-C)}{f}(i)\)
Where l₁ = 15; f = 5;\(\frac{N}{4}=7.5\); C=3;i=10
\(Q_1=10+\frac{(7.5-3)}{5}(10)=24\)
\(Q_3=\frac{3(n)}{4}\ observation\frac{3(30)}{4}=22.5\ observation\)
Q₃ class is 45-55
Then Quartile of a continuous series can be calculated by the below interpolation formula.
\(Q_3=l_1+\frac{(\frac{3N}{4}-C)}{f}(i)\)
Where l₁ = 45, f= 5;\(\frac{3N}{4}=22.5\);C = 22, i = 10
\(Q_3=45+\frac{(22.5-20)}{5}(10)=50\)​​​​​​​

\(Q_1=\frac{n}{4}\ observation\frac{95}{4}\ observation\)

Q₁ class is 20-30
\(Q_1-l_1\frac{(\frac{N}{4}-C)}{f}(i)\)
Where l₁=20;f=18;\(\frac{N}{2}=23.75\);C=20;i=10
\(Q_1=\frac{(23.75-20)}{18}=22.08\)
\(Q_3\frac{3(n)}{4}\ observation\frac{3(95)}{4}=71.25 \ observation\)
Q₃ class is 50-60
Then Quartile of a continuous series can be calculated by the below interpolation formula.
\(Q_3l_1\frac{(3\frac{N}{4}-C)}{f}(i)\)
Where l₁=50;f=15;\(\frac{3N}{2}=71.25\);C=20;i=10
\(Q_1=\frac{(71.25-70)}{15}=50.83\)

\(Q_1=\frac{n+1}{4}\ observation\frac{11+1}{4}\ observation=32\)
\(Q_3=\frac{3(n+1)}{4}th\ observation\frac{3(11+1)}{4}=9th\ observation=72\)

\(Median=\frac{n+1}{2}th\ observation\frac{6+1}{2}\ observation=\frac{3rd\ item+4\ item}{2}=\frac{140+145}{2}=142.5\)

From the definition of median, we should be able to tell that the first step is to rearrange the given set of numbers in order of increasing magnitude, i.e. from the lowest to the highest

\(Median=\frac{n+1}{2}th\ observation \frac{11+1}{2}\) observation=57.