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.

According to Connor, "The median is that value of the variable which divides the group into two equal parts, one part comprising all parts greater, and the other values less than median."
According to Horace Secrist, "Median of a series is the value of the item actual or estimated when a series is arranged in order of magnitude which divides the
distribution into two parts."
According to Croxton and Cowden, ''the median is generally defined as the value which divides the distribution so that an equal number of items is on either side of it."
Merits:
(a) The median is useful in case of frequency distribution with open-end classes.
(b) The median is recommended if distribution has unequal classes.
(c) Extreme values do not affect the median as strongly as they affect the mean.
(d) It is the most appropriate average in dealing with qualitative data.
(e) The value of median can be determined graphically where as the value of mean cannot be determined graphically.
(f) It is easy to calculate and understand.
Demerits:
(a) For calculating median it is necessary to arrange the data, where as other averages do not need arrangement.
(b) Since it is a positional average its value is not determined by all the observations in the series.
(c) Median is not capable for further algebraic calculations.
(d) The sampling stability of the median is less as compared to mean.

We can find cumulative frequency as follows:

It is known that total frequencies are 82.
56 + F1 + F2 = 82; F1 + F2= 26............(i)
Median is 41. Therefore, median lies between 40-50
\(Median=l_1+\frac{(\frac{N}{2}-C)}{f}(i)\)
Where l₁=40;f=20;\(\frac{N}{2}=41\);C=25+F1;i=10
\(41=40+\frac{(41-25-F1)}{20}(10)\)
On solving the equation,
F1 = 80 + 41 - 25 - 82 = 14
F1 = 14
On putting this value I equation (i), we get,
14 + F2 = 26
F2 = 12
Cumulative frequency now is:

D₇ = Size of 7(82)/10 item = 57.2 item
It lies in 40-50. It can be calculated by applying the formula:
\(D_7=\frac{(\frac{7N}{10}-C)}{f}(h)\)
Where​​​​​​​ l₁=40;f=20;\(\frac{N}{2}=57.5\);C=39;i=10
\(D_7=40+\frac{(57.5-39)}{20}(10)=49.25\)
\(P_{90}=size\ of\frac{90N}{100}th\ item\ of\ the\ series\)
90 (82)/100 = 72.8th item
It lies in the class 60-70
Interpolation formula for continuous series:
\(P_{90}=l_1+\frac{(\frac{90N}{100}-C)}{f}(h)\)
Where​​​​​​​ l₁=60;f= frequency corresponding to Pi class​​​​​​​
\(\frac{iN}{100}=72.8;\)C=71;f=11;h=10
\(P_{90}=60+\frac{(72.8-71)}{11}(10)=61.63\)​​​​​​​​​​​​​​

\(Q_1=\frac{n}{4}th\ observation\frac{25}{4}=6.25th\ observation\)
Q₁ class is 10-30
\(Q_1=l_1+\frac{(\frac{N}{4}-C)}{f}(i)\)
Where l₁=10;f=9;\(\frac{N}{4}=6.25\);C = 5; i = 20
\(Q_1=10+\frac{(6.25-2)}{9}(20)=21.66\)
\(Q_3=\frac{3(n)}{4}th\ observation\frac{3(25)}{4}=18.75\ observation\)
Q₃ class is 100-500
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=100;f=5;\(\frac{3N}{4}=18.75\);C = 14; i = 400
\(Q_3=100+\frac{(18.75-14)}{5}(400)=480\)​​​​​​​

\(Q_1=\frac{n}{4}th\ observation\frac{50}{4}=12.5th\ observation\)
Q₁ class is 10-20
\(Q_1=l_1+\frac{(\frac{N}{4}-C)}{f}(i)\)
Where l₁=10;f=15;\(\frac{N}{4}=12.5\);C = 0; i = 10
\(Q_1=10+\frac{(12.5-0)}{15}(10)=18.33\)
\(Q_3=\frac{3(n)}{4}\ observation\frac{3(50)}{4}=37.50\ observation\)
Q₃ class is 50-60
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=50;f=8;\(\frac{3N}{4}=37.5\);C=32;i=10
\(Q_3=50+\frac{(37.5-32)}{8}(10)=56.875\)​​​​​​​

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

Median = size of \(\frac{N}{2}\) Item
Median = size of \(\frac{30}{2}=15\) item 15th item lies in 200-300
We can find median by using the formula equal to
\(Median=l_1+\frac{(\frac{N}{2}-C)}{f}(i)\)
Where l₁=200;f=12;\(\frac{N}{2}=15\);C=14;i=100
\(Median=200+\frac{15-14}{12}(100)=208.33\)​​​​​​​