Advantages of Quartiles, Deciles and Percentiles:
(a) These averages can be directly determined in case of open end class intervals without knowing the lower limit of lowest class and upper limit of the largest class.
(b) These averages can be calculated easily in absence of some data in a series.
(c) These averages are helpful in the calculation of measures of dispersion.
(d) These averages are not affected very much by the extreme items.
(e) These averages can be located graphically.
Disadvantages of Quartiles, Deciles and Percentiles:
(a) These averages are not easily understood by a common man. These are not well defined and easy to calculate.
(b) These averages are not based on all the observations of a series.
(c) These averages cannot be computed if items are not given in ascending or descending order.
(d) These averages are affected very much by the fluctuation of sampling.
(e) The computation of these averages is not so easy in case of continuous series as the formula of interpolation is to be used.

There will be no mode because it is a bimodal series in which we can't determine mode.

(a) The sum of the deviations of the items from median, ignoring signs is the least.
i.e. 0 is lesat.
(b) Median can be located graphically.

But in a moderately skewed distribution,
Mode = 3Median - 2Mean
Mode = 3 (20) - 2 (25)
Mode = 10

\(Q_1=\frac{n+1}{4}\ observation\frac{41+1}{4}\ observation=25\)
\(Q_3=\frac{3(n+1)}{n}\ observation\frac{3(41+1)}{4}\ observation=40\)

\(Q_1=\frac{n+1}{4}th\ observation\frac{7+1}{4}=2nd\ observation=21\)
\(Q_3=\frac{3(n+1)}{4}th\ observation\frac{3(7+1)}{4}=6th\ observation=65\)

The values of the variable in ascending order are:
100, 400, 500, 600, 700, 800, 1000, 1200, N = 8
\(Q_3=size\ of\frac{(N+1)}{4}\) th item of the series
\(=size\ of\frac{(8+1)}{4}\) th item of the series = size of 2.25th item
= Size of [Second item + 0.25 (Third item - Second item)]
= 400 + 0.25 (500 - 400) = 400 + 25 = 425
\(Q_3=size\ of\frac{3(N+1)}{4}\) th item of the series​​​​​​​
\(=size\ of\frac{3(8+1)}{4}\) th item of the series = size of 6.75th item​​​​​​​
= Size of [6^th item + 0.75 7th item - 6th item)]
= 800 + 0.75 (1000 - 800) = 800 + 150 = 950
Required Q₁ and Q₃ are 425 and 950 respectively​​​​​​​.

Values of the variable are in ascending order:
i.e. 4, 5, 6, 7, 8, 9, 10, 12, 13, 15,20, So N = 11 (No. of Values)
\(Q_1=size\ of\frac{i(N+1)}{4}\) th item of the series = size of 3^rd item = 6
\(Q_3=size\ of\frac{3(N+1)}{4}\) th item of the series = size of 9^th item = 13​​​​​​​
\(\therefore\) Required Q₁ and Q₃ are 6 and 13 respectively​​​​.

Median=size of \(\frac{N}{2}\) Item
Median=size of \(\frac{50}{2}=50\) item 25th item lies in 20-30
We can find median by using the formula equal to
\(Median=l_1+\frac{(\frac{N}{2}-C)}{f}(i)\)

I₁ = 20; f = 10; \(\frac{N}{2}=25\)
C=10;i=10
\(Median=20+\frac{25-15}{8}(10)=28.625\)

Median = size of \(\frac{N}{2}\) Item
Median = size of \(\frac{30}{2}\)  15th item lies in 30-40
We can find median by using the formula equal to
\(Median=l_1+\frac{(\frac{N}{2}-C)}{f}(i)\)
I₁ = 30; f = 8;\(\frac{N}{2}=15\)
C = 10;i = 10
\(Median=30+\frac{15-10}{8}(10)=36.25\)

\(M=\frac{\frac{N}{2}-cf}{f}\times c;M=\frac{47.5-38}{12}\times10\)
\(M=30+\frac{95}{12}\) 
M = 30 + 7.91
Median = 37.91

Median = size of \(\frac{N+1}{2}\)
Size of \(\frac{60+1}{2}=30.5\)^th observation.

Median = size of \(\frac{N+1}{2}\)
size of \(\frac{41+1}{2}=Item\)
Item 19 to 28 have a value equal to 35. Therefore, median = 35.

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