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

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
\(Q_1=\frac{(71.25-70)}{15}=50.83\)

We start off by rearranging the data in order from the smallest to the largest.

\(Q_1=\frac{n+1}{4}th\ observation\frac{14+1}{4}=3.75th\ item\ observation\)

= size of {Third item + 0.75 (Forth item - Third item)}
\(Q_3=size\ of\frac{3(N+1)}{4}th\ item\ of \ the\ series\frac{3(14+1)}{4}=10.25th\ observation\)
= size of {Tenth item + 0.25 (Eleventh item - Tenth item)}
= 7 + 0.25 (8 - 7) = 7.25

\(Median=size\ of\frac{N}{2}\)
Median = size of \(\frac{25}{2}=12.5^{th}\) item 12.5th item lies in 30-100
We can find median by using the formula equal to
\(Median=l_1+\frac{(\frac{N}{2}-C)}{f}(i)\)
Where l₁=30;f=3;\(\frac{N}{2}=12.5\);C = 11; i = 70​​​​​​​
\(Median=30+\frac{(12.5-11)}{3}(70)=65\)​​​​​​​

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.

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

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