The quartile deviation for the data

$\begin{array}{l}x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\\ f\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 3\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}8 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}$ is 

Computation of $Q_1$ and $Q_3:$

   $\begin{array}{ccc} & & \\ x& f& c.f.\\ 2& 3& 3\\ 3& 4& 7\\ 4& 8& 15\\ 5& 4& 19\\ 6& 1& 20\\ & N=20& \end{array}$

We have, $N=20$

$\therefore \frac{N}{4}=5$ and $\frac{3N}{4}=15$

We observe that the cumulative frequencies just greater than $\frac{N}{4}(=5)$ and $\frac{3N}{4}(=15)$ are $7$ and $19$ respectively. The

corresponding values of $X$ are $3$ and $5$ respectively.

$\therefore Q_1=3$ and $Q_3=5$

Hence, $\text{ Q.D. }=\frac{1}{2}\left(Q_3-Q_1\right)=1$