The mean deviation of the following distribution is

x	10	11	12	13	14
f	3	12	18	12	3

$μ_{i}$	$f_{i}$	$μ$i$f_{i}$	$μ$i-$f_{i}$	$|\bar{x}-\bar{u}|f_{i}$
10	3	30	2	6
11	12	132	1	12
12	18	216	0	0
14	3	42	2	6
13	12	156	1	12
	=48	576		=36

$\begin{aligned} \bar{μ} & =\frac{\sum u_{i} f_{i}}{\sum f_{i}} \\ & =\frac{576}{48} \\ & =12\end{aligned}$

$\begin{array}{l} \text { mean deriation }=\frac{\sum f_{i}\left|u_{i}-\bar{\mu}\right|}{\sum f_{i}} \\ =\frac{36}{48} \\ =0.75\end{array}$