Question

Using an appropriate method, find the mean of the following frequency distribution:
$\begin{array}{ccccccc}\text{Class}& & & & & & \\ \text{interval}& 84-90& 90-96& 96-102& 102-108& 108-114& 114-120\\ \text{Frequency}& 8& 10& 16& 23& 12& 11\end{array}$ Which method did you use and why?

Answer 1

The given data is shown as follows:

Class Interval	Frequency ($f_i$)	Mid-value ($x_i$)	$u_i=\frac{x_i-A}{h}$ $u_i=\frac{x_i-99}{6}$	$\left(f_i{u}_i\right)$
84 - 90	8	87	-2	-16
90 - 96	10	93	-1	-20
96 - 102	16	99 = A	0	0
102 - 108	23	105	1	23
108 - 114	12	111	2	24
114 - 120	11	117	3	33
Total	∑($f_i$)=80			∑$\left(f_i{u}_i\right)$=44

A = 99 , h = 6, ∑($f_i$) = 80, ∑$\left(f_i{u}_i\right)$ = 44

Mean = A + $h\times \frac{\sum f_i{u}_i}{\sum f_i}$

= 99 + $6\times \frac{44}{80}$ = 99 + 3.3

= 102.3

Thus, the mean of the following data is 102.3

Since the values of ($f_i$) , ($x_i$) are larger, we use step-deviation method.