Using an appropriate method, find the mean of following frequency distribution: Class interval 84-90 90-96 96-102 102-108 108-114 114-120 Frequency 8 10 16 23 12 11 Which method did you use, and why?

Class interval Frequency f_i Mid-value x_i u_i= x_i-A h = x_i-99 6 f_i u_i 84-90 8 87 -2 -16 90-96 10 93 -1 -20 960-102 16 99 = A 0 0 102-108 23 105 1 23 108-114 12 111 2 24 114-120 11 117 3 33 Total _i=80 _iu_i=44 Thus, A=99, h=6, _i=80 and _iu_i=44 Mean =A+ h _iu_i _i =99+ 6 44 80 =99+3.3 =102.3 Since the values of x_i' s f_i' s are larger, we use step-deviation method.

Using an appropriate method, find the mean of following frequency…

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Question

Using an appropriate method, find the mean of following frequency distribution:

Class interval	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	8	10	16	23	12	11

Which method did you use, and why?

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Answer

Class interval	Frequency $f_i$	Mid-value $x_i$	$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{A}}{\text{h}}=\frac{\text{x}_\text{i}-99}{6}$	$f_i \times u_i$
84-90	8	87	-2	-16
90-96	10	93	-1	-20
960-102	16	99 = A	0	0
102-108	23	105	1	23
108-114	12	111	2	24
114-120	11	117	3	33
Total	$\sum\text{f}_\text{i}=80$			$\sum\text{f}_\text{i}\text{u}_\text{i}=44$

Thus, $\text{A}=99,\ \text{h}=6,\ \sum\text{f}_\text{i}=80$ and $\sum\text{f}_\text{i}\text{u}_\text{i}=44$
Mean $=\text{A}+\Big\{\text{h}\times\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big\}$
$=99+\Big\{6\times\frac{44}{80}\Big\}$
$=99+3.3$
$=102.3$
Since the values of x_i' s f_i' s are larger, we use step-deviation method.