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Mean, Median, Mode of Grouped Data — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsMean, Median, Mode of Grouped Data5 Marks
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?
✓

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.

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