Answer :
Class Interval |
Frequency |
Cumulative Frequency |
1-4 |
6 |
6 |
4-7 |
30 |
36 |
7-10 |
40 |
76 |
10-13 |
16 |
92 |
13-16 |
4 |
96 |
16-19 |
4 |
100 |
We have n = 100
\(\Rightarrow \frac{n}{2} \ = \ 50 \)
The cumulative frequency just greater than \( \frac{n}{2} \) is 76 and the corresponding class is 7-10 .
Thus 7-10 is the median class such that
\( \frac{n}{2} \ = \ 50 , l = 7, cf = 36 , f = 40 and h = 3 \)
Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)
\( = \ 7 \ + \ ( \frac{50 \ - \ 36}{40}) \ × \ 3 \)
\( = \ 7 \ + \ \frac{14}{40} \ × \ 3 \)
\(= \ 7 \ + \ 1.05 \ = \ 8.05 \)
Now, to calculate the mean,
Class Interval |
\( f_i \) |
\( x_i \) |
\( f_i x_i \) |
1-4 |
6 |
2.5 |
15 |
4-7 |
30 |
5.5 |
165 |
7-10 |
40 |
8.5 |
340 |
10-13 |
16 |
11.5 |
184 |
13-16 |
4 |
14.5 |
51 |
16-19 |
4 |
17.5 |
70 |
|
\( \sum f_i \ = \ 100 \) |
|
\( \sum f_i x_i \ = \ 825 \) |
\(\therefore \) Mean = \( \overline{x} \ = \ \frac{ \sum f_i x_i}{ \sum f_i} \)
\( = \ \frac{832}{100} \)
Therefore, Mean = \( 8.32 \)
Calculation of mode :
The class 7-10 has the maximum frequency therefore, this is the modal class.
Here l = 7 , h = 3, f
1 = 40 , f
0 = 30 and f
2 = 16
Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)
\(= \ 7 \ + \ \frac{40 \ - \ 30}{80 \ - \ 30 \ - \ 16} \ × \ 3 \)
\(= \ 7 \ + \ \frac{10}{34} \ × \ 3 \)
\( = \ 7 \ + \ 0.88 \ = \ 7.88 \)
\(\therefore \)
Median = 8.05 , Mean = 8.32 and Mode = 7.88