Question 6:
The table shows the heights in cm of some seedlings in a certain garden.
Ages (years) |
15 – 18 |
19 – 22 |
23 – 26 |
27 – 30 |
31 - 34 |
35 – 38 |
Frequency |
40 |
33 |
25 |
10 |
8 |
4 |
Using an assumed mean of 24.5, calculate the mean of the distribution.
Observation
The Chief Examiner reported that this question was very popular among the candidates. They also performed very well. They presented the frequency table as follows:
Ages (years) |
Frequency (f) |
Midpoint |
d = (x – 24.5) |
fd |
15 – 18 |
40 |
16.5 |
-8 |
-320 |
19 – 22 |
33 |
20.5 |
-4 |
-132 |
23 – 26 |
25 |
2. 45 |
0 |
0 |
27 – 30 |
10 |
28.5 |
4 |
40 |
31 – 34 |
8 |
32.5 |
8 |
64 |
35 – 38 |
4 |
36.5 |
12 |
48 |
|
|
|
|
|
= 120
= -300Mean = 24.5 - 
= 24.5 – 2.5 = 22.