Example:
The following table shows the frequency distribution of the diameters of 40 bottles. (Lengths have been measured to the nearest millimeter) Find the mean of the data.
Diameter (mm) 
35 – 39 
40 – 44 
45 – 49 
50 – 54 
55 – 60 
Frequency 
6 
12 
15 
10 
7 
Solution:
Step 1: Find the midpoint of each interval.
Midpoint of interval = (Lower class limit + Upper class limit)
= (39 + 35) = 37
Diameter (mm) 
35 – 39 
40 – 44 
45 – 49 
50 – 54 
55 – 60 
Frequency (f) 
6 
12 
15 
10 
7 
Midpoint (x) 
37 
42 
47 
52 
57 
Diameter (mm) 
35 – 39 
40 – 44 
45 – 49 
50 – 54 
55 – 60 
Frequency (f) 
6 
12 
15 
10 
7 
Midpoint (x) 
37 
42 
47 
52 
57 
f × x 
222 
504 
705 
520 
399 
Total 

Diameter (mm) 
35 – 39 
40 – 44 
45 – 49 
50 – 54 
55 – 60 

Frequency (f) 
6 
12 
15 
10 
7 
50 
Midpoint (x) 
37 
42 
47 
52 
57 

f × x 
222 
504 
705 
520 
399 
2350 
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