Use class midpoints while acknowledging lost individual values.
Extended · Syllabus E9.3
Notes and key methods
A grouped table gives intervals rather than exact observations. Estimate its mean by using each class midpoint as a representative value, multiplying by class frequency and dividing the resulting total by total frequency.
The estimated mean is not usually exact because the individual positions within each class are unknown. For equal-width classes the modal class is the one with greatest frequency. With unequal-width histograms, compare frequency density when discussing the tallest bar.
Check the conditions
Use the stated units and precision.
Show the method as well as the final answer.
Three worked examples
Example 1 · Grouped mean estimates and modal classes
A measurement x is grouped as follows.
Class
Frequency
\(\displaystyle 0\le x<8\)
7
\(\displaystyle 8\le x<16\)
13
\(\displaystyle 16\le x<24\)
3
Find the midpoint of the second class.
Find the total frequency.
Estimate the mean using class midpoints.
Find the lower boundary of the modal class.
Show the reasoning
(a)
Average the class boundaries.
The mean of its two boundaries is \(12\).
Answer: \(12\).
(b)
Count every observation once using the disjoint classes.
Adding all three frequencies gives 23.
Answer: \(23\).
(c)
Use each class midpoint times frequency, then divide by total frequency.
The midpoint-weighted total is \(244\); divide by 23 to obtain \(\displaystyle \frac{244}{23}\).
Answer: \(\displaystyle \frac{244}{23}\).
(d)
Identify the greatest-frequency class, then give its requested boundary.
The middle equal-width class has greatest frequency 13, and lower boundary 8.
Answer: \(8\).
Example 2 · Grouped mean estimates and modal classes
A measurement x is grouped as follows.
Class
Frequency
\(\displaystyle 0\le x<4\)
3
\(\displaystyle 4\le x<8\)
9
\(\displaystyle 8\le x<12\)
4
Find the midpoint of the second class.
Find the total frequency.
Estimate the mean using class midpoints.
Find the lower boundary of the modal class.
Show the reasoning
(a)
Average the class boundaries.
The mean of its two boundaries is \(6\).
Answer: \(6\).
(b)
Count every observation once using the disjoint classes.
Adding all three frequencies gives 16.
Answer: \(16\).
(c)
Use each class midpoint times frequency, then divide by total frequency.
The midpoint-weighted total is \(100\); divide by 16 to obtain \(\displaystyle \frac{25}{4}\).
Answer: \(\displaystyle \frac{25}{4}\).
(d)
Identify the greatest-frequency class, then give its requested boundary.
The middle equal-width class has greatest frequency 9, and lower boundary 4.
Answer: \(4\).
Example 3 · Grouped mean estimates and modal classes
A measurement x is grouped as follows.
Class
Frequency
\(\displaystyle 0\le x<5\)
4
\(\displaystyle 5\le x<10\)
10
\(\displaystyle 10\le x<15\)
5
Find the midpoint of the second class.
Find the total frequency.
Estimate the mean using class midpoints.
Find the lower boundary of the modal class.
Show the reasoning
(a)
Average the class boundaries.
The mean of its two boundaries is \(\displaystyle \frac{15}{2}\).
Answer: \(\displaystyle \frac{15}{2}\).
(b)
Count every observation once using the disjoint classes.
Adding all three frequencies gives 19.
Answer: \(19\).
(c)
Use each class midpoint times frequency, then divide by total frequency.
The midpoint-weighted total is \(\displaystyle \frac{295}{2}\); divide by 19 to obtain \(\displaystyle \frac{295}{38}\).
Answer: \(\displaystyle \frac{295}{38}\).
(d)
Identify the greatest-frequency class, then give its requested boundary.
The middle equal-width class has greatest frequency 10, and lower boundary 5.
Answer: \(5\).
Common mistakes
Keep intermediate values exact until the final step.