# The Mean, Median and Mode of each Frequency Table

MEASURES OF THE CENTRE FROM OTHER SOURCES

When the Same data values appear Several times we often summarise the data in a frequency table.

Consider the data in the given table:
We can find the measures of the centre directly from the table.
The mode
The data value 7 has the highest frequency.
The mode is therefore 7. The mean
Adding a ‘Product’ column to the table helps to add all scores. For example, there are 15 data of value 7 and these add to 7×15=105.
Rememberin that where k is the number of different data values. This formula is often abbreviated as In this case the mean =278/40=6.95.

The median
since (n+1)/2=41/2=20.5, the median is the average of the 20th and 21st data values, when they are listed in order.
In the table, the blue numbers show us accumulated frequency values, or the cumulative frequency.
We can see that the 20th and 21st data values (in order) are both 75.

∴ the median =(7+7)/2=7

Notice that we have a skewed distribution even though the mean, median and mode are nearly equal. So, We must be careful if we use measures of the middle to call a distribution symmetric. Median

The median is the data point in the middle when all of the data points are arranged in order (high to low or low to high). To find where it is, we take into account the number of data points. If the number of data points is odd, divide the number of data points by 2 and then round up to the next integer, the resulting integer is the location of the median.
If the number of data points is even, there are two middle values. We take the number of data points and divide by 2, this integer is the first of the two middles, the next one is also a middle. Now we average these two middle values to get the median.

Example 1. An odd number of data points with no frequency distribution.

3, 4.5, 7, 8.5, 9, 10, 15

There are 7 data points and 7/2=3.5 so the median is the 4th number, 8.5.

Example 2. An odd number of data points with a frequency distribution.

Age Frequency
18 12
19 5
20 3
21 9
22 2
Total 31

There are 31 data points and 31/2=15.5 so the median is the 16th number. Start counting, 18 occurs 12 times, then 19 occurs 5 times getting us up to entry 17 (12+5), so the 16th entry must be a 19. This data set has a median of 19.

Example 3. An even number of data points with no frequency distribution.

3, 4.5, 7, 8.5, 9, 10, 15, 15.5

There are 8 data points and 8/2=4 so the median is the average of the 4th and 5th data point, (8.5+9)/2=8.75. This data set has a median of 8.75.

Example 4. An even number of data points with a frequency distribution.

Age Frequency
18 11
19 5
20 3
21 9
22 2
Total 30

There are 30 data points and 30/2=15 so the median is the average of the 15th and 16th number. Start counting, 18 occurs 11 times, then 19 occurs 5 times getting us up to entry 16 (11+5), so both the 15th and 16th entries must be a 19. This data set has a median of 19.
Example 5. A second case of an even number of data points with a frequency distribution.

Age Frequency
18 10
19 5
20 4
21 9
22 2
Total 30

There are 30 data points and 30/2=15 so the median is the average of the 15th and 16th number. Start counting, 18 occurs 10 times, then 19 occurs 5 times getting us up to entry 15 (10+5), so both the 15th entries is a 19 and the 16th entry must be 20, so the median is the average of these two datapoints, (19+20)/2=19.5. This data set has a median of 19.5.

Summary median: middle score of an ordered list of data
mode: the most frequent score

Q1. Find the mode and the median of the following frequency distribution. solution:
Since the frequency for x=14 is maximum.
So Mode =14. = frequency of the 15th term
According to the table it can be observed that the value of x from the 13th term to the 17th term is 13. So the median =13.

Q2. At a shooting competition the score of a competitor were as given below: (i) What was his modal score?
(ii) What was his median score?
(iii) What was his total score?
(iv) What was his mean score?
solution: (i) Modal score =4 as it has maximum frequency 7.
(ii) Median =(25+1)/2=13th term =3
(iii) Total score =80
(iv) Mean =80/25=3.2

Q3. The ages of 37 students in a class are given in the following table: Find the median. Number of terms =37.
Median =((37+1))/2th term=19th term
Median =14

Q4. The distribution, given below, shows the marks obtained by 25 students in an aptitude test. Find th mean distribution. solution: Number of terms =25
(i) Mean =171/25=6.84
(ii) Median =(25+1)/2 th term=13th term=7
(iii) Mode =6 as it has maximum frequencies i.e. 6

Q5. The table below shows the number of aces served by tennis players in their first sets of a tournament. Determine the: a) mean b) median c) mode for this data. b) There are 55 data values, so n=55. (n+1)/2=28, so the median is the 28th data Value.
From the cumulative frequency column, the data Values 16 to 33 are 3 aces.
∴ the 28th data Value is 3 aces.
∴ the median is 3 aces.

c) Looking down the frequency column, the highest frequency is 18. This corresponds to 3 aces, so the mode is 3 aces.