Page 287 - Start Up Mathematics_7

Example 5: Organize the marks scored in a class test in the form of a frequency table and find
its arithmetic mean.

7, 6, 4, 2, 5, 3, 5, 4, 5, 2, 7, 6, 2, 8, 5, 6, 9, 5, 3, 6
Solution: Marks (x ) Tally marks Frequency (f) f x
i
i
i i
2 3 6
3 2 6
4 2 8

5 5 25
6 4 24

7 2 14

8 1 8
9 1 9

20 100
∑f x 100
Thus, Arithmetic Mean = i i = = 5 marks.
∑f i 20
Median
Consider the following data which gives the distance travelled (in km) by 10 students to reach
school from their homes: 5, 1, 9, 3, 7, 1, 8, 7, 5, 4
The mean distance covered by these students is 5 km. Dilip, a new admission who lives 93 km
away from school joins them.
The average distance travelled by the students once Dilip joins them, i.e., 5, 1, 9, 3, 7, 1, 8, 7, 5, 4,
93 increases to 13 km. The average has become more than double because of one extreme value of
93 km. Hence, mean cannot be taken as a satisfactory central value for all types of data.

Let’s find a central tendency which is not affected by extreme values. First arrange the above data
in ascending order: 1, 1, 3, 4, 5, 5 , 7, 7, 8, 9, 93
In the above data, the number of observations n is 11 which is odd, therefore we have single middle
n + 1
term occurring in the th position, when the data is arranged in ascending or descending order.
2
Thus, Median = Value of the n + 1 th term = 11 + 1 = 12 = 6th term.
2 2 2
Hence, the value of the median = T = 5
6
Thus, we have learnt that median is the value which lies in the middle of the data with half the
observations having greater or equal values lying above it and the other half with smaller or equal
values lying below it.
Example 6: The weights of 6 students (in kg) are: 31, 32, 25, 24, 26, 30. Find the median.
Solution: Arranging the above data in ascending order, we get: 24, 25, 26, 30, 31, 32

The number of observations n is 6 which is even, hence no single value is the median.
n
On careful thinking, we find two middle terms, i.e., 26 and 30. These are 2 th term
n
and + 1 th term in the arranged data.
2
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