Page 112 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Unit 6: Measures of Central Tendency

For a grouped frequency distribution, P is that value of the variate such that the area under the Notes
k
histogram to the left of the ordinate at P is k% and the area to its right is (100 – k)% . The formula
k
for the kth percentile can be written as
kN
C
100
P k L k P , h (k = 1, 2, ...... 99)
f
k P
Example: Locate Median, Q , Q , D , D P P and P from the following data:
1 3 4 7, 15, 60 90
( ) 75 76 77 78 79 80 81 82 83 84 85
15 20 32 35 33 22 20 10 8 3 2

Solution:
First we calculate the cumulative frequencies, as in the following table:

75 76 77 78 79 80 81 82 83 84 85
15 20 32 35 33 22 20 10 8 3 2
. . 15 35 67 102 135 157 177 187 195 198 200

N
1. Determination of Median: Here = 100. From the cumulative frequency column, we
2
note that there are 102 (greater than 50% of the total) observations that are less than or
equal to 78 and there are 133 observations that are greater than or equal to 78. Therefore,
M = 78.
d
N
2. Determination of Q and Q : First we determine which is equal to 50. From the
1 3 4
cumulative frequency column, we note that there are 67 (which is greater than 25% of the
total) observations that are less than or equal to 77 and there are 165 (which is greater than
75% of the total) observations that are greater than or equal to 77. Therefore, Q = 77.
1
Similarly, Q = 80.
3
3. Determination of D and D : From the cumulative frequency column, we note that there
4 7
are 102 (greater than 40% of the total) observations that are less than or equal to 78 and
there are 133 (greater than 60% of the total) observations that are greater than or equal to
78. Therefore, D = 78. Similarly, D = 80.
4 7
4. Determination of P , P and P : From the cumulative frequency column, we note that
15 60 90
there are 35 (greater than 15% of the total) observations that are less than or equal to 76 and
there are 185 (greater than 85% of the total) observations that are greater than or equal to
76. Therefore, P = 76. Similarly, P = 79 and P = 82.
15 60 90

Example: The following incomplete table gives the number of students in different age
groups of a town. If the median of the distribution is 11 years, find out the missing frequencies.
Age Group : 0 - 5 5 -10 10 -15 15 - 20 20 - 25 25 - 30 Total
- No. of Students : 15 125 ? 66 ? 4 300

Solution:

Let x be the frequency of age group 10 - 15. Then the frequency of the age group 20 - 25 will be
300 – (15 + 125 + x + 66 + 4) = 90 – x.

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