Median

b.) Median = in the midpoint of the distribution. Half of the values in a distribution fall below median and the other half fall above it.

When to use the median:
1. When the exact midpoint f the distribution is wanted the, 50% point.
2. When there are extreme scores which would markedly affect the mean. Extreme scores do not disturb the median.
3. When it is desired that certain scores should influence the central tendency, but all that is known about them is that they are above or below the median.
- determine of whether the cases fall within the lower halve or the upper halve of a distribution (appropriate locator of central tendency).

Finding the median from ungrouped data:
1. When N is odd, the median is the middle score.
Ex. 20 15 13 11 8 7 6
There are 7 scores and the median is 11

2. When N is even, the median is the average of the two middle score.
Ex. 21 18 15 14 11 8 8 7
There are 8 scores and the median is (14+11)/2 = 12.5

3. When several scores have the same value as the midscore.
Ex. 15 15 14 11 9 9 9 6 5
Median is 9
Ex. 1. Find the median of the following set of observations.
8 4 1 3 & 7
Sol. Array the set of observations and find the median
1 3 4 7 8
4 is the middle item

2. Compute for the median from the following set of data
12 9 6 10 7 & 14
Array the data and computer for the median
6 7 9 10 12 14
Median = (9 + 10 )/2 = 9.5

Finding the Median from the Grouped Data

Md = L + [N/2 – F2 / f2] C

Where: L = lower class boundary of the interval where the median lies
N = No. of scores or sum of frequency
F2 = cumulative frequency less than up to the class immediately preceding
the median class (F<)
f2 = frequency of the median class
C= class size
Steps:
1. Prepare 3 columns (Class intervals, class frequency and cumulative frequency less than)
2. Determine the Median class. The median class is that class interval where n/2 lies.
3. Substitute the data to the formula.

Ex. Find the median of the frequency distribution

Weekly wages,No. of Workers(f) ,F< (cumulative frequency less than)
(in peso)
870 – 899, 4, 4
900 – 929 ,6, 10
930 – 959 ,10, 20 – F<
960 – 989, 13, 33 Median class
990 – 1019 ,8 , 41
1020 – 1049, 7, 48
1050 – 1079, 2, 50
N=50
To determine the median class:

Solve for N/2 = 50/2 = 25th

25th items fall in 960 – 989 class interval therefore it is the median class

Md = L + [(N/2 – F2)/f2] C

= 959.5 + [(50/2 – 20) /13] 30

Md = Php 971.04