Median of ungrouped data, we first arrange the data values of the observations in ascending order. Then , if n is odd, the median is the th observations. And, if n is even, then the median will be average of the and the observations.
Example: A survey regarding the height ( in cm ) of 51 girls of Class X of a school was conducted and the following data was obtained:
Height ( in cm ) | Number of girls |
Less than 140 | 4 |
Less than 145 | 11 |
Less than 150 | 29 |
Less than 155 | 40 |
Less than 160 | 46 |
Lesss than 165 | 51 |
Find the median height.
Solution: To calculate the median height, we need to find the class intervals and their corresponding frequencies.
The given distribution being of the less than type, 140, 145, 150,..., 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140-145, 145-150,....,160-165. Observe that from the given distribution , we find that there are 4 girls with height less than 140, i.e., the frequency of class interval below 140 is 4. Now , there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140-145 is 11-4=7. Similarly, the frequency of 145-150 is 29-11=18, for 150-155, it is 40-29=11, and so on. So. our frequency distribution table with the given cumulative frequencies becomes:
Class intervals | Frequency | Cumulative frequency |
Below 140 | 4 | 4 |
140-145 | 7 | 11 |
145-150 | 18 | 29 |
150-155 | 11 | 40 |
155-160 | 6 | 46 |
160-165 | 5 | 51 |
Now . This observation lies in the class 145-150. Then,
l ( the lower limit ) = 145
cf ( the cumulative frequency of the class preceding 145-150) = 11.
f ( the frequency of the median class 145-150 ) = 18.
h ( the class size ) = 5.
Using the formula, Median we have
Median
Median is that measure of central tendency which divides the total frequency into _________________ equal parts. | |||
Right Option : C | |||
View Explanation |
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