Question

A survey regarding the height (in cm) of 51 girls of class X of a school was conducted
and the following data was obtained:
$$\begin{array}{cc}& \\ \text{Height in cm}& \text{Number of Girls}\\ \text{Less than 140}& 4\\ \text{Less than 145}& 11\\ \text{Less than 150}& 29\\ \text{Less than 155}& 40\\ \text{Less than 160}& 46\\ \text{Less than 165}& 51\end{array}$$
Find the median height.

Answer 1

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, 155, 160, 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140-145, 145-150, 150-155, 155-160, 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 given cumulative frequencies becomes:

Class interval	Frequency	Cumulative frequency
Below 140 140 – 145 145 – 150 150 – 155 155 – 160 160 – 165	4 7 18 11 6 5	4 11 29 40 46 51

Now n = 51. an odd number that is n/ 2 = 51/ 2 =25.5 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

So, the median height of the girls is 149.03 cm

This means that the height of about 50% of the girls in less than this height, and 50% are taller than this height.