Question

Question 4
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

$\begin{array}{cc}Length\left(inmm\right)& Numberofleaves{f}_i\\ 118-126& 3\\ 127-135& 5\\ 136-144& 9\\ 145-153& 12\\ 154-162& 5\\ 163-171& 4\\ 172-180& 2\end{array}$

Find the median length of the leaves.

Answer 1

The given data does not have continuous class intervals. We can observe that difference between two class intervals is 1. So, we have to add and subtract $\frac{1}{2}=0.5$ from upper-class limits and lower class limits respectively.

Now, continuous class intervals with respective cumulative frequencies can be represented as below:

$\begin{array}{ccc}Length\left(inmm\right)& Numberofleaves{f}_i& Cumulativefrequency\\ 117.5-126.5& 3& 3\\ 126.5-135.5& 5& 3+5=8\\ 135.5-144.5& 9& 8+9=17\\ 144.5-153.5& 12& 17+12=29\\ 153.5-162.5& 5& 29+5=34\\ 162.5-171.5& 4& 34+4=38\\ 171.5-180.5& 2& 38+2=40\end{array}$

From the table, we observe that cumulative frequency just greater then $\frac{n}{2}(i.e.\frac{40}{2}=20)$ is 29, belonging to class interval 144.5 - 153.5.

Median class = 144.5 - 153.5

Lower limit $l$ of median class = 144.5

Class size $h$ = 9

Frequency $f$ of median class = 12

Cumulative frequency $cf$ of class preceding median class = 17, $Median=l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h=144.5+\left(\frac{20-17}{12}\right)\times 9=144.5+\frac{9}{4}=146.75$

So, median length of leaves is 146.75 mm.