Statistics | Exercise 13.3

Question 4

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

Find the median length of the leaves.

(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . . , 171.5 - 180.5.)

Solution

Understand the Question

The given class intervals are in an inclusive (discontinuous) format (e.g., $118-126, 127-135$ ). To use the median formula, the classes must be continuous. We make them continuous by subtracting $0.5$ from each lower limit and adding $0.5$ to each upper limit.
Find the cumulative frequencies ( $cf$ ) and determine the median class using $\dfrac{N}{2} = \dfrac{40}{2} = 20$ .
Apply the median formula: $\text{Median} = l + \left(\dfrac{\frac{N}{2} - cf}{f}\right) \times h$

Step 1 · Convert to Continuous Classes

Subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of each class interval: Diagram 1

\begin{array}{|c|c|} \hline \text{Length (in mm) (Continuous Classes)} & \text{Number of leaves } (f) \\ \hline 117.5 - 126.5 & 3 \\ \hline 126.5 - 135.5 & 5 \\ \hline 135.5 - 144.5 & 9 \\ \hline 144.5 - 153.5 & 12 \\ \hline 153.5 - 162.5 & 5 \\ \hline 162.5 - 171.5 & 4 \\ \hline 171.5 - 180.5 & 2 \\ \hline \end{array}

Step 2 · Calculate Cumulative Frequencies

Compute cumulative frequencies ( $cf$ ) to find the median class: Diagram 2

\begin{array}{|c|c|c|} \hline \text{Length (in mm) (Continuous Classes)} & \text{Number of leaves } (f) & \text{Cumulative Frequency } (cf) \\ \hline 117.5 - 126.5 & 3 & 3 \\ \hline 126.5 - 135.5 & 5 & 8 \\ \hline 135.5 - 144.5 & 9 & 17 \\ \hline 144.5 - 153.5 & 12 & 29 \\ \hline 153.5 - 162.5 & 5 & 34 \\ \hline 162.5 - 171.5 & 4 & 38 \\ \hline 171.5 - 180.5 & 2 & 40 \\ \hline \end{array}

Step 3 · Identify the Median Class

Total frequency $N = 40$ .

\begin{aligned} \dfrac{N}{2} &= \dfrac{40}{2} \\[0.6em] &= 20 \end{aligned}

The cumulative frequency just greater than $20$ is $29$ , which corresponds to the class interval $144.5 - 153.5$ .

Therefore, the median class is $144.5 - 153.5$ .

Step 4 · Calculate the Median Length

From the median class $144.5 - 153.5$ :

Lower limit, $l = 144.5$
Cumulative frequency of preceding class, $cf = 17$
Frequency of median class, $f = 12$
Class size, $h = 153.5 - 144.5 = 9$

Using the median formula: $\text{Median} = l + \left(\dfrac{\frac{N}{2} - cf}{f}\right) \times h$

\begin{aligned} \text{Median} &= 144.5 + \left(\dfrac{20 - 17}{12}\right) \times 9 \\[0.6em] &= 144.5 + \left(\dfrac{3}{12}\right) \times 9 \\[0.6em] &= 144.5 + \left(\dfrac{1}{4}\right) \times 9 \\[0.6em] &= 144.5 + 0.25 \times 9 \\[0.6em] &= 144.5 + 2.25 \\[0.6em] &= 146.75 \text{ mm} \end{aligned}

Answer

$146.75 \text{ mm}$

Common Mistakes

Skipping Class Continuity: Using the original limits (e.g., $l = 145$ and $h = 153 - 145 = 8$ ) instead of converting to continuous classes ( $l = 144.5$ and $h = 9$ ).
Incorrect $cf$ Selection: Using $cf = 29$ (the cumulative frequency of the median class) instead of $cf = 17$ (the cumulative frequency of the class preceding the median class).

Question 4

More questions in Exercise 13.3