Question

The given table shows the frequency according to the class intervals.

$\begin{array}{cc}\text{Class Interval}& \text{Frequency}\\ 0-10& 10\\ 10-20& 5\\ 20-30& 8\\ 30-40& 12\\ 40-50& 16\end{array}$

Find the Median value using Step jump method.

• A. 30
• B. 37.6
• C. 32.9
• D. 27.67

Answer 1

The correct option is C

32.9

The Cumulative frequency table for the given data can be drawn as :

$\begin{array}{cc}\text{Less than}& \text{Cumulative Frequency}\\ 10& 10\\ 20& 15\\ 30& 23\\ 40& 35\\ 50& 51\end{array}$

N = 51 (An odd number)

Median = $\frac{(51+1)}{2}th$ observation = 26th observation.

So, the Median class is 30 - 40

We know that there are 12 observations from the 23rd observation to the 35th observation which come between 30 and 40.

However, we do not know their individual heights.

We will divide 10 from 30 to 40 into 12 equal parts and assume that one observation is in each subdivision.

We will assume that the value of each observation in a subdivision is the mid - value of the subdivision.

So, the the height of the 24th observation is the mid value of 30 and $30\frac{10}{12}$ft, that is $30\frac{10}{24}$

So, from the 24th observation to the 30th observation, there are 6 observations.

If the 24th term of the AP is $30\frac{10}{24}$ and the common difference is $\frac{10}{12}$,

Then the 26th term = $30\frac{10}{24}$ + $6\times \frac{10}{12}$

= $30+\frac{10+120}{24}$

= 30 + 5.417

= 35.417