Question

199 students were asked how many hours they exercise each week.

The following table shows the data :

$\begin{array}{cc}\text{Time(t hours)}& \text{No. of students}\\ 0-5& 12\\ 5-10& 15\\ 10-15& 23\\ 15-20& 30\\ 20-25& 40\\ 25-30& 35\\ 30-35& 25\\ 35-40& 20\end{array}$

Find the Median of the data.

• A. 20.33
• B. 22.35
• C. 25.5
• D. 27.35

Answer 1

The correct option is B

22.35

The Cumulative frequency table is given by :

$\begin{array}{cc}\text{Time(t hours)}& \text{No. of students}\\ <5& 12\\ <10& 27\\ <15& 50\\ <20& 80\\ <25& 120\\ <30& 155\\ <35& 180\\ <40& 200\end{array}$

N = 200 (An even number)

Median = Average of $\frac{200}{2}th$ and $(\frac{200}{2}+1)th$ observation

= Average of 100th and 101st observation.

So, the Median class is 20 - 25

We know that there are 40 students from the 80th student to the 120th student who exercise between 20 and 25 hours

However, we do not know their individual hours of exercise.

We will divide 5 hrs from 20 to 25 into 40 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 exercise hours of the 81st student is the mid value of 20 and $20\frac{5}{40}$, that is $20\frac{5}{80}$

So, from the 81st student to the 100th student, there are 19 students and from the 81st student to the 101th student, there are 20 students.

If the 81st term of the AP is $20\frac{5}{80}$ and the common difference is $\frac{5}{40}$,

Then the 100th and 101st terms are

100th = $20\frac{5}{80}$ + $19\times \frac{5}{40}$

= $20+\frac{5+190}{85}$

= 20 + 2.29

= 22.29

101st = $20\frac{5}{80}$ + $20\times \frac{5}{40}$

= $20+\frac{5+200}{85}$

= 20 + 2.41

= 22.41

Median = Average of 100th and 101st observation

= $\frac{22.29+22.41}{2}$

= 22.35