Question

Below is given frequency distribution of I.Q. (Intelligent Quotient) of 80 candidates.

I.Q.	70-80	80-90	90-100	100-110	110-120	120-130	130-140
No. of candidates	7	16	20	17	11	7	2

Find median I.Q. of candidates.
A. 100.5
B. 98.5
C. 94.5
D. None of these

Answer 1

Hint: We here have to find the median of the given data. For this, we will first make a table of the data along with the cumulative frequency of all the classes. Then we will use the formula for the median given by: $Median={{l}_{m}}+\left( \dfrac{\dfrac{n}{2}-cf}{{{f}_{m}}} \right)\times h$
Where,
${{l}_{m}}$ = lower limit of median class
n= total frequency
h= interval gap of the grouped data
${{f}_{m}}$ = frequency of the median class
cf= cumulative frequency of the class above the median class
Then, we will calculate all these values from the table and then put those values in this formula for median and hence we will get our required answer.

Complete step-by-step answer:
We here have to find the median of the grouped data presented as:

I.Q.	70-80	80-90	90-100	100-110	110-120	120-130	130-140
No. of candidates	7	16	20	17	11	7	2

We now know that the median of a grouped data is given by the formula:
$Median={{l}_{m}}+\left( \dfrac{\dfrac{n}{2}-cf}{{{f}_{m}}} \right)\times h$
Where,
${{l}_{m}}$ = lower limit of median class
n= total frequency
h= interval gap of the grouped data
${{f}_{m}}$ = frequency of the median class
cf= cumulative frequency of the class above the median class
Now, we know that the cumulative frequency of each class is calculated as adding the frequency of that class and all the classes above that.
Hence, the tabular representation of the given data along with their cumulative frequency is given as:

I.Q.	Number of candidates	Cumulative frequency
70-80	7	7
80-90	16	23
90-100	20	43
100-110	17	60
110-120	11	71
120-130	7	78
130-140	2	80

 Now here, we have been given that n=80
Thus, $\dfrac{n}{2}=40$
This lies in the class group of 90-100.
Thus, the median class is the class group of 90-100.
Hence, we get our required values as:
${{l}_{m}}=90$
h=10
${{f}_{m}}=20$
cf= 23
$\dfrac{n}{2}=40$
Thus, putting these values in the formula for median we get:
$\begin{align}
  & Median={{l}_{m}}+\left( \dfrac{\dfrac{n}{2}-cf}{{{f}_{m}}} \right)\times h \\
 & \Rightarrow Median=90+\left( \dfrac{40-23}{20} \right)\times 10 \\
 & \Rightarrow Median=90+\dfrac{17}{2} \\
 & \Rightarrow Median=90+8.5 \\
 & \therefore Median=98.5 \\
\end{align}$
Thus, the median of the given data is 98.5.

So, the correct answer is “Option (B)”.

Note: We here have been given the continuous form of data but if we are given with the discontinuous data, we have to convert it into continuous data and then use the above mentioned formula. We will convert it by subtracting from the lower limit and adding in the upper limit of all the classes the half of the difference of the lower limit of one class and the upper limit of the previous class.