Question

Following is the distribution of IQ of 100 students. Find the median IQ.

I.Q	55-64	65-74	75-84	85-94	95-104	105-114	115-124	125-134	135-144
Number of students	1	2	9	22	33	22	8	2	1

Answer 1

Hint: In this question, we are given an inclusive series and we have to find the median of the data. For this, we will first change the series to an exclusive series and then draw the frequency distribution table having columns as class, frequency and cumulative frequency. After that, we will apply the following formula for calculating median.
\[\text{Median}={{l}_{1}}+\dfrac{\dfrac{N}{2}-c.f}{f}\times i\]
Where, ${{l}_{1}}$ is the lower limit of median class, N is the sum of frequencies, c.f. is cumulative frequency of class preceding the median class, f is the simple frequency of median class and i is the class interval for finding median class, we will use cumulative frequency and $\dfrac{N}{2}$.

Complete step-by-step answer:
Let us first convert inclusive series to exclusive series by adding $\dfrac{n}{2}$ to upper limit and subtracting $\dfrac{n}{2}$ to lower limit where, n is the difference between upper limit of first class and lower limit of next class. Here, n = 65-64 = 1.
Hence, adding 0.5 to upper limit and subtracting 0.5 from lower limit. Series becomes:

Class	Frequency (f)	Cumulative Frequency (c.f.)
54.5-64.5	1	1
64.5-74.5	2	3
74.5-84.5	9	12
84.5-94.5	22	34
94.5-104.5	33	67
104.5-114.5	22	89
114.5-124.5	8	97
124.5-134.5	2	99
134.5-144.5	1	100

Here, cumulative frequency for every class is calculated by taking the sum of all the previous frequencies.
Now, the sum of frequencies will be equal to the value of the last cumulative frequency. Hence, N = 100.
Now, we have to find ${{\left( \dfrac{N}{2} \right)}^{th}}$ item.
${{\left( \dfrac{N}{2} \right)}^{th}}$ item = ${{\left( \dfrac{100}{2} \right)}^{th}}$ item = ${{50}^{th}}$ item.
Since 67 is the cumulative frequency just greater than 50, therefore, the required median class is the class corresponding to cumulative frequency as 67. Hence, the median class is 94.5-104.5.
Lower limit of class ${{l}_{1}}=94.5$
f is frequency of 94.5-104.5 which is 33. f = 33
i is the class interval which is difference between upper limit and lower limit.
$i=104.5-94.5=10$.
c.f. is the cumulative frequency of the class preceding median class, therefore, c.f. = 34.
Now, median is given as
\[\text{Median}={{l}_{1}}+\dfrac{\dfrac{N}{2}-c.f}{f}\times i\]
Putting all the value obtained we get:
\[\begin{align}
  & \text{Median}=94.5+\dfrac{50-34}{33}\times 10 \\
 & \Rightarrow 94.5+4.88 \\
 & \Rightarrow 99.5 \\
\end{align}\]
Hence, the value of the median for given data is 99.5.

Note: Students should note that, while taking median class, we have to take cumulative frequency which is either equal or just greater than the ${{\left( \dfrac{N}{2} \right)}^{th}}$ term. But while using the formula, we have to take cumulative frequency of the class preceding the median class. As there are a lot of terms involved in the formula, students should try to avoid mistakes in using formula.