Question

The following data gives distribution of height of students-

Height(in cm)	160	150	152	161	156	154	155
Number of students	12	8	4	4	3	3	7

Find the median of distribution-
A)154 B)155 C)160 D)161

Answer 1

Hint: First we have to arrange the data in ascending order of height. Make three tables, in the third table calculate cumulative frequency. If the total frequency is even number use formula
Median distribution= ${\dfrac{{\text{n}}}{2}^{{\text{th}}}}{\text{term}}$ where n is total frequency.
 If n is odd number use the formula,
Median distribution=${\dfrac{{\left( {{\text{n + 1}}} \right)}}{2}^{{\text{th}}}}{\text{term}}$ where n is total frequency. Find the term.

Complete step-by-step answer:
Here, we are given the height of students and number of students. First we will arrange the heights in ascending order and find the cumulative frequency. For this we will make three tables instead of two and calculate the data in the following manner- as $150$ is the smallest number here,it will come first then we will go in ascending order upto $161$.

Height(in cm)	Frequency	Cumulative frequency
150	8	8
152	4	8 + 4 = 12
154	3	12 + 3 = 15
155	7	15 + 7 = 22
156	3	22 + 3 = 25
160	12	25 + 12 = 37
161	4	37 + 4 = 41

 If the total frequency is even number use formula-
Median distribution= ${\dfrac{{\text{n}}}{2}^{{\text{th}}}}{\text{term}}$ where n is total frequency. If n is odd number use the formula,
Median distribution=${\dfrac{{\left( {{\text{n + 1}}} \right)}}{2}^{{\text{th}}}}{\text{term}}$ where n is total frequency.

Since here, the total number of students is $41$ which is an odd number. So we will use the formula-
$ \Rightarrow $ Median distribution=${\dfrac{{\left( {{\text{n + 1}}} \right)}}{2}^{{\text{th}}}}{\text{term}}$ where n is total frequency. Here n is the total number of students. So on putting the values, we get-
$ \Rightarrow $ Median distribution=${\dfrac{{\left( {{\text{41 + 1}}} \right)}}{2}^{{\text{th}}}}{\text{term}}$
On solving and simplifying, we get-
$ \Rightarrow $ Median distribution=${\dfrac{{\left( {{\text{42}}} \right)}}{2}^{{\text{th}}}}{\text{term = 2}}{{\text{1}}^{{\text{th}}}}{\text{term}}$
Here, we have to find the ${21^{{\text{th}}}}{\text{term}}$. So the corresponding term is $155$ as it is between ${15^{{\text{th}}}}{\text{ and }}{22^{{\text{th}}}}{\text{term}}$ .

Hence, the correct answer is ‘B’.

Note: The student may obtain the wrong answer is he/she does not arrange the heights in ascending order. This is because there are different numbers of students of different heights so when you calculate cumulative frequency without arranging in ascending order, you may get the wrong term as answer.