Mean & Median of Ungrouped Data & Frequency Polygon — Case-Study Based Questions

Case Study:
A school organized a Health Check Up Camp for its students. The weights (in kg) of the students of a class were recorded as below :

41, 40, 36, 52, 50, 48, 47, 45,
40, 41, 42, 49, 50, 51, 38, 41,
40, 45, 40, 39, 49, 51, 48, 46,
44, 50, 57, 38, 41, 51

Based on the above information, answer the following questions:

1. The range of the data is :
   (a) 20 kg
   (b) 21 kg
   (c) 22 kg
   (d) 23 kg

2. Mean weight of the data is :
   (a) 45 kg
   (b) 44.5 kg
   (c) 42.5 kg
   (d) 40.5 kg

3. Median of the data is :
   (a) 44 kg
   (b) 45 kg
   (c) 45.5 kg
   (d) 46 kg

4. One student weighing 45 kg, was absent on that day. Next day, the teacher added his name in the list. The average weight of the new group is :
   (a) 47 kg
   (b) 46.4 kg
   (c) 46 kg
   (d) 45 kg

5. The median weight of the students after adding the absentee in the list is :
   (a) 45 kg
   (b) 46 kg
   (c) 44.5 kg
   (d) 44.2 kg

Answer

1. Range = Highest value - Lowest value

= 57 - 36 = 21 kg.

Hence, option (b) is the correct option.

2. Mean = $\dfrac{\text{Total Weight}}{\text{Total number of observations}}$

= $\dfrac{1350}{30}$ = 45 kg.

Hence, option (a) is the correct option.

3. By arranging data in the ascending order, we get :

36, 38, 38, 39, 40, 40, 40, 40, 41, 41, 41, 41, 42, 44, 45, 45, 46, 47, 48, 48, 49, 49, 50, 50, 50, 51, 51, 51, 52, 57

Number of observations (n) = 30, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{30}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{30}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{15th term} + \text{16th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{45} + \text{45}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{90}{2} \\[1em]$

∴ Median = 45 kg.

Hence, option (b) is the correct option.

4. Earlier total sum = 1350

After adding the new student (weighing 45 kg) total sum = 1350 + 45 = 1395

∴ New mean = $\dfrac{1395}{31}$ = 45 kg.

Hence, option (d) is the correct option.

5. After adding the absentee ( weighing 45 kg) the observations are :

36, 38, 38, 39, 40, 40, 40, 40, 41, 41, 41, 41, 42, 44, 45, 45, 45, 46, 47, 48, 48, 49, 49, 50, 50, 50, 51, 51, 51, 52, 57

Number of observations (n) = 31, which is odd.

Median = $\dfrac{n + 1}{2}$ th observation

= $\dfrac{31 + 1}{2}$ th observation

= $\dfrac{32}{2}$th observation

= 16th observation

Median = 45 kg.

Hence, option (a) is the correct option.