The marks obtained by 19 students of a class are given below:
27,36,22,31,25,26,33,24,37,32,29,28,36,35,27,26,32,35 and 28. Find
(i) Median (ii) Lower quartile
(iii) Upper quartile (iv) Inter quartile range

Answer

To solve the problem step by step, we will find the median, lower quartile, upper quartile, and interquartile range of the given marks.

### Step 1: Arrange the data in ascending order
The marks obtained by the 19 students are:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35, 28.

Arranging these marks in ascending order, we get:
22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 31, 32, 32, 33, 35, 35, 36, 36, 37.

### Step 2: Find the Median
The median is the middle value of a data set. Since there are 19 values (which is odd), we can find the median using the formula:
\[ \text{Median} = \frac{n + 1}{2} \]
where \( n \) is the number of observations.

Calculating:
\[ \text{Median position} = \frac{19 + 1}{2} = \frac{20}{2} = 10 \]
The 10th term in the ordered list is 29. Therefore, the median is:
\[ \text{Median} = 29 \]

### Step 3: Find the Lower Quartile (Q1)
The lower quartile (Q1) is the median of the first half of the data. To find Q1, we use the formula:
\[ Q1 = \frac{n + 1}{4} \]
Calculating:
\[ Q1 \text{ position} = \frac{19 + 1}{4} = \frac{20}{4} = 5 \]
The 5th term in the ordered list is 26. Therefore, the lower quartile is:
\[ Q1 = 26 \]

### Step 4: Find the Upper Quartile (Q3)
The upper quartile (Q3) is the median of the second half of the data. We use the formula:
\[ Q3 = \frac{3(n + 1)}{4} \]
Calculating:
\[ Q3 \text{ position} = \frac{3(19 + 1)}{4} = \frac{3 \times 20}{4} = 15 \]
The 15th term in the ordered list is 35. Therefore, the upper quartile is:
\[ Q3 = 35 \]

### Step 5: Find the Interquartile Range (IQR)
The interquartile range is calculated as:
\[ IQR = Q3 - Q1 \]
Substituting the values we found:
\[ IQR = 35 - 26 = 9 \]

### Final Results
(i) Median = 29
(ii) Lower Quartile (Q1) = 26
(iii) Upper Quartile (Q3) = 35
(iv) Interquartile Range (IQR) = 9