The quartile deviation of the income of a certain person given in rupees for 12 months in a year : 129, 150,151,151,157,158,160,161,162,162,173,175

Options

A. `4*5`
B. `5*5`
C. `6*2`
D. none of these

Answer

To find the quartile deviation of the income data provided, we will follow these steps:

### Step 1: Organize the Data
First, we need to arrange the income data in ascending order. The given data is:
129, 150, 151, 151, 157, 158, 160, 161, 162, 162, 173, 175

The data is already sorted.

### Step 2: Calculate Q1 (First Quartile)
To find the first quartile (Q1), we use the formula:
\[ Q1 = \text{value at } \frac{N + 1}{4} \text{th term} \]
where \( N \) is the number of observations.

Here, \( N = 12 \):
\[ \frac{N + 1}{4} = \frac{12 + 1}{4} = \frac{13}{4} = 3.25 \]

This means Q1 is located between the 3rd and 4th terms. We can calculate Q1 using interpolation:
\[ Q1 = \text{3rd term} + 0.25 \times (\text{4th term} - \text{3rd term}) \]
The 3rd term is 151 and the 4th term is also 151:
\[ Q1 = 151 + 0.25 \times (151 - 151) = 151 + 0 = 151 \]

### Step 3: Calculate Q3 (Third Quartile)
Next, we calculate the third quartile (Q3) using the formula:
\[ Q3 = \text{value at } \frac{3(N + 1)}{4} \text{th term} \]

Calculating:
\[ \frac{3(N + 1)}{4} = \frac{3(12 + 1)}{4} = \frac{39}{4} = 9.75 \]

This means Q3 is located between the 9th and 10th terms. We can calculate Q3 using interpolation:
\[ Q3 = \text{9th term} + 0.75 \times (\text{10th term} - \text{9th term}) \]
The 9th term is 162 and the 10th term is also 162:
\[ Q3 = 162 + 0.75 \times (162 - 162) = 162 + 0 = 162 \]

### Step 4: Calculate the Quartile Deviation (QD)
Now, we can calculate the quartile deviation (QD) using the formula:
\[ QD = \frac{Q3 - Q1}{2} \]
Substituting the values:
\[ QD = \frac{162 - 151}{2} = \frac{11}{2} = 5.5 \]

### Conclusion
The quartile deviation of the income data is **5.5**.