Suppose a population A has 100 observations `101,102,................,200` and another population B has 100 observations `151,152,........250.` If `V_A and V_B` represent the variances of the two populations respectively, then `V_A/V_B` is

Options

A. 1
B. `9//4`
C. `4//9`
D. `2//3`

Explanation

To solve the problem, we need to find the variances of two populations A and B, and then calculate the ratio \( \frac{V_A}{V_B} \).

### Step 1: Identify the populations
Population A consists of the integers from 101 to 200. Population B consists of the integers from 151 to 250.

### Step 2: Calculate the mean of Population A
The mean \( \mu_A \) of Population A can be calculated using the formula for the mean of a uniform distribution of consecutive integers:

\[ \mu_A = \frac{\text{first term} + \text{last term}}{2} = \frac{101 + 200}{2} = \frac{301}{2} = 150.5 \]

### Step 3: Calculate the mean of Population B
Similarly, the mean \( \mu_B \) of Population B is:

\[ \mu_B = \frac{151 + 250}{2} = \frac{401}{2} = 200.5 \]

### Step 4: Calculate the variance of Population A
The variance \( V_A \) is calculated using the formula:

\[ V_A = \frac{\sum (x_i - \mu_A)^2}{n} \]

However, since we are dealing with consecutive integers, we can use the property of variance for a uniform distribution of consecutive integers. The variance for a set of \( n \) consecutive integers starting from \( a \) is given by:

\[ V = \frac{(n^2 - 1)}{12} \]

For Population A:
- The number of observations \( n_A = 100 \)
- The variance \( V_A \) is:

\[ V_A = \frac{(100^2 - 1)}{12} = \frac{9999}{12} = 833.25 \]

### Step 5: Calculate the variance of Population B
For Population B:
- The number of observations \( n_B = 100 \)
- The variance \( V_B \) is also:

\[ V_B = \frac{(100^2 - 1)}{12} = \frac{9999}{12} = 833.25 \]

### Step 6: Calculate the ratio of variances
Now, we can find the ratio \( \frac{V_A}{V_B} \):

\[ \frac{V_A}{V_B} = \frac{833.25}{833.25} = 1 \]

### Final Answer
Thus, the value of \( \frac{V_A}{V_B} \) is \( 1 \).

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