Suppose population A has 100 observations` 101,102,...,200 and ` another population B has 100 observatons 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. `9/4`
B. `4/9`
C. `2/3`
D. `1`

Answer

To solve the problem, we need to calculate the variances of populations A and B and then find the ratio of these variances.

### Step 1: Define the populations

Population A consists of the observations from 101 to 200.
Population B consists of the observations from 151 to 250.

### Step 2: Calculate the mean of each population

**Mean of Population A (μ_A):**
\[ \mu_A = \frac{\text{Sum of observations in A}}{\text{Number of observations in A}} = \frac{101 + 102 + ... + 200}{100} \]
The sum of an arithmetic series can be calculated using the formula:
\[ \text{Sum} = \frac{n}{2} \times (\text{first term} + \text{last term}) \]
For Population A:
\[ \text{Sum}_A = \frac{100}{2} \times (101 + 200) = 50 \times 301 = 15050 \]
Thus,
\[ \mu_A = \frac{15050}{100} = 150.5 \]

**Mean of Population B (μ_B):**
\[ \mu_B = \frac{151 + 152 + ... + 250}{100} \]
For Population B:
\[ \text{Sum}_B = \frac{100}{2} \times (151 + 250) = 50 \times 401 = 20050 \]
Thus,
\[ \mu_B = \frac{20050}{100} = 200.5 \]

### Step 3: Calculate the variance of each population

**Variance of Population A (V_A):**
\[ V_A = \frac{\sum (x_i - \mu_A)^2}{n} \]
Where \( x_i \) are the observations in Population A. We can also use the formula:
\[ V_A = \frac{\sum x_i^2}{n} - \mu_A^2 \]
First, we need to calculate \( \sum x_i^2 \):
\[ \sum_{i=101}^{200} i^2 = \frac{n}{6} (2a + (n - 1)d)(2a + (n - 1)d) = \frac{100}{6}(2 \times 101 + 99 \times 1)(2 \times 101 + 99 \times 1) = \frac{100}{6}(202 + 99)(202 + 99) = \frac{100}{6}(301)(301) \]
Calculating \( \sum x_i^2 \):
\[ \sum_{i=101}^{200} i^2 = \frac{100}{6} \times 301^2 = \frac{100 \times 90601}{6} = 1510166.67 \]
Now calculate variance:
\[ V_A = \frac{1510166.67}{100} - (150.5)^2 = 15101.67 - 22650.25 = -7548.58 \]

**Variance of Population B (V_B):**
Using the same method:
\[ V_B = \frac{\sum x_i^2}{n} - \mu_B^2 \]
Calculating \( \sum x_i^2 \) for B:
\[ \sum_{i=151}^{250} i^2 = \frac{100}{6}(2 \times 151 + 99)(2 \times 151 + 99) = \frac{100}{6}(302 + 99)(302 + 99) = \frac{100}{6}(401)(401) \]
Calculating \( \sum x_i^2 \):
\[ \sum_{i=151}^{250} i^2 = \frac{100}{6} \times 401^2 = \frac{100 \times 160801}{6} = 2680016.67 \]
Now calculate variance:
\[ V_B = \frac{2680016.67}{100} - (200.5)^2 = 26800.17 - 40200.25 = -13400.08 \]

### Step 4: Calculate the ratio of the variances

Now we can find the ratio of the variances:
\[ \frac{V_A}{V_B} = \frac{-7548.58}{-13400.08} \approx 0.562 \]

### Final Answer
\[ \frac{V_A}{V_B} \approx 0.562 \]