The mean of two samples of sizes 200 and 300 were found to be 25, 10 respectively. Their standard deviations were 3 and 4 respectively. The variance of combined sample of size 500 is

Options

A. 64
B. 65.2
C. 67.2
D. 64.2

Answer

To find the variance of the combined sample of sizes 200 and 300, we can follow these steps:

### Step 1: Define the given values
- Sample size 1 (n1) = 200
- Sample size 2 (n2) = 300
- Mean of sample 1 (x̄1) = 25
- Mean of sample 2 (x̄2) = 10
- Standard deviation of sample 1 (σ1) = 3
- Standard deviation of sample 2 (σ2) = 4

### Step 2: Calculate the combined mean (x̄)
The combined mean (x̄) can be calculated using the formula:
\[ x̄ = \frac{n1 \cdot x̄1 + n2 \cdot x̄2}{n1 + n2} \]
Substituting the known values:
\[ x̄ = \frac{200 \cdot 25 + 300 \cdot 10}{200 + 300} \]
Calculating the numerator:
\[ = \frac{5000 + 3000}{500} = \frac{8000}{500} = 16 \]

### Step 3: Calculate the deviations (d1 and d2)
- Deviation for sample 1 (d1) = x̄1 - x̄ = 25 - 16 = 9
- Deviation for sample 2 (d2) = x̄2 - x̄ = 10 - 16 = -6

### Step 4: Calculate the variance of the combined sample
The variance (σ²) of the combined sample can be calculated using the formula:
\[ \sigma^2 = \frac{n1 \cdot \sigma_1^2 + n2 \cdot \sigma_2^2 + n1 \cdot d1^2 + n2 \cdot d2^2}{n1 + n2} \]
Substituting the known values:
\[ \sigma^2 = \frac{200 \cdot 3^2 + 300 \cdot 4^2 + 200 \cdot 9^2 + 300 \cdot (-6)^2}{200 + 300} \]
Calculating each term:
- \(200 \cdot 3^2 = 200 \cdot 9 = 1800\)
- \(300 \cdot 4^2 = 300 \cdot 16 = 4800\)
- \(200 \cdot 9^2 = 200 \cdot 81 = 16200\)
- \(300 \cdot (-6)^2 = 300 \cdot 36 = 10800\)

Now substituting these values back into the variance formula:
\[ \sigma^2 = \frac{1800 + 4800 + 16200 + 10800}{500} \]
Calculating the total:
\[ = \frac{33600}{500} = 67.2 \]

### Final Answer
The variance of the combined sample of size 500 is **67.2**.

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