The means of two samples of size 40 and 50 were found to be 54 and 63 respectively. Their standard deviations were 6 and 9 respectively. The variance of the combined sample of size 90 is

Options

A. 90
B. 7
C. 9
D. 81

Answer

To find the variance of the combined sample, we can follow these steps:

### Step 1: Identify the given data
- Sample 1:
- Size (n1) = 40
- Mean (M1) = 54
- Standard Deviation (SD1) = 6
- Variance (σ1²) = SD1² = 6² = 36

- Sample 2:
- Size (n2) = 50
- Mean (M2) = 63
- Standard Deviation (SD2) = 9
- Variance (σ2²) = SD2² = 9² = 81

### Step 2: Calculate the sum of all sample units
- For Sample 1:
\[ \text{Sum of Sample 1} = M1 \times n1 = 54 \times 40 = 2160 \]

- For Sample 2:
\[ \text{Sum of Sample 2} = M2 \times n2 = 63 \times 50 = 3150 \]

### Step 3: Calculate the combined sum of all sample units
\[ \text{Combined Sum} = \text{Sum of Sample 1} + \text{Sum of Sample 2} = 2160 + 3150 = 5310 \]

### Step 4: Calculate the combined mean
\[ \text{Combined Mean} (M) = \frac{\text{Combined Sum}}{n1 + n2} = \frac{5310}{90} = 59 \]

### Step 5: Calculate the differences from the combined mean
- For Sample 1:
\[ D1 = M1 - M = 54 - 59 = -5 \]

- For Sample 2:
\[ D2 = M2 - M = 63 - 59 = 4 \]

### Step 6: Calculate the combined variance using the formula
The combined variance formula is:
\[ \sigma^2 = \frac{n1 \cdot \sigma_1^2 + n2 \cdot \sigma_2^2 + n1 \cdot D1^2 + n2 \cdot D2^2}{n1 + n2} \]

Substituting the values we have:
\[ \sigma^2 = \frac{40 \cdot 36 + 50 \cdot 81 + 40 \cdot (-5)^2 + 50 \cdot 4^2}{90} \]

Calculating each term:
- \(40 \cdot 36 = 1440\)
- \(50 \cdot 81 = 4050\)
- \(40 \cdot 25 = 1000\) (since \((-5)^2 = 25\))
- \(50 \cdot 16 = 800\) (since \(4^2 = 16\))

Now substituting these values back into the formula:
\[ \sigma^2 = \frac{1440 + 4050 + 1000 + 800}{90} = \frac{7290}{90} = 81 \]

### Final Answer
The variance of the combined sample is **81**.

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