Question

The mean of two samples of series 50 and 100 are 53.7 and 52.3 respectively. Find the mean of combine sample.

• A. 50.65
• B. 52.766
• C. 52.0
• D. 53.05

Hint:

To perform combined mean calculations quickly without a calculator:
- Simplify the sample sizes into ratios:
\[ n_1 : n_2 = 50 : 100 = 1 : 2 \] - Calculate using the simplified weights:
\[ \bar{x}_{12} = \frac{1(53.7) + 2(52.3)}{1 + 2} = \frac{53.7 + 104.6}{3} = \frac{158.3}{3} \approx 52.767 \] This simplified method avoids working with large numbers.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When an overall average must be calculated from two or more distinct groups with known sizes and averages, we use the combined arithmetic mean.
The combined mean accounts for the different sizes (weights) of each sample group.

Step 2: Key Formula or Approach:
The formula for the combined arithmetic mean (\(\bar{x}_{12}\)) of two samples is:
\[ \bar{x}_{12} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \] Where:
- \(n_1, n_2\) are the sizes of the two samples.
- \(\bar{x}_1, \bar{x}_2\) are the means of the two samples.

Step 3: Detailed Explanation:
Let us extract the values from the problem and perform the calculation:
1. For the first sample:
- Size \(n_1 = 50\)
- Mean \(\bar{x}_1 = 53.7\)
2. For the second sample:
- Size \(n_2 = 100\)
- Mean \(\bar{x}_2 = 52.3\)
3. Calculate the sum of the observations for the first sample:
\[ n_1\bar{x}_1 = 50 \times 53.7 = 2685 \] 4. Calculate the sum of the observations for the second sample:
\[ n_2\bar{x}_2 = 100 \times 52.3 = 5230 \] 5. Calculate the total number of observations in the combined sample:
\[ n_1 + n_2 = 50 + 100 = 150 \] 6. Substitute these values into the combined mean formula:
\[ \bar{x}_{12} = \frac{2685 + 5230}{150} = \frac{7915}{150} \] 7. Performing the division:
\[ \bar{x}_{12} = 52.7666\dots \approx 52.766 \] Therefore, the mean of the combined sample is approximately 52.766.

Step 4: Final Answer:
The combined mean of the two sample series is 52.766, matching Option (B).