Understanding the Problem: Calculating Combined Statistics

This question asks us to find the average monthly wage and the variance of the distribution of wages for all workers when two groups (Firm A and Firm B) are combined. We are given the number of workers, the average wage, and the variance of wages for each firm separately.

Calculating the Combined Average Wage

To find the combined average wage, we need to consider the total wages paid by both firms and divide by the total number of workers. The total wage for a firm is the number of workers multiplied by the average wage.

• Number of workers in Firm A ($n_A$): 500
• Average monthly wage in Firm A ($\bar{x}_A$): Rs. 1860
• Total wage paid by Firm A: $n_A \times \bar{x}_A = 500 \times 1860 = 930000$ Rs.

• Number of workers in Firm B ($n_B$): 600
• Average monthly wage in Firm B ($\bar{x}_B$): Rs. 1750
• Total wage paid by Firm B: $n_B \times \bar{x}_B = 600 \times 1750 = 1050000$ Rs.

Total number of workers ($N = n_A + n_B$): $500 + 600 = 1100$

Total wage paid by both firms: $930000 + 1050000 = 1980000$ Rs.

The combined average wage ($\bar{x}_{AB}$) is:

$$ \bar{x}_{AB} = \frac{\text{Total wage paid}}{\text{Total number of workers}} = \frac{1980000}{1100} $$

$$ \bar{x}_{AB} = 1800 $$

The combined average monthly wage for all workers is Rs. 1800.

Calculating the Combined Variance of Wages

Calculating the combined variance requires considering both the variance within each group and the variance between the group means and the overall mean. The formula for combined variance ($\sigma_{AB}^2$) for two groups is:

$$ \sigma_{AB}^2 = \frac{n_A \sigma_A^2 + n_B \sigma_B^2 + n_A (\bar{x}_A - \bar{x}_{AB})^2 + n_B (\bar{x}_B - \bar{x}_{AB})^2}{n_A + n_B} $$

Where:

• $n_A, n_B$ are the number of workers in Firm A and Firm B
• $\sigma_A^2, \sigma_B^2$ are the variances of wages in Firm A and Firm B
• $\bar{x}_A, \bar{x}_B$ are the average wages in Firm A and Firm B
• $\bar{x}_{AB}$ is the combined average wage

We have the following values:

• $n_A = 500$
• $\sigma_A^2 = 81$
• $\bar{x}_A = 1860$
• $n_B = 600$
• $\sigma_B^2 = 100$
• $\bar{x}_B = 1750$
• $\bar{x}_{AB} = 1800$ (calculated above)

Now, let's calculate the terms:

• $n_A \sigma_A^2 = 500 \times 81 = 40500$
• $n_B \sigma_B^2 = 600 \times 100 = 60000$
• $(\bar{x}_A - \bar{x}_{AB}) = 1860 - 1800 = 60$
• $(\bar{x}_A - \bar{x}_{AB})^2 = 60^2 = 3600$
• $n_A (\bar{x}_A - \bar{x}_{AB})^2 = 500 \times 3600 = 1800000$
• $(\bar{x}_B - \bar{x}_{AB}) = 1750 - 1800 = -50$
• $(\bar{x}_B - \bar{x}_{AB})^2 = (-50)^2 = 2500$
• $n_B (\bar{x}_B - \bar{x}_{AB})^2 = 600 \times 2500 = 1500000$

Now, substitute these values into the combined variance formula:

$$ \sigma_{AB}^2 = \frac{40500 + 60000 + 1800000 + 1500000}{500 + 600} $$

$$ \sigma_{AB}^2 = \frac{100500 + 3300000}{1100} $$

$$ \sigma_{AB}^2 = \frac{3400500}{1100} $$

$$ \sigma_{AB}^2 = \frac{34005}{11} \approx 3091.36 $$

The combined variance of wages is approximately 3091.36.

Comparing Results with Options

We calculated the combined average monthly wage as Rs. 1800 and the combined variance as approximately 3091.36.

Let's look at the given options:

1. Rs. 1860, 100
2. Rs. 1750, 100
3. Rs. 1800, 81
4. None of the above

Our calculated average wage (Rs. 1800) matches the average in option 3, but our calculated variance (approx. 3091.36) does not match the variance in option 3 (81). None of the other options match our calculated values for both average and variance.

Therefore, the correct answer is "None of the above".

Revision Table: Key Formulas for Combined Statistics

When combining data from multiple groups, different formulas are used for calculating measures of central tendency and dispersion.

Additional Information: Why Combined Variance is Not Just the Average of Variances

It might seem intuitive that the combined variance is simply the average of the individual variances, perhaps weighted by the number of workers. However, this is generally incorrect. Variance measures the spread of data points around the mean. When combining groups, the data points not only spread around their own group mean but also around the overall combined mean. If the group means are different, this difference contributes to the overall spread (variance) of the combined data.

The term $n_A (\bar{x}_A - \bar{x}_{AB})^2 + n_B (\bar{x}_B - \bar{x}_{AB})^2$ in the numerator of the combined variance formula accounts for this "between-group" variability. This term is zero only if the group means are all equal to the combined mean. If the group means are different, this term is positive, increasing the combined variance compared to a simple weighted average of the individual variances.

For example, if Firm A had an average wage of 1000 and Firm B had an average wage of 5000, even if the wages within each firm had very small variance, combining them would result in a large overall variance because of the big difference in the average wages between the two firms.