$n_1 = 50, \bar{z}_1 = 630, \sigma_1 = 90$ and $n_2 = 40, \bar{z}_2 = 540, \sigma_2 = 60$
Step-by-Step Solution
Key Concept: When combining two distributions, deviations from the combined mean are calculated as $d_i = \bar{z}_i - \bar{x}$ for each group's original mean.
For two groups, the combined mean is $\bar{x} = \frac{n_1\bar{z}_1 + n_2\bar{z}_2}{n_1 + n_2} = \frac{50(630) + 40(540)}{90} = \frac{3150 + 2160}{90} = \frac{5310}{90} = 590 - 40 = 550$. Wait, recalculating: $\bar{x} = \frac{50 \times 630 + 40 \times 540}{50 + 40} = \frac{31500 + 21600}{90} = \frac{53100}{90} = 590$. Now, $d_1 = \bar{z}_1 - \bar{x} = 630 - 590 = 40$ and $d_2 = \bar{z}_2 - \bar{x} = 540 - 590 = -50$. The answer is $d_2 = -50$, but the question asks for $d_1 - d_2 = 40 - (-50) = 90$. However, based on the shown working, $d_2 = -50$ appears to be requested, giving $-50$.
Correct Answer: -40