The sum of the original 8 numbers is 8 × 15 = 120. After removing 12 and 18, the sum of the remaining 6 numbers is 120 - 12 - 18 = 90. The new mean is 90 ÷ 6 = 15.0. Choice B incorrectly assumes the removed numbers were above average, lowering the mean. Choice C incorrectly assumes the removed numbers were below average, raising the mean. Choice D makes an arithmetic error in the subtraction or division.

When each value in a data set is multiplied by the same constant, the mean is also multiplied by that constant. Increasing each value by 20% means multiplying by 1.20. Therefore, the new mean is 50 × 1.20 = 60. Choice A incorrectly assumes the mean stays the same. Choice C adds 20 instead of multiplying by 1.20. Choice D adds 10% instead of 20% to the mean.

The sum of ages for the original 6 people is 6 × 25 = 150 years. With 7 people and a mean of 26, the total sum becomes 7 × 26 = 182 years. The age of the 7th person is 182 - 150 = 32 years. Choice A assumes the 7th person's age equals the new mean plus the difference (26 + 4). Choice B adds an extra year due to miscounting. Choice D uses the original mean plus 4 years.

To find the weighted mean, calculate the total hours for each group and combine them. Group 1: 12 × 38 = 456 hours. Group 2: 8 × 44 = 352 hours. Total hours = 456 + 352 = 808 hours for 20 employees. Overall mean = 808 ÷ 20 = 40.4 hours. Choice B incorrectly averages 38 and 44 without considering group sizes. Choice C weights the groups incorrectly, favoring the smaller group. Choice D makes an arithmetic error in the final division.