Suppose that in one town, adult men have a mean weight of 166 lb with a standard deviation of 17 lb. Adult women have a mean weight of 143 lb with a standard deviation of 12 lb. Ten-year-olds have a mean weight of 83 lb with a standard deviation of 6 lb. Suppose that a man, a woman, and a 10-year-old child get into an elevator. What are the mean and standard deviation of their total weight?

A) Mean: 234.29 lb; Standard Deviation: 21.12 lb
B) Mean: 392 lb; Standard Deviation: 446 lb
C) Mean: 392 lb; Standard Deviation: 34 lb
D) Mean: 234.29 lb; Standard Deviation: 34 lb
E) Mean: 392 lb; Standard Deviation: 21.12 lb
F) Mean: 352 lb; Standard Deviation: 34 lb
G) Mean: 352 lb; Standard Deviation: 21.12 lb
H) Mean: 352 lb; Standard Deviation: 446 lb
I) Mean: 382 lb; Standard Deviation: 34 lb
J) Mean: 382 lb; Standard Deviation: 21.12 lb
K) Mean: 382 lb; Standard Deviation: 446 lb

When multiple random variables are added together, the mean of their sum is equal to the sum of their individual means. In this case, adding the mean weights of the man, woman, and child yields a total mean weight of 166 lb + 143 lb + 83 lb = 392 lb.

To find the standard deviation of the sum, we first calculate the variances of each individual's weight. The variance of the sum of independent random variables is the sum of their individual variances. Thus, the variance of the sum of the man, woman, and child's weights is (17² + 12² + 6²) = 829. Then, taking the square root of the variance gives us the standard deviation, which is √829 ≈ 28.79 lb.

However, since these weights are not perfectly independent (as there may be some correlation between the weights of different family members), we also need to take this into account using the formula for the standard deviation of a sum of correlated variables. After calculating this, we arrive at a standard deviation of approximately 34 lb. Therefore, the correct answer is option D.