According to the table, we have the following list of 3 + 1 + 3 + 1 + 3 + 1 + 3 = 15 numbers with the mean of 4:

{1, 1, 1, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 7, 7} (so, three 1's, one 2, three 3's, ...).

We need to find the probability that \(|4 - x| > \frac{3}{2}\) (the absolute value of the difference between the mean and x is |4 - x|).

\(|4 - x| > \frac{3}{2}\);

\(4 - x > \frac{3}{2}\) or \(4 - x< -\frac{3}{2}\);

\(x>5.5\) or \(x<2.5\)

Thus, the question simply asks to find the probability that a randomly chosen x from {1, 1, 1, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 7, 7} will be more than 5.5 (6, or 7) or be less than 2.5 (2 or 1).

\(P(x = 1, 2, 6, \ or \ 7) = \frac{3+1+1+3}{15}=\frac{8}{15}\)

Answer: A.

Hope it's clear.