$\sum_{x=0}^{16} 6(x ; 19,0.7)-\sum_{x=0}^{15} 6(x ; 19,0.7)$

Step 1 ：The problem is about the sum of a function \(6(x ; 19,0.7)\) from \(x=0\) to \(x=16\) minus the sum of the same function from \(x=0\) to \(x=15\).

Step 2 ：We assume that the function is a binomial distribution function, then the function would be \(6(x ; 19,0.7) = 6 \cdot \binom{19}{x} \cdot (0.7)^x \cdot (0.3)^{19-x}\), where \(\binom{19}{x}\) is the binomial coefficient.

Step 3 ：The difference between the sum from \(x=0\) to \(x=16\) and the sum from \(x=0\) to \(x=15\) would be the value of the function at \(x=16\), i.e., \(6(16 ; 19,0.7)\).

Step 4 ：Substituting the values \(x = 16\), \(n = 19\), and \(p = 0.7\) into the function, we get the result as 0.5216838974954823.

Step 5 ：Final Answer: \(\boxed{0.5216838974954823}\)