Let $X$ be the number of seeds that germinate. We want to find $P(X \ge 116)$. We can use the normal approximation to the binomial distribution to find this probability.The mean of the binomial distribution is $np = 165(0.7) = 115.5$, and the standard deviation is $\sqrt{np(1-p)} = \sqrt{165(0.7)(0.3)} = 8.4$.We use the continuity correction and find the probability that a normal random variable with mean 115.5 and standard deviation 8.4 is greater than or equal to 115.5.Using a standard normal table or a calculator, we find that the probability that a standard normal random variable is greater than or equal to 0 is 0.5. Therefore, the probability that a normal random variable with mean 115.5 and standard deviation 8.4 is greater than or equal to 115.5 is approximately 0.5.Thus, the probability that at least 116 of the 165 seeds will germinate is approximately 0.5.The final answer is $\boxed{0.500}$.