1. The problem asks to find the value of $k$ such that the area to the right of $k$ under the standard normal distribution curve is $P(z>k) = 0.9991$. 2. This means we want to find $k$ where the right-tail probability is 0.9991. Since the total area under the curve is 1, the left-tail area is $P(z \leq k) = 1 - 0.9991 = 0.0009$. 3. We use the standard normal distribution table or a calculator to find the $z$-score corresponding to a cumulative probability of 0.0009. 4. Looking up 0.0009 in the cumulative distribution function (CDF) table for the standard normal distribution, or using an inverse normal function, we find $k \approx -3.12$. 5. Therefore, the value of $k$ such that $P(z>k) = 0.9991$ is $\boxed{-3.12}$. This matches the option -3.12.