1. The problem asks to find the value of $k$ such that the area to the right of $k$ under the standard normal curve is $P(z > k) = 0.0096$. 2. This means we want to find the $z$-score $k$ where the right-tail probability is 0.0096. 3. Recall that the total area under the standard normal curve is 1, so the area to the left of $k$ is $P(z \leq k) = 1 - 0.0096 = 0.9904$. 4. We use the standard normal distribution table or a calculator to find the $z$-score corresponding to a cumulative probability of 0.9904. 5. Looking up 0.9904 in the z-table or using an inverse normal function, we find: $$k \approx 2.33$$ 6. Therefore, the value of $k$ such that $P(z > k) = 0.0096$ is approximately $2.33$. This means that about 0.96% of the data lies above $z = 2.33$ in a standard normal distribution.