To find the value of $c$ such that $P(Z > c) = 0.6915$ where $Z$ follows the standard normal distribution, follow these steps:

1. Recognize that $P(Z > c) = 0.6915$. This is the probability that the standard normal variable is greater than $c$.

2. To use the standard normal distribution table (which usually gives the cumulative probability from the left, $P(Z < c)$), convert the given probability:

   $P(Z < c) = 1 - P(Z > c) = 1 - 0.6915 = 0.3085$

3. Now, look for the value of $c$ where $P(Z < c) = 0.3085$ in the standard normal distribution table.

4. Checking the table or using a calculator, $P(Z < c) = 0.3085$ corresponds to $c \approx -0.50$.

Thus, the value of $c$ rounded to two decimal places is: $\boxed{-0.50}$

Would you like further details or have any other questions?

Here are five related questions:

1. How do you find a critical value for a two-tailed test using the standard normal distribution?
2. What does the value $P(Z > c)$ represent in terms of the standard normal curve?
3. How do you convert between left-tail and right-tail probabilities?
4. What is the significance of the mean and standard deviation in a standard normal distribution?
5. How would you find $c$ if $P(Z < c)$ were given instead?

Tip: Always remember that standard normal tables typically provide left-tail probabilities (i.e., $P(Z < c)$). For right-tail probabilities, use the complement: $P(Z > c) = 1 - P(Z < c)$.