To solve for the value of $c$ such that $P(-c \leq Z \leq c) = 0.9500$ where $Z$ follows the standard normal distribution, we follow these steps:

1. Understand the Problem:

   • The probability $P(-c \leq Z \leq c) = 0.9500$ implies that the middle 95% of the standard normal distribution lies between $-c$ and $c$.
   • Since the standard normal distribution is symmetric about 0, the probability to the left of $-c$ and to the right of $c$ is equal, each accounting for $(1 - 0.9500)/2 = 0.0250$.

2. Use the Standard Normal Distribution:

   • We need to find the z-value $c$ such that the cumulative probability to the left of $c$ is $0.9750$ (since 0.9500 + 0.0250 = 0.9750).

Let's calculate the value of $c$.

The value of $c$ such that $P(-c \leq Z \leq c) = 0.9500$ is $c = 1.96$ (rounded to two decimal places).

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. What is the value of $c$ for $P(-c \leq Z \leq c) = 0.99$?
2. How would you interpret the value $c = 1.96$ in a confidence interval context?
3. What is $P(Z \leq -1.96)$ in the standard normal distribution?
4. How does $c$ change if the probability $P(-c \leq Z \leq c)$ decreases to 0.90?
5. How do you use $c$ to find critical values for a two-tailed test?

Tip: The value $c = 1.96$ is often used in statistics for constructing 95% confidence intervals, a common scenario in hypothesis testing.