To solve the problem, we are given that $Z$ follows the standard normal distribution, and we are asked to find the value of $c$ such that:

$P(-1.24 \leq Z \leq c) = 0.8746$

Step 1: Find $P(Z \leq -1.24)$

Using a standard normal distribution table or a calculator, we first find the probability for $Z = -1.24$. From the standard normal table:

$P(Z \leq -1.24) = 0.1075$

Step 2: Set up the equation

We are told that the probability between $Z = -1.24$ and $Z = c$ is 0.8746. This can be written as:

$P(-1.24 \leq Z \leq c) = P(Z \leq c) - P(Z \leq -1.24)$

Substitute the known probability for $Z \leq -1.24$:

$0.8746 = P(Z \leq c) - 0.1075$

Step 3: Solve for $P(Z \leq c)$

Add 0.1075 to both sides:

$P(Z \leq c) = 0.8746 + 0.1075 = 0.9821$

Step 4: Find $c$

Now, we look up the value of $c$ such that $P(Z \leq c) = 0.9821$ using the standard normal distribution table or a calculator.

From the table, $Z = 2.08$ corresponds to $P(Z \leq 2.08) = 0.9821$.

Thus, $c = 2.08$.

Final Answer:

The value of $c$ is approximately 2.08.

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Would you like more details or have any questions? Here are some related questions you might find helpful:

1. How do you use the standard normal table to find probabilities?
2. What is the relationship between cumulative probability and the Z-score?
3. How does changing the limits of integration affect the cumulative probability in a normal distribution?
4. What is the significance of the value of $c$ in this problem?
5. How do you calculate probabilities for non-standard normal distributions?

Tip: When working with the standard normal distribution, always confirm whether you are working with left-tailed or right-tailed probabilities to avoid errors.