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The problem involves finding the value of \( c \) such that the probability of \( Z \), a standard normal variable, falling between 1.18 and \( c \) is 0.0798. We are tasked with solving this:

\[
P(1.18 \leq Z \leq c) = 0.0798
\]

### Steps to Solve:
1. **Convert probabilities to Z-scores using the standard normal table:**
   - The probability corresponding to \( P(Z \leq c) \) can be split as:
     \[
     P(1.18 \leq Z \leq c) = P(Z \leq c) - P(Z \leq 1.18)
     \]
   - From the standard normal table, find the cumulative probability \( P(Z \leq 1.18) \).

2. **Solve for \( P(Z \leq c) \):**
   - Using the known values, determine \( P(Z \leq c) \).
   - Use the cumulative probability table or a calculator to find the Z-score \( c \).

I will now proceed to calculate the value of \( c \).The value of \( c \) that satisfies the equation \( P(1.18 \leq Z \leq c) = 0.0798 \) is approximately \( c = 1.76 \) when rounded to two decimal places.

Would you like any further details on the steps? 

Here are 5 related questions to deepen your understanding:
1. What is the meaning of the standard normal distribution in probability?
2. How do you interpret cumulative probabilities for a Z-score?
3. Can you explain why we subtract probabilities in this case?
4. What are other methods to calculate probabilities in the standard normal distribution?
5. How would the answer change if the given probability \( P(1.18 \leq Z \leq c) \) was different?

**Tip:** The cumulative distribution function (CDF) for the standard normal distribution helps find the area under the curve to the left of a given Z-score.

Suppose Z follows the standard normal distribution. Determine the value of c such that P(1.18 ≤ Z ≤ c) = 0.0798.

Learn how to calculate the value of c in a standard normal distribution such that P(1.18 ≤ Z ≤ c) equals 0.0798. This problem demonstrates the use of cumulative probabilities and the standard normal table. Suitable for advanced high school or college-level statistics students.

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