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The problem asks us to find the value of \( c \) such that:

\[
P(-c \leq Z \leq c) = 0.9729
\]

where \( Z \) follows a standard normal distribution (mean 0, standard deviation 1). This is essentially asking for the symmetric range around 0 where the total probability is 0.9729.

### Step-by-Step Solution:

1. **Understanding the Probability**: 
   The probability \( P(-c \leq Z \leq c) = 0.9729 \) implies that the probability outside this range is \( 1 - 0.9729 = 0.0271 \). Since the normal distribution is symmetric, half of this remaining probability (i.e., 0.01355) is in each tail.

   Therefore, the probability that \( Z \) is greater than \( c \) or less than \( -c \) is 0.01355 in each tail.

2. **Using the Z-Table or Inverse Standard Normal**:
   We need to find the Z-score that corresponds to a cumulative probability of \( 1 - 0.01355 = 0.98645 \) (this is the cumulative probability to the left of \( c \)).

   Using a standard normal Z-table or a calculator:
   - For \( P(Z \leq c) = 0.98645 \), we look up the corresponding Z-score.
   - From the table, \( Z = 2.17 \) corresponds to a cumulative probability of 0.9864.

3. **Conclusion**:
   Thus, \( c = 2.17 \).

Therefore, the value of \( c \) is approximately **2.17**.

Would you like further details on any part of this solution or have questions? Here are five related questions you might find helpful:

1. What is the Z-score and how is it used in normal distribution?
2. How do we interpret the probability in standard normal distribution problems?
3. What are the key characteristics of the standard normal distribution?
4. How do cumulative probability tables work in finding Z-scores?
5. How can I calculate the Z-score without using a table?

**Tip**: In standard normal problems, knowing how to use Z-tables or the inverse normal function on calculators is essential for finding specific probabilities.

Suppose Z follows the standard normal distribution. Find the value of c such that P(-c ≤ Z ≤ c) = 0.9729.

Learn how to find the value of c such that P(-c ≤ Z ≤ c) = 0.9729 in a standard normal distribution. This problem involves concepts of Z-scores and symmetry in probability distributions. Step-by-step explanation provided.

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