Find the value of z such that 0.5762
 of the area lies between −z
 and z. Round your answer to two decimal places.

To find the value of \( z \) such that 0.5762 of the area lies between \(-z\) and \(z\), we need to work with the standard normal distribution.

1. **Step 1: Understanding the area requirement**  
   The area between \(-z\) and \(z\) is 0.5762, which represents the central area under the standard normal curve. This leaves a combined area of \( 1 - 0.5762 = 0.4238 \) in the two tails (the areas outside \(-z\) and \(z\)).

2. **Step 2: Divide the tail area**  
   Since the normal distribution is symmetric, half of the remaining area lies in each tail.  
   Thus, the area in each tail is \( \frac{0.4238}{2} = 0.2119 \).

3. **Step 3: Find the z-score**  
   To find \( z \), we need the z-score that corresponds to a cumulative area of \( 1 - 0.2119 = 0.7881 \) (the area to the left of \(z\)).

   Using a standard normal distribution table or a calculator:
   \[
   z \approx 0.80
   \]

Thus, the value of \( z \) is approximately **0.80**.

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Would you like more details or have any questions?

Here are 5 related questions:
1. How do you interpret areas under the standard normal curve?
2. What is the meaning of a z-score in a normal distribution?
3. How do you find the cumulative area using a z-table?
4. Can you calculate the area to the right of a given z-score?
5. How would the value of \( z \) change if a different area was specified?

**Tip:** A z-score represents how many standard deviations a point is from the mean in a normal distribution.

Find the value of z such that 0.5762 of the area lies between −z and z. Round your answer to two decimal places.

Learn how to find the z-score such that 0.5762 of the area lies between −z and z in a standard normal distribution. This detailed solution involves understanding z-scores, calculating tail areas, and using cumulative area tables. Ideal for students in Grades 10-12.

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