Find the indicated area under the standard normal curve.
To the left of equals negative 0.88 and to the right of z equals0.88

To solve this, we need to calculate two areas under the standard normal curve:

1. **The area to the left of \( z = -0.88 \)**
2. **The area to the right of \( z = 0.88 \)**

The total area under the standard normal curve equals 1 (or 100%).

### 1. Area to the left of \( z = -0.88 \)
From standard normal distribution tables or using a calculator, the cumulative probability to the left of \( z = -0.88 \) is approximately 0.1894.

### 2. Area to the right of \( z = 0.88 \)
The cumulative probability to the left of \( z = 0.88 \) is 0.8106. So, the area to the right of \( z = 0.88 \) is:

\[
1 - 0.8106 = 0.1894
\]

### Total Area
Since the areas to the left of \( z = -0.88 \) and to the right of \( z = 0.88 \) are equal, we can calculate the total area by summing these two areas:

\[
\text{Total Area} = 0.1894 + 0.1894 = 0.3788
\]

Thus, the total area under the standard normal curve to the left of \( z = -0.88 \) and to the right of \( z = 0.88 \) is approximately **0.3788**.

Would you like more details or have any questions about the steps?

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Here are five related questions you might find helpful:
1. What is the area between two z-scores, like between \( z = -1 \) and \( z = 1 \)?
2. How do you calculate the area under the normal curve for any given z-value?
3. How does the symmetry of the normal distribution simplify these types of problems?
4. Can you find the area for z-values beyond 3 standard deviations?
5. How do z-scores relate to probabilities in a normal distribution?

**Tip:** Using a z-table or calculator is a quick way to find cumulative areas under the normal curve.

Find the indicated area under the standard normal curve. To the left of z = -0.88 and to the right of z = 0.88.

Learn how to find the area under the standard normal curve to the left of z = -0.88 and to the right of z = 0.88. This problem covers key concepts like Z-scores, normal distribution, and probability. Suitable for students in Grades 11-12 and college level.

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