For a standard normal distribution, find: P(z<-2.31 or z > -1.09)

Round your answer to at least three decimal places.


To solve \( P(z < -2.31 \text{ or } z > -1.09) \) for a standard normal distribution, we can break it down into two parts:

1. \( P(z < -2.31) \)
2. \( P(z > -1.09) \)

### Step 1: Find \( P(z < -2.31) \)
Using the standard normal distribution table or a calculator, we can find the cumulative probability for \( z = -2.31 \):

\[
P(z < -2.31) \approx 0.0104
\]

### Step 2: Find \( P(z > -1.09) \)
For \( P(z > -1.09) \), we first find the cumulative probability for \( z = -1.09 \):

\[
P(z < -1.09) \approx 0.1379
\]

Since \( P(z > -1.09) = 1 - P(z < -1.09) \):

\[
P(z > -1.09) = 1 - 0.1379 = 0.8621
\]

### Step 3: Combine the two parts
Now, we combine the two probabilities:

\[
P(z < -2.31 \text{ or } z > -1.09) = P(z < -2.31) + P(z > -1.09)
\]

\[
P(z < -2.31 \text{ or } z > -1.09) = 0.0104 + 0.8621 = 0.8725
\]

### Final Answer:
\[
P(z < -2.31 \text{ or } z > -1.09) \approx 0.873
\]

Would you like more details, or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you interpret a standard normal distribution curve?
2. What is the significance of the z-score in probability?
3. How do you calculate the cumulative probability for any z-score?
4. What is the probability for the complement event \( P(z > 2.31 \text{ and } z < -1.09) \)?
5. How does the empirical rule (68-95-99.7) relate to this problem?

**Tip:** You can use z-tables, a calculator, or statistical software to find probabilities for any z-score in a standard normal distribution.

For a standard normal distribution, find: P(z<-2.31 or z > -1.09). Round your answer to at least three decimal places.

Learn how to calculate the probability P(z < -2.31 or z > -1.09) for a standard normal distribution. This detailed solution explains key concepts such as z-scores, the complement rule, and provides step-by-step instructions to find the cumulative probabilities. Suitable for high school students in Grades 10-12.

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