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From the image, you are working on a question that involves calculating probabilities using the standard normal distribution. Let's walk through each part of the question.

You already got parts **a**, **b**, and **c** correct:

- **a. \( P(z < 0) = 0.500 \)** is correct.
- **b. \( P(z < -2.06) = 0.980 \)** is correct.
- **c. \( P(-0.96 < z < 1.31) = 0.736 \)** is correct.

However, part **d** appears to be incorrect. Let's correct it:

**d. \( P(z > 1.59) \)**

To calculate \( P(z > 1.59) \), we need to first find the cumulative probability for \( z = 1.59 \) using the standard normal distribution table or a calculator.

- The cumulative probability for \( z = 1.59 \) is approximately **0.9441**.
- Since we want \( P(z > 1.59) \), we need to subtract this from 1:

\[
P(z > 1.59) = 1 - 0.9441 = 0.0559
\]

So, the correct answer for part **d** is:

\[
P(z > 1.59) \approx 0.056
\]

Would you like any further details on how this was calculated or clarification on the standard normal distribution? 

Here are five related questions you might find useful:
1. How do you find probabilities for negative \(z\)-scores in a normal distribution?
2. What is the total area under the standard normal curve?
3. How do you calculate \(P(a < z < b)\) for different \(z\)-values?
4. How can you use a calculator or statistical software to compute normal probabilities?
5. What are some applications of the standard normal distribution in real-world problems?

**Tip:** Always remember that the total area under the normal distribution curve equals 1, so for \(P(z > x)\), you subtract the cumulative probability from 1.

Evaluate the following probabilities based on the standard normal distribution:

a. P(z < 0) 
b. P(z < -2.06) 
c. P(-0.96 < z < 1.31) 
d. P(z > 1.59)

Round all answers to at least 3 decimal places.

Learn how to evaluate probabilities using the standard normal distribution and z-scores. This example involves calculating cumulative probabilities and using the complement rule for P(z > x), with a step-by-step solution for each part. Suitable for students in Grades 9-12.

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