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The problem asks to evaluate \( P(Z > 2.33) \), where \( Z \) follows a standard normal distribution. This represents the probability that a standard normal variable is greater than 2.33. 

Using a standard normal distribution table or a calculator:

1. Find the cumulative probability \( P(Z < 2.33) \), which corresponds to the area under the curve to the left of \( Z = 2.33 \). From the standard normal distribution table, \( P(Z < 2.33) = 0.9901 \).

2. To find \( P(Z > 2.33) \), subtract this value from 1:
\[
P(Z > 2.33) = 1 - 0.9901 = 0.0099
\]

Thus, \( P(Z > 2.33) = 0.0099 \), rounded to four decimal places.

Would you like more details or explanations on this? Here are 5 related questions for further practice:

1. What is \( P(Z > 1.96) \)?
2. What is \( P(Z < 1.64) \)?
3. How do you interpret \( P(Z > 0) \)?
4. How do you calculate \( P(a < Z < b) \) for any two values \( a \) and \( b \)?
5. What does the value of \( P(Z > 2.33) \) tell you about the likelihood of such an event?

**Tip:** When using the normal distribution table, always remember to subtract from 1 when dealing with "greater than" probabilities.

Use tech to evaluate: P(Z > 2.33). Round your answer to FOUR decimal places.

Learn how to evaluate the probability P(Z > 2.33) using a standard normal distribution table or calculator. This problem demonstrates key statistical concepts like Z-scores and cumulative probability, suitable for students in Grades 11-12.

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