Let's solve each part of the problem using the standard normal distribution.

(a) $P(Z > 1.04)$

1. Find the cumulative probability $P(Z \leq 1.04)$ using the standard normal distribution table or calculator.
2. Subtract this value from 1 to get $P(Z > 1.04)$.

From the standard normal table:

• $P(Z \leq 1.04) \approx 0.8508$

Therefore:

• $P(Z > 1.04) = 1 - 0.8508 = 0.1492$

(b) $P(Z \leq -0.86)$

1. Find the cumulative probability $P(Z \leq -0.86)$.

From the standard normal table:

• $P(Z \leq -0.86) \approx 0.1949$

(c) $P(1.04 < Z < 2.16)$

1. Find the cumulative probabilities $P(Z \leq 2.16)$ and $P(Z \leq 1.04)$.
2. Subtract the smaller cumulative probability from the larger one.

From the standard normal table:

• $P(Z \leq 2.16) \approx 0.9846$
• $P(Z \leq 1.04) \approx 0.8508$

Thus:

• $P(1.04 < Z < 2.16) = 0.9846 - 0.8508 = 0.1338$

Final Answers:

• (a) $P(Z > 1.04) = 0.149$
• (b) $P(Z \leq -0.86) = 0.195$
• (c) $P(1.04 < Z < 2.16) = 0.134$

Would you like any further explanations or have additional questions?

---

Relative Questions:

1. What are the steps to calculate $P(Z > a)$ for any value $a$?
2. How can you determine $P(Z < a)$ directly using a calculator?
3. Why is the subtraction method used for calculating $P(1.04 < Z < 2.16)$?
4. What do the values of $Z$ signify in real-world applications?
5. How does symmetry of the standard normal distribution help in finding probabilities?
6. Can you explain the importance of the standard deviation in these calculations?
7. How would these probabilities change if the mean were not zero?
8. What might cause a shift in the standard normal distribution curve?

Tip: When dealing with two-tailed probabilities (like in part c), it's essential to calculate the cumulative probabilities separately and subtract the lower from the higher to ensure accuracy.