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Let's break down and solve each part of this question related to probabilities for standard and non-standard normal distributions.

### a. \( P(z > -1.11) \)
This is the probability that the standard normal variable \( z \) is greater than \(-1.11\).

To find this:
1. First, look up the cumulative probability for \( z = -1.11 \) in the standard normal distribution table, which gives \( P(Z < -1.11) \).
2. Use the fact that \( P(Z > -1.11) = 1 - P(Z < -1.11) \).

From the table, \( P(Z < -1.11) \approx 0.1335 \).

Thus, \( P(Z > -1.11) = 1 - 0.1335 = 0.8665 \).

### b. \( P(z < 0.43) \)
This is the probability that the standard normal variable \( z \) is less than 0.43.

1. Look up the cumulative probability for \( z = 0.43 \).

From the table, \( P(Z < 0.43) \approx 0.6664 \).

Thus, \( P(Z < 0.43) = 0.6664 \).

### c. \( P(-1.35 < z < 1.23) \)
This is the probability that the standard normal variable \( z \) lies between \(-1.35\) and \( 1.23 \).

1. Look up the cumulative probabilities for \( z = -1.35 \) and \( z = 1.23 \).
   - \( P(Z < -1.35) \approx 0.0885 \)
   - \( P(Z < 1.23) \approx 0.8907 \)

2. Subtract: 
   \[
   P(-1.35 < z < 1.23) = P(Z < 1.23) - P(Z < -1.35) = 0.8907 - 0.0885 = 0.8022
   \]

### d. \( P(38 < x < 83) \) where \( \mu = 60 \) and \( \sigma = 12 \)
This requires converting the non-standard normal variable \( x \) into the standard normal variable \( z \).

1. First, convert \( x = 38 \) and \( x = 83 \) into their corresponding \( z \)-scores:
   - For \( x = 38 \), \( z = \frac{38 - 60}{12} = \frac{-22}{12} \approx -1.83 \)
   - For \( x = 83 \), \( z = \frac{83 - 60}{12} = \frac{23}{12} \approx 1.92 \)

2. Look up the cumulative probabilities:
   - \( P(Z < -1.83) \approx 0.0336 \)
   - \( P(Z < 1.92) \approx 0.9726 \)

3. Subtract:
   \[
   P(38 < x < 83) = P(Z < 1.92) - P(Z < -1.83) = 0.9726 - 0.0336 = 0.9390
   \]

### Final answers:
- a. \( P(z > -1.11) = 0.8665 \)
- b. \( P(z < 0.43) = 0.6664 \)
- c. \( P(-1.35 < z < 1.23) = 0.8022 \)
- d. \( P(38 < x < 83) = 0.9390 \)

Would you like further details or explanations on any part?

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Here are five questions related to the content:
1. How do you interpret a \( z \)-score in terms of standard deviation?
2. What is the relationship between the cumulative probability and the standard normal distribution curve?
3. How can you use symmetry properties of the normal distribution to simplify probability calculations?
4. Why is it necessary to convert non-standard normal variables to standard \( z \)-scores?
5. What is the significance of the total area under the standard normal curve?

**Tip:** Always remember that the area under the standard normal distribution curve represents probability and sums to 1!

Find probabilities for the standard and non-standard normal distribution.

Learn how to find probabilities for both standard and non-standard normal distributions, including converting to Z-scores and using the cumulative probability. Includes solutions for P(Z > -1.11), P(Z < 0.43), P(-1.35 < Z < 1.23), and P(38 < X < 83). Suitable for high school students.

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