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### Information Given:
- Mean blood pressure \( \mu = 130 \) mmHg
- Standard deviation \( \sigma = 22 \) mmHg
- Blood pressure is normally distributed.

### Solutions:

1. **State the random variable:**
   The random variable here represents the **blood pressure of a person in the country**.
   
   Answer: \( \text{The blood pressure of a person in the country.} \)

2. **Find \( P(X \geq 119) \):**
   - Convert 119 to a z-score:
     \[
     z = \frac{119 - 130}{22} = \frac{-11}{22} = -0.5
     \]
   - Use the standard normal distribution table to find \( P(Z \geq -0.5) \).
     - \( P(Z \geq -0.5) = 1 - P(Z \leq -0.5) \approx 1 - 0.3085 = 0.6915 \).

   Answer: \( P(X \geq 119) \approx 0.6915 \).

3. **Find \( P(X \leq 141) \):**
   - Convert 141 to a z-score:
     \[
     z = \frac{141 - 130}{22} = \frac{11}{22} = 0.5
     \]
   - Use the standard normal distribution table to find \( P(Z \leq 0.5) \).
     - \( P(Z \leq 0.5) \approx 0.6915 \).

   Answer: \( P(X \leq 141) \approx 0.6915 \).

4. **Find \( P(133 \leq X \leq 135) \):**
   - Convert 133 and 135 to z-scores:
     \[
     z_{133} = \frac{133 - 130}{22} = \frac{3}{22} \approx 0.1364
     \]
     \[
     z_{135} = \frac{135 - 130}{22} = \frac{5}{22} \approx 0.2273
     \]
   - Use the standard normal distribution table to find probabilities:
     - \( P(Z \leq 0.2273) \approx 0.5910 \)
     - \( P(Z \leq 0.1364) \approx 0.5548 \)
   - Find the probability:
     \[
     P(133 \leq X \leq 135) = 0.5910 - 0.5548 = 0.0362
     \]

   Answer: \( P(133 \leq X \leq 135) \approx 0.0362 \).

5. **Is it unusual for a person to have a blood pressure of at least 169 mmHg?**
   - Convert 169 to a z-score:
     \[
     z = \frac{169 - 130}{22} = \frac{39}{22} \approx 1.7727
     \]
   - Use the standard normal distribution table to find \( P(Z \geq 1.7727) \).
     - \( P(Z \geq 1.7727) \approx 0.0384 \), which is about 3.84%.
   - Since this probability is less than 5%, it is unusual.

   Answer: Yes, because the probability is less than 5%.

6. **What blood pressure do 90% of all people have less than?**
   - We need to find the blood pressure corresponding to the 90th percentile.
   - The z-score for the 90th percentile is approximately 1.28.
   - Use the z-score formula:
     \[
     X = \mu + z \cdot \sigma = 130 + 1.28 \cdot 22 \approx 130 + 28.16 = 158.16
     \]

   Answer: Approximately 158.2 mmHg.

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Would you like further details on any of these solutions or have any questions?

### Related Questions:
1. How would the probabilities change if the standard deviation were different?
2. What is the probability that a person has a blood pressure between 125 and 135 mmHg?
3. How can we interpret the 90th percentile in practical terms?
4. What blood pressure do the top 5% of people have more than?
5. Why is the z-score used in calculating probabilities for normal distributions?

### Tip:
When working with normal distributions, converting values to z-scores is essential for finding probabilities and percentiles, as it standardizes the distribution regardless of mean or standard deviation.

Suppose the mean blood pressure for people in a country is 130 mmHg with a standard deviation of 22 mmHg. Assume that blood pressure is normally distributed.

State the random variable.
- What is the probability that a person in the country has blood pressure of 119 mmHg or more?
- What is the probability that a person in the country has blood pressure of 141 mmHg or less?
- What is the probability that a person in the country has blood pressure between 133 and 135 mmHg?
- Is it unusual for a person in the country to have a blood pressure of at least 169 mmHg?
- What blood pressure do 90% of all people in the country have less than?

Learn to calculate probabilities for different blood pressure levels in a country where the mean is 130 mmHg and standard deviation is 22 mmHg. This example includes z-score calculations and explains why a blood pressure of at least 169 mmHg might be considered unusual.

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