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Let's break down and solve the problem based on the image you have provided. The problem deals with a normal distribution of heights where:

- \( X \sim N(165, 15) \), meaning the mean height (\( \mu \)) is 165 cm, and the standard deviation (\( \sigma \)) is 15 cm.
- The questions involve calculating probabilities related to this normal distribution.

### a) What is the \( P(X \geq 195) \)?

This is asking for the probability that a randomly selected person's height is greater than or equal to 195 cm.

- First, calculate the z-score for \( X = 195 \):
  \[
  z = \frac{X - \mu}{\sigma} = \frac{195 - 165}{15} = 2
  \]
- Using a z-table or a calculator, the probability to the left of \( z = 2 \) (i.e., \( P(X \leq 195) \)) is approximately 0.9772.
- Therefore, \( P(X \geq 195) \) is the complement of this:
  \[
  P(X \geq 195) = 1 - P(X \leq 195) = 1 - 0.9772 = 0.0228
  \]
So, the probability that a person is taller than or equal to 195 cm is approximately 0.0228, or 2.28%.

### b) What is \( P(X \leq 195) \)?

This was already calculated in part a:

\[
P(X \leq 195) \approx 0.9772
\]

### c) What is \( P(X = 165) \)?

Since this is a continuous distribution, the probability of any specific value (like exactly 165 cm) is 0. So:

\[
P(X = 165) = 0
\]

### d) What is \( P(135 \leq X \leq 195) \)?

To find this probability, we need the area under the normal curve between 135 and 195 cm.

- First, calculate the z-score for \( X = 135 \):
  \[
  z = \frac{135 - 165}{15} = -2
  \]
- From the z-table, \( P(X \leq 135) \) (i.e., the area to the left of \( z = -2 \)) is approximately 0.0228.
- From part b, \( P(X \leq 195) \approx 0.9772 \).
- Therefore, \( P(135 \leq X \leq 195) = P(X \leq 195) - P(X \leq 135) \):
  \[
  P(135 \leq X \leq 195) = 0.9772 - 0.0228 = 0.9544
  \]
So, the probability that a person's height is between 135 cm and 195 cm is approximately 0.9544, or 95.44%.

---

Let me know if you'd like further details on any part of the solution!

### Five related questions:
1. How is the 68-95-99.7 rule applied to approximate probabilities in a normal distribution?
2. What does the z-score represent in a normal distribution, and how is it used?
3. How can the normal distribution be used to model real-world data such as heights?
4. What is the probability that a randomly selected person is shorter than 120 cm, given the same distribution?
5. How would the probabilities change if the mean and standard deviation were different?

**Tip:** Always check if a distribution is continuous or discrete. For continuous distributions, the probability of a specific point is always zero.

Let X be the height of a randomly selected person in cm, where X ~ N(165, 15). For the following questions, draw the distribution and mark the area under the curve that represents your answer. Then calculate the answer.
a) What is the P(X ≥ 195)?
b) What is the P(X ≤ 195)?
c) What is the P(X = 165)?
d) What is P(135 ≤ X ≤ 195)?

Learn how to calculate probabilities using a normal distribution with mean 165 cm and standard deviation 15 cm. Find the probability for P(X ≥ 195), P(X ≤ 195), and P(135 ≤ X ≤ 195) using z-scores and the 68-95-99.7 rule. Perfect for high school students studying probability.

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