To solve these problems, we use the properties of the standard normal distribution, where $Z$ follows a normal distribution with mean $\mu = 0$ and standard deviation $\sigma = 1$. We can find the probabilities using a standard normal table or a calculator. Below are the answers:

(a) $P(Z > 200)$

• Since $Z = 200$ is far beyond the typical range of a standard normal distribution, the probability that $Z$ is greater than 200 is extremely small, essentially zero.

  $P(Z > 200) \approx 0$

(b) $P(Z \leq 1.44)$

• To find this, we use the cumulative distribution function (CDF) for the standard normal distribution at $Z = 1.44$.

  $P(Z \leq 1.44) \approx 0.925$

(c) $P(-1.09 < Z < 220)$

• First, $P(Z < 220)$ is extremely close to 1, so we focus on the range from $-1.09$ to $220$.

  $P(-1.09 < Z < 220) \approx P(Z > -1.09) \approx 1 - P(Z \leq -1.09)$ Using the CDF for $Z = -1.09$: $P(Z \leq -1.09) \approx 0.1379$ So, $P(-1.09 < Z < 220) \approx 1 - 0.1379 = 0.8621$

Let me know if you want details on these calculations or have any questions!

---

Here are 8 related questions you might want to explore:

1. What is $P(Z > 1.96)$?
2. How do you calculate $P(Z < -2.33)$?
3. What is the 95th percentile of the standard normal distribution?
4. How do you interpret the z-score in real-world terms?
5. What is $P(-0.5 < Z < 0.5)$?
6. How do you find $P(Z > 0.84)$?
7. How does the standard normal distribution differ from other normal distributions?
8. How do you convert a normal variable to a standard normal variable?

Tip: The cumulative distribution function (CDF) is crucial for finding probabilities in a normal distribution, giving the area under the curve to the left of a specific z-score.