Find the value of the standard normal random variable z, called z0 such that:
(a) P(z≤z0)=0.8472 z0 =
(b) P(−z0≤z≤z0)=0.0484 z0 =
(c) P(−z0≤z≤z0)=0.095 z0 =
(d) P(z≥z0)=0.1402 z0 =
(e) P(−z0≤z≤0)=0.0434 z0 =
(f) P(−2.01≤z≤z0)=0.6121 z0 =

To solve these problems, we will use properties of the **standard normal distribution** \( Z \), which has a mean \( \mu = 0 \) and a standard deviation \( \sigma = 1 \). 

We will find the critical value \( z_0 \) by using a **standard normal table** (Z-table) or the **inverse cumulative distribution function** of the standard normal distribution (sometimes written as \( \Phi^{-1}(p) \) where \( \Phi(z_0) \) is the cumulative probability up to \( z_0 \)).

Let's go through each part:

### (a) \( P(z \leq z_0) = 0.8472 \)
Here, you are looking for the value of \( z_0 \) where the cumulative probability is 0.8472. This is the inverse cumulative distribution function for the standard normal distribution.

From the Z-table or using \( \Phi^{-1}(0.8472) \), the value of \( z_0 \) is approximately:
\[
z_0 \approx 1.02
\]

### (b) \( P(-z_0 \leq z \leq z_0) = 0.0484 \)
This is the probability that the random variable lies between \( -z_0 \) and \( z_0 \). First, we need to find the total probability outside this range, which is \( 1 - 0.0484 = 0.9516 \). Since the distribution is symmetric, the area in each tail is \( \frac{1 - 0.0484}{2} = 0.4758 \).

From the Z-table or using \( \Phi^{-1}(0.4758) \), the value of \( z_0 \) is approximately:
\[
z_0 \approx 0.06
\]

### (c) \( P(-z_0 \leq z \leq z_0) = 0.095 \)
Again, the total area outside this range is \( 1 - 0.095 = 0.905 \). The area in each tail is \( \frac{1 - 0.095}{2} = 0.4525 \).

From the Z-table or using \( \Phi^{-1}(0.4525) \), the value of \( z_0 \) is approximately:
\[
z_0 \approx 0.11
\]

### (d) \( P(z \geq z_0) = 0.1402 \)
This is the upper tail probability. The cumulative probability up to \( z_0 \) is \( 1 - 0.1402 = 0.8598 \).

From the Z-table or using \( \Phi^{-1}(0.8598) \), the value of \( z_0 \) is approximately:
\[
z_0 \approx 1.08
\]

### (e) \( P(-z_0 \leq z \leq 0) = 0.0434 \)
This represents the area between \( -z_0 \) and 0. Since the distribution is symmetric, the cumulative area up to \( z_0 \) would be \( 0.5 - 0.0434 = 0.4566 \).

From the Z-table or using \( \Phi^{-1}(0.4566) \), the value of \( z_0 \) is approximately:
\[
z_0 \approx 0.11
\]

### (f) \( P(-2.01 \leq z \leq z_0) = 0.6121 \)
First, find the probability that corresponds to \( z = -2.01 \). From the Z-table, \( P(z \leq -2.01) \approx 0.0222 \).

Now, the probability up to \( z_0 \) is \( 0.0222 + 0.6121 = 0.6343 \).

From the Z-table or using \( \Phi^{-1}(0.6343) \), the value of \( z_0 \) is approximately:
\[
z_0 \approx 0.35
\]

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Let me know if you need any more details on the process! Here are 5 related questions to explore:

1. How is the standard normal distribution used in hypothesis testing?
2. What is the significance of the symmetry of the standard normal distribution?
3. How do you use the Z-table to find probabilities for non-standard normal distributions?
4. How can you transform a normal distribution to a standard normal one?
5. What is the relationship between z-scores and percentiles?

**Tip:** The area under the standard normal curve is always 1, and it helps to break down problems using the symmetry of the distribution when finding probabilities.

Solve problems related to finding the z-score (z0) for various probabilities in the standard normal distribution, including cumulative and tail probabilities. Learn how to use Z-tables or inverse CDF functions to find these values. Suitable for advanced high school or college-level statistics students.

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