Find the percentage of all values in a normal distribution for the following​ z-score.
z greater than or equals 1.52z≥1.52

To find the percentage of all values in a normal distribution for the given z-score (where \( z \geq 1.52 \)), we'll follow these steps:

### Step 1: Find the cumulative probability for \( z = 1.52 \)
Using the standard normal distribution table or a calculator, we find the cumulative probability (area to the left) for \( z = 1.52 \).

The cumulative probability \( P(Z < 1.52) \) is approximately **0.9357**.

### Step 2: Calculate the area to the right
Since we want the area to the right of \( z = 1.52 \), we subtract the cumulative probability from 1:

\[
P(Z \geq 1.52) = 1 - P(Z < 1.52) = 1 - 0.9357 = 0.0643
\]

### Step 3: Convert to a percentage
To convert this to a percentage, multiply by 100:

\[
P(Z \geq 1.52) = 0.0643 \times 100 = 6.43\%
\]

### Final Answer:
Approximately **6.43%** of the values in a normal distribution have a z-score greater than or equal to 1.52.

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Would you like further details on how the standard normal distribution table works, or do you have any other questions?

Here are 5 related questions you might find interesting:

1. What is the percentage for a z-score less than or equal to 1.52?
2. How would you find the percentage for \( z \geq 2.33 \)?
3. Can you calculate the percentage for a z-score between two values, such as \( 1.52 \) and \( 2.33 \)?
4. What is the area under the curve for the entire normal distribution?
5. How is the z-score related to standard deviation in a normal distribution?

**Tip:** When working with normal distributions, remember that the total area under the curve is always 1 (or 100%). This means you can use the cumulative probabilities and symmetry properties to find areas in any section of the distribution.

Learn how to calculate the percentage of values in a normal distribution with z-score greater than or equal to 1.52. Understand cumulative probability and conversion to percentages with step-by-step guidance. Suitable for advanced high school students and anyone interested in statistics.

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