Suppose X∼N(9,1.5), and x=13.5. Find and interpret the z-score of the standardized normal random variable.

Provide your answer below:

The  $z$ -score when  $x=13.5$ is . The mean is .

This $z$ -score tells you that $x=13.5$  is  standard deviations to the **right** of the mean.

To find the $ z $-score, we use the formula:

\[
z = \frac{x - \mu}{\sigma}
\]

where:
- $ x = 13.5 $
- $ \mu = 9 $ (mean)
- $ \sigma = 1.5 $ (standard deviation)

Substitute these values into the formula:

\[
z = \frac{13.5 - 9}{1.5} = \frac{4.5}{1.5} = 3
\]

So, the $ z $-score is 3.

### Interpretation
The $ z $-score of 3 means that $ x = 13.5 $ is 3 standard deviations to the **right** of the mean.

### Final Answer
The $ z $-score when $ x = 13.5 $ is **3**. The mean is **9**. This $ z $-score tells you that $ x = 13.5 $ is **3** standard deviations to the **right** of the mean.

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Do you want more details or have any questions?

Here are 5 related questions:
1. What is the probability of $ X $ being greater than 13.5 in this distribution?
2. How would the $ z $-score change if the standard deviation were increased to 2?
3. What does a negative $ z $-score indicate in terms of position relative to the mean?
4. How would you interpret a $ z $-score of 0?
5. If the mean were 12, what would the $ z $-score be for \( x = 13

Suppose X∼N(9,1.5), and x=13.5. Find and interpret the z-score of the standardized normal random variable.

Learn how to calculate the z-score for a normal distribution where X∼N(9,1.5) and x=13.5. This problem involves understanding normal distribution, the z-score formula, and interpreting the result. Suitable for students in Grades 10-12.

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