To solve these problems, we'll use the properties of the normal distribution, which is defined by its mean (μ) and standard deviation (σ). Given:

• Mean (μ) = 12
• Standard Deviation (σ) = 4

We'll calculate the Z-scores for the specific values and then use the standard normal distribution table (or a calculator) to find the corresponding probabilities.

(1) Percentage of cases between 8 and 16:

To find the percentage of cases between 8 and 16, we'll calculate the Z-scores for both 8 and 16.

• Z-score for 8:
  $Z = \frac{X - \mu}{\sigma} = \frac{8 - 12}{4} = \frac{-4}{4} = -1$

• Z-score for 16:
  $Z = \frac{X - \mu}{\sigma} = \frac{16 - 12}{4} = \frac{4}{4} = 1$

Now, we look up the Z-scores in the standard normal distribution table:

• The probability corresponding to $Z = -1$ is approximately 0.1587.
• The probability corresponding to $Z = 1$ is approximately 0.8413.

The percentage of cases between 8 and 16 is the difference between these probabilities: $0.8413 - 0.1587 = 0.6826$ So, 68.26% of the cases fall between 8 and 16.

(2) Percentage of cases above 18:

First, we calculate the Z-score for 18:

• Z-score for 18:
  $Z = \frac{X - \mu}{\sigma} = \frac{18 - 12}{4} = \frac{6}{4} = 1.5$

The probability corresponding to $Z = 1.5$ is approximately 0.9332.

The percentage of cases above 18 is: $1 - 0.9332 = 0.0668$ So, 6.68% of the cases lie above the score of 18.

(3) Percentage of scores below 6:

Finally, we calculate the Z-score for 6:

• Z-score for 6:
  $Z = \frac{X - \mu}{\sigma} = \frac{6 - 12}{4} = \frac{-6}{4} = -1.5$

The probability corresponding to $Z = -1.5$ is approximately 0.0668.

So, 6.68% of the scores lie below the score of 6.

Summary:

1. 68.26% of cases fall between 8 and 16.
2. 6.68% of cases lie above the score 18.
3. 6.68% of scores lie below the score 6.

Would you like more details on any of these calculations or have any other questions?

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Here are five related questions for further exploration:

1. What percentage of scores lie between 10 and 14?
2. How would the percentages change if the standard deviation were 5 instead of 4?
3. What score corresponds to the 90th percentile in this distribution?
4. What is the Z-score for a score of 20?
5. How do you interpret the Z-score in a real-world context?

Tip: Understanding how to read and interpret Z-scores is crucial for working with normal distributions, as they allow you to standardize any normal distribution to compare different datasets.