To complete the table, we will use the properties of the normal distribution. The z-score formula is:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the exam score
• $\mu = 72$ is the mean
• $\sigma = 7$ is the standard deviation

To determine percentiles, we will reference a Z-table.

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Step-by-step:

1. Calculate missing Z-scores:

We solve for $Z$ using the z-score formula for scores that don't have a corresponding $Z$.

2. Calculate missing exam scores:

Rearranging the formula to: $X = Z \cdot \sigma + \mu$

3. Match Z-scores with percentiles from a Z-table:

Using a Z-table or equivalent resources.

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Fill in the blanks in the table:

Exam Score	Z-score	Percentile (%)
54.5	−2.5	0.6
66.75	−0.75	22.7
88.1	2.3	98.9
74.2	0.31	62.1

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Explanation for each missing value:

1. Percentile for 66.75 (Z = −0.75):

Using a Z-table:

• $Z = -0.75$ corresponds to a percentile of 22.7%.

2. Exam score for Z = 2.3:

Using $X = Z \cdot \sigma + \mu$: $X = 2.3 \cdot 7 + 72 = 88.1$

3. Percentile for Z = 2.3:

From the Z-table:

• $Z = 2.3$ corresponds to a percentile of 98.9%.

4. Percentile for Z = 0.31:

From the Z-table:

• $Z = 0.31$ corresponds to a percentile of 62.1%.

5. Exam score for Z = −0.75:

Already given as 66.75.

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Would you like further clarification or additional examples?

5 Related Questions:

1. What is the process of converting between raw scores and Z-scores?
2. How does the normal distribution apply to real-world data analysis?
3. Why does the Z-score determine the percentile?
4. How would the results differ if the standard deviation changed to 5?
5. What does the area under the curve represent in the Z-table?

Tip:

Always ensure the Z-table corresponds to the type of distribution (left-tailed, right-tailed, or cumulative) to avoid errors.