Using a normal curve table, give the percentage of scores between the mean and Z score of (a) 0.53, (b) 0.63, (c) 1.37, (d) 1.47, (e) -0.53.  (A) find the percentage of scores between the mean and a Z score of 0.53

To find the percentage of scores between the mean and a Z-score of 0.53 using the standard normal distribution, we can look up the Z-score in a standard normal table or use statistical software.

For Z = 0.53, the corresponding value from the Z-table (representing the cumulative area from the left of the curve up to Z = 0.53) is approximately **0.7019**.

Since we are looking for the percentage of scores between the mean (Z = 0) and Z = 0.53:

1. The area under the normal curve from the mean (Z = 0) to the right of the curve equals 50% (0.5000).
2. The percentage between the mean and Z = 0.53 is the difference between 0.7019 and 0.5000.

Thus, the percentage of scores between the mean and Z = 0.53 is:

\[
0.7019 - 0.5000 = 0.2019
\]

Therefore, the percentage of scores between the mean and Z = 0.53 is **20.19%**.

Would you like further details or explanations?

Here are 5 related questions:
1. What is the cumulative probability for a Z-score of 1.00?
2. How do you calculate probabilities for negative Z-scores?
3. What percentage lies beyond a Z-score of 2.00?
4. How do we find the probability for Z-scores between two given values?
5. Can we use Z-tables for distributions that are not normal?

**Tip:** Always subtract 0.5000 from the Z-table value when calculating percentages from the mean to any positive Z-score.

Using a normal curve table, give the percentage of scores between the mean and Z score of (a) 0.53, (b) 0.63, (c) 1.37, (d) 1.47, (e) -0.53.

Calculate the percentage of scores between the mean and given Z-scores (e.g., 0.53, 0.63, 1.37, 1.47, -0.53) using a standard normal distribution table. This problem explores key statistical concepts such as Z-scores and the normal distribution, suitable for students in Grades 10-12.

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