The average score on the Stats midterm was 7272 points with a standard deviation of 55 ​points, and FranzFranz​'s ​z-score was 22. How many points did he​ score?

The \( z \)-score formula is used to relate an individual score to the mean and standard deviation of the data set. The formula is:

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

Where:
- \( z \) is the \( z \)-score,
- \( x \) is the individual score (Franz's score in this case),
- \( \mu \) is the mean score,
- \( \sigma \) is the standard deviation.

In this problem, we are given:
- \( z = 22 \),
- \( \mu = 7272 \) (mean score),
- \( \sigma = 55 \) (standard deviation).

We need to solve for \( x \), Franz's score:

\[
22 = \frac{x - 7272}{55}
\]

To solve for \( x \), multiply both sides by 55:

\[
22 \times 55 = x - 7272
\]

\[
1210 = x - 7272
\]

Now, add 7272 to both sides:

\[
x = 1210 + 7272
\]

\[
x = 8482
\]

Therefore, Franz scored 8482 points.

Would you like further details or have any questions?

Here are 5 related questions for further practice:
1. What would Franz's score be if his \( z \)-score were 15 instead of 22?
2. How would the \( z \)-score change if the mean was 7300 but the score remained 8482?
3. If the standard deviation was 60 points, how would this affect Franz's score for the same \( z \)-score?
4. What is the meaning of a high \( z \)-score like 22 in the context of this exam?
5. How would you calculate the percentage of students who scored below Franz using the normal distribution?

**Tip**: A \( z \)-score indicates how many standard deviations a data point is from the mean.

The average score on the Stats midterm was 7272 points with a standard deviation of 55 points, and Franz's z-score was 22. How many points did he score?

Learn how to calculate Franz's score using a z-score of 22, a mean of 7272 points, and a standard deviation of 55. This problem demonstrates the use of the z-score formula in statistics. Suitable for high school students in Grades 10-12.

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