the mean mark was 22 and with a standard deviation of 6.25 and i scored 22, what portion of students had a lower mark than me

Given that the mean mark is 22 with a standard deviation of 6.25, and you scored exactly 22, we can use the concept of the **z-score** to determine what portion of students had a lower mark than you.

The **z-score** is calculated using the formula:

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

Where:
- \( X \) is your score (22).
- \( \mu \) is the mean (22).
- \( \sigma \) is the standard deviation (6.25).

Substituting the values:

\[
z = \frac{22 - 22}{6.25} = 0
\]

A z-score of 0 means that your score is exactly at the mean.

In a standard normal distribution, a z-score of 0 corresponds to the 50th percentile. This means that **50%** of the students scored lower than you.

### Summary:
- **50%** of the students had a lower mark than you.

Would you like more details on this, or do you have any other questions?

Here are 5 related questions you might find interesting:
1. How do you interpret z-scores that are negative?
2. What portion of students scored higher than me?
3. How would the portion of students change if I scored 28 instead of 22?
4. What does a standard deviation tell us about the distribution of marks?
5. How would the analysis change if the distribution of marks is not normal?

**Tip:** The z-score is a powerful tool for understanding how your score compares to others in a normal distribution!

Understand how to calculate the portion of students with lower marks than you using the z-score in statistics. Given a mean mark of 22 and a standard deviation of 6.25, and scoring exactly 22, find out how your score compares. Learn about normal distribution and the central limit theorem.

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