To find the lowest score someone can get to be in the top 30% of the distribution, we need to determine the score that corresponds to the 70th percentile in a normal distribution (since the top 30% begins above the lower 70%).

Key Data:

• Mean ($\mu$) = 25
• Standard deviation ($\sigma$) = 4
• Cumulative probability ($P(X \leq x)$) = 0.70 (for the 70th percentile)

The formula for a $z$-score in a normal distribution is:

$z = \frac{x - \mu}{\sigma}$

1. Find the z-score for the 70th percentile:
   Using a standard normal distribution table or a calculator, the $z$-score corresponding to $P(X \leq x) = 0.70$ is approximately:

   $z = 0.524$

2. Convert the z-score to the raw score ($x$):
   Rearrange the $z$-score formula to find $x$:

   $x = \mu + z \cdot \sigma$

   Substituting the values:

   $x = 25 + (0.524 \cdot 4)$

   $x = 25 + 2.096 = 27.096$

3. Interpret the result:
   The lowest score required to earn a certificate is approximately 27.10 (rounding to two decimal places).

---

Do you want to see a detailed explanation of z-scores or the derivation?

Here are some follow-up questions for deeper understanding:

1. What is the probability of scoring above a given $z$-score in a normal distribution?
2. How does the mean and standard deviation affect the shape of the distribution?
3. What does a $z$-score represent in practical terms?
4. Can we calculate the proportion of scores between two values in this distribution?
5. What is the difference between cumulative probability and percentile?

**Tip