To find the probability that a gorilla will live longer than 23.9 years using the Empirical Rule, we first need to determine how many standard deviations away from the mean this lifespan is.
The mean lifespan \( \mu \) is 20.8 years, and the standard deviation \( \sigma \) is 3.1 years. We can calculate the z-score for 23.9 years using the formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
where \( X \) is the value for which we are finding the z-score. Plugging in the values, we get:
\[ z = \frac{(23.9 - 20.8)}{3.1} = \frac{3.1}{3.1} = 1 \]
A z-score of 1 indicates that 23.9 years is 1 standard deviation above the mean.
According to the Empirical Rule (or the 68-95-99.7 rule), we can summarize the distribution of a normal distribution as follows:
- Approximately 68% of the data falls within 1 standard deviation of the mean (from \( \mu - \sigma \) to \( \mu + \sigma \)).
- Approximately 95% falls within 2 standard deviations.
- Approximately 99.7% falls within 3 standard deviations.
Given a z-score of 1:
- The range from \( \mu - \sigma \) (17.7 years) to \( \mu + \sigma \) (23.9 years) contains 68% of the gorillas' lifespans. This means:
- 34% of gorillas live between the mean (20.8 years) and one standard deviation above the mean (23.9 years).
- 34% of gorillas live between the mean and one standard deviation below the mean (17.7 years).
This leaves 32% of the gorillas living outside of the range of one standard deviation from the mean (100% - 68%).
Since we are interested in the probability that a gorilla lives longer than 23.9 years, we look at the right tail of the distribution:
\[ P(X > 23.9) = \frac{32%}{2} = 16% \]
Thus, the probability that a gorilla will live longer than 23.9 years is 0.16, or 16%.