Yes, your answer of 84% is correct.

Here's how we can use the Empirical Rule to verify this:

1. Mean (μ): 13.1 years
2. Standard Deviation (σ): 1.5 years
3. We need to find the probability of a meerkat living less than 14.6 years.

Applying the Empirical Rule:

The empirical rule states that:

• About 68% of data falls within 1 standard deviation (μ ± σ).
• About 95% falls within 2 standard deviations (μ ± 2σ).
• About 99.7% falls within 3 standard deviations (μ ± 3σ).

Calculating the Z-score:

To find how many standard deviations 14.6 is from the mean:

$Z = \frac{14.6 - 13.1}{1.5} = \frac{1.5}{1.5} = 1$

This means that 14.6 years is 1 standard deviation above the mean.

Using the Empirical Rule for 1 Standard Deviation:

• The empirical rule tells us that about 68% of the data lies within 1 standard deviation (13.1 ± 1.5 years).
• This leaves 32% of the data outside this range, divided equally between the lower and upper tails.
• Thus, 16% of the data lies above 14.6 years (1 standard deviation above the mean).
• Consequently, the remaining data, or 84%, lies below 14.6 years.

So, your estimation of 84% probability for a meerkat living less than 14.6 years is indeed correct.

Would you like more details or have any other questions? Here are a few related questions to explore:

1. What is the probability of a meerkat living more than 14.6 years?
2. How do you find probabilities for values not exactly at 1, 2, or 3 standard deviations?
3. What is the probability of a meerkat living between 11.6 and 14.6 years?
4. How does the empirical rule differ from using the standard normal table?
5. What if the standard deviation was different—how would it change the probabilities?

Tip: When using the empirical rule, it helps to draw a normal distribution curve to visualize the mean and standard deviations, making it easier to see how the percentages are distributed.