Question

the lifespans of meerkats in a particular zoo are normally distributed. the average meerkat lives 13.1 years; the standard deviation is 1.5 years. use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a meerkat living longer than 14.6 years.

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 14.6$, $\mu=13.1$, and $\sigma = 1.5$. So $z=\frac{14.6 - 13.1}{1.5}=\frac{1.5}{1.5}=1$.

Step2: Recall the empirical rule

The empirical rule for a normal distribution states that about 68% of the data lies within 1 standard - deviation of the mean, i.e., $P(\mu-\sigma

Step3: Calculate the probability

The probability of a value being greater than $\mu+\sigma$ is $\frac{1 - 0.68}{2}=\frac{0.32}{2}=0.16$. To convert this to a percentage, we multiply by 100, so the probability is $16\%$.

Answer:

$16$