Question

the lifespans of gorillas in a particular zoo are normally distributed. the average gorilla lives 20.8 years; the standard deviation is 3.1 years. use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a gorilla living longer than 23.9 years.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 23.9$, $\mu=20.8$ and $\sigma = 3.1$. So $z=\frac{23.9 - 20.8}{3.1}=\frac{3.1}{3.1}=1$.

Step2: Apply the empirical rule

The empirical rule states that for a normal distribution, about 68% of the data lies within 1 standard - deviation of the mean ($\mu\pm\sigma$), about 95% lies within 2 standard - deviations ($\mu\pm2\sigma$) and about 99.7% lies within 3 standard - deviations ($\mu\pm3\sigma$). The area within 1 standard - deviation of the mean is 68%. So the area outside of $\mu\pm1\sigma$ is $100 - 68=32\%$. Since the normal distribution is symmetric, the area above $\mu + 1\sigma$ is $\frac{32}{2}=16\%$.

Answer:

16