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 between 11.5 and 27 years.

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 20.8$ (mean), $\sigma=3.1$ (standard deviation).
For $x = 11.5$, $z_1=\frac{11.5 - 20.8}{3.1}=\frac{-9.3}{3.1}=- 3$.
For $x = 27$, $z_2=\frac{27 - 20.8}{3.1}=\frac{6.2}{3.1}=2$.

Step2: Apply the empirical rule

The empirical rule states that about 68% of the data lies within 1 standard - deviation of the mean ($z=\pm1$), about 95% lies within 2 standard - deviations ($z = \pm2$), and about 99.7% lies within 3 standard - deviations ($z=\pm3$).
The area between $z=-3$ and $z = 2$ is the area between $z=-3$ and $z = 3$ minus the area between $z = 2$ and $z = 3$.
The area between $z=-3$ and $z = 3$ is 99.7%. The area between $z=-2$ and $z = 2$ is 95%, so the area between $z = 2$ and $z = 3$ and $z=-3$ and $z=-2$ is $\frac{99.7 - 95}{2}=2.35\%$ each.
The area between $z=-3$ and $z = 2$ is $99.7\%-2.35\%=97.35\%$.

Answer:

97.35%