Question

the lifespans of lions in a particular zoo are normally distributed. the average lion lives 10 years; the standard deviation is 1.4 years. use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a lion living longer than 7.2 years.

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 7.2$, $\mu=10$, and $\sigma = 1.4$. So $z=\frac{7.2 - 10}{1.4}=\frac{- 2.8}{1.4}=-2$.

Step2: Apply the empirical rule

The empirical rule for a normal distribution states that about 95% of the data lies within $z=-2$ and $z = 2$. So the percentage of data outside of $z=-2$ and $z = 2$ is $100\%-95\% = 5\%$. Since the normal distribution is symmetric, the percentage of data less than $z=-2$ is $\frac{5\%}{2}=2.5\%$.

Step3: Find the probability of a lion living longer than 7.2 years

The probability of a lion living longer than 7.2 years (i.e., $z>-2$) is $100\%-2.5\% = 97.5\%$.

Answer:

97.5%