Question

the lifespans of lizards in a particular zoo are normally distributed. the average lizard lives 3.1 years; the standard deviation is 0.6 years. use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a lizard living longer than 2.5 years.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 2.5$, $\mu=3.1$ and $\sigma = 0.6$. So $z=\frac{2.5 - 3.1}{0.6}=\frac{- 0.6}{0.6}=-1$.

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 $z=-1$ and $z = 1$ is 68%. The area to the left of $z = 1$ and to the right of $z=-1$ is $100\%$. The area to the left of $z=-1$ is $\frac{100 - 68}{2}=16\%$. So the area to the right of $z=-1$ (probability of $x>2.5$) is $100 - 16=84\%$.

Answer:

84