Question

the lifespans of tigers in a particular zoo are normally distributed. the average tiger lives 22.4 years; the standard deviation is 2.7 years. use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a tiger living less than 14.3 years. % show calculator

Explanation:

Step1: Calculate number of standard - deviations

First, find how many standard - deviations 14.3 is from the mean. Let $\mu = 22.4$ be the mean and $\sigma=2.7$ be the standard deviation. Calculate the z - score $z=\frac{x - \mu}{\sigma}$, where $x = 14.3$. So $z=\frac{14.3 - 22.4}{2.7}=\frac{- 8.1}{2.7}=-3$.

Step2: Apply empirical rule

The empirical rule for a normal distribution states that about 99.7% of the data lies within $z=-3$ and $z = 3$. The total area under the normal curve is 100%. The area outside of $z=-3$ and $z = 3$ is $100 - 99.7=0.3\%$. Since the normal distribution is symmetric, the area to the left of $z=-3$ is $\frac{0.3}{2}=0.15\%$.

Answer:

0.15