Question

the lifespans of zebras in a particular zoo are normally distributed. the average zebra lives 20.5 years; the standard deviation is 3.9 years. use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a zebra living less than 32.2 years. % show calculator

Explanation:

Step1: Calculate number of standard - deviations

First, find how many standard - deviations 32.2 is from the mean. Let $\mu = 20.5$ (mean) and $\sigma=3.9$ (standard deviation). Calculate $z=\frac{x - \mu}{\sigma}$, where $x = 32.2$. So $z=\frac{32.2-20.5}{3.9}=\frac{11.7}{3.9}=3$.

Step2: Apply empirical rule

The empirical rule for a normal distribution states that about 99.7% of the data lies within 3 standard - deviations of the mean, i.e., between $\mu - 3\sigma$ and $\mu+3\sigma$. The area to the left of $\mu + 3\sigma$ is $0.5+\frac{0.997}{2}$. Since the total area under the normal curve is 1 and the distribution is symmetric about the mean. $0.5+\frac{0.997}{2}=0.5 + 0.4985=0.9985$.

Answer:

99.85