Question

the lifespans of meerkats in a particular zoo are normally distributed. the average meerkat lives 10.4 years; the standard deviation is 1.9 years. use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a meerkat living longer than 16.1 years.

Explanation:

Step1: Calculate number of standard - deviations

First, find out how many standard deviations 16.1 is away from the mean. Let $\mu = 10.4$ (mean) and $\sigma=1.9$ (standard deviation). Calculate $z=\frac{x - \mu}{\sigma}$, where $x = 16.1$. So $z=\frac{16.1 - 10.4}{1.9}=\frac{5.7}{1.9}=3$.

Step2: Apply the 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., $P(\mu - 3\sigma

Step3: Calculate the probability

$\frac{1 - 0.997}{2}=\frac{0.003}{2}=0.0015$. To convert to a percentage, multiply by 100. So the probability as a percentage is $0.15\%$.

Answer:

$0.15$