We have been given that the lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. We are asked to find the probability of a lion living longer than 10.1 years using empirical rule.
First of all, we will find the z-score corresponding to sample score 10.1.
, where,
z = z-score,
x = Random sample score,
= Mean
= Standard deviation.
![z=\frac{10.1-12.5}{2.4}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B10.1-12.5%7D%7B2.4%7D)
![z=\frac{-2.4}{2.4}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B-2.4%7D%7B2.4%7D)
![z=-1](https://tex.z-dn.net/?f=z%3D-1)
Since z-score of 10.1 is
. Now we need to find area under curve that is below one standard deviation from mean.
We know that approximately 68% of data points lie between one standard deviation from mean.
We also know that 50% of data points are above mean and 50% of data points are below mean.
To find the probability of a data point with z-score
, we will subtract half of 68% from 50%.
![\frac{68\%}{2}=34\%](https://tex.z-dn.net/?f=%5Cfrac%7B68%5C%25%7D%7B2%7D%3D34%5C%25)
![50\%-34\%=16\%](https://tex.z-dn.net/?f=50%5C%25-34%5C%25%3D16%5C%25)
Therefore, the probability of a lion living longer than 10.1 years is approximately 16%.