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.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
z = z-score,
x = Random sample score,
[tex]\mu[/tex] = Mean
[tex]\sigma[/tex] = Standard deviation.
[tex]z=\frac{10.1-12.5}{2.4}[/tex]
[tex]z=\frac{-2.4}{2.4}[/tex]
[tex]z=-1[/tex]
Since z-score of 10.1 is [tex]-1[/tex]. 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 [tex]-1[/tex], we will subtract half of 68% from 50%.
[tex]\frac{68\%}{2}=34\%[/tex]
[tex]50\%-34\%=16\%[/tex]
Therefore, the probability of a lion living longer than 10.1 years is approximately 16%.
