Question

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard

Answer 1

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.

$z=\frac{x-\mu}{\sigma}$, where,

z = z-score,

x = Random sample score,

$\mu$ = Mean

$\sigma$ = Standard deviation.

$z=\frac{10.1-12.5}{2.4}$

$z=\frac{-2.4}{2.4}$

$z=-1$

Since z-score of 10.1 is $-1$. 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 $-1$, we will subtract half of 68% from 50%.

$\frac{68\%}{2}=34\%$

$50\%-34\%=16\%$

Therefore, the probability of a lion living longer than 10.1 years is approximately 16%.