the lifespans of lions in a particular zoo are normally distributed. the average lion lives 10 years; the…

Explanation:

Step1: Calculate the number of standard - deviations

The mean $\mu = 10$ years and the standard deviation $\sigma=1.4$ years. We want to find the number of standard - deviations for $x = 7.2$ years. The formula for the z - score is $z=\frac{x-\mu}{\sigma}$. So, $z=\frac{7.2 - 10}{1.4}=\frac{- 2.8}{1.4}=-2$.

Step2: Apply the empirical rule

The empirical rule for a normal distribution states that about 95% of the data lies within 2 standard deviations of the mean, i.e., $P(\mu - 2\sigma<X<\mu + 2\sigma)\approx95%$. The total area under the normal curve is 100%. The area outside of $\mu\pm2\sigma$ is $100 - 95=5%$. This 5% is split evenly between the two tails, so the area in the left - tail (less than $\mu - 2\sigma$) is $\frac{5%}{2}=2.5%$.

Step3: Find the probability of $X>7.2$

The probability of a lion living longer than 7.2 years is $P(X > 7.2)=1 - P(X\leq7.2)$. Since $P(X\leq7.2)$ is the area in the left - tail which is 2.5%, then $P(X > 7.2)=100%-2.5% = 97.5%$.

Answer:

97.5