Question

The lifespans of lions in a particular zoo
are normally distributed. The average lion
lives 10 years, the standard deviation is
1.4 years
Use the empirical rule
(68 - 95 - 99.7%) to estimate the
probability of a lion living less than 7.2
years.

Answer 1

Answer:

From the empirical rule we know that we have within one deviation from the mean 68% of the values, within two deviations 95% and within 3 deviations 99.7%. We want to find the following probability:

$P( X<7.2)$

We can use the z score formula given by:

$z = \frac{X-\mu}{\sigma}$

And replacing we got:

$z = \frac{7.2-10}{1.4}= -2$

So we want to find the probability that the data lies below 2 deviations from the mean and using the empirical rule we got:

$(1-0.95)/2 = 0.025$

Step-by-step explanation:

For this problem we know that the average lion lives with a mean of $\mu =10 years$ and the deviation is $\sigma =1.4$

From the empirical rule we know that we have within one deviation from the mean 68% of the values, within two deviations 95% and within 3 deviations 99.7%. We want to find the following probability:

$P( X<7.2)$

We can use the z score formula given by:

$z = \frac{X-\mu}{\sigma}$

And replacing we got:

$z = \frac{7.2-10}{1.4}= -2$

So we want to find the probability that the data lies below 2 deviations from the mean and using the empirical rule we got:

$(1-0.95)/2 = 0.025$