Question

the lifespans of lizards in a particular zoo are normally distributed. the average lizard lives 3.1 years; the standard deviation is 0.6 years. use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a lizard living between 2.5 and 4.3 years.

Explanation:

Step1: Calculate number of standard - deviations from the mean

The mean $\mu = 3.1$ years and the standard deviation $\sigma=0.6$ years.
For $x_1 = 2.5$ years, the z - score $z_1=\frac{x_1-\mu}{\sigma}=\frac{2.5 - 3.1}{0.6}=\frac{- 0.6}{0.6}=-1$.
For $x_2 = 4.3$ years, the z - score $z_2=\frac{x_2-\mu}{\sigma}=\frac{4.3 - 3.1}{0.6}=\frac{1.2}{0.6}=2$.

Step2: Apply the empirical rule

The empirical rule for a normal distribution states that:

• Approximately 68% of the data lies within 1 standard deviation of the mean ($z=\pm1$), i.e., $\mu\pm\sigma$.
• Approximately 95% of the data lies within 2 standard deviations of the mean ($z = \pm2$), i.e., $\mu\pm2\sigma$.
• Approximately 99.7% of the data lies within 3 standard deviations of the mean ($z=\pm3$), i.e., $\mu\pm3\sigma$.

The probability of data between $z=-1$ and $z = 2$:
The probability between $z=-1$ and $z = 1$ is 68%. The probability between $z = 1$ and $z = 2$ is half of the difference between the probability of data within 2 standard - deviations and the probability of data within 1 standard - deviation.
The probability of data within 2 standard - deviations is 95%, and within 1 standard - deviation is 68%. The difference is $95 - 68=27\%$, and half of it is 13.5%.
So the probability between $z=-1$ and $z = 2$ is $68\%+13.5\% = 81.5\%$.

Answer:

81.5