Question

a) draw the model for auto fuel economy. clearly label it, showing what the 68 - 95 - 99.7 rule predicts.
b) in what interval would you expect the central 95% of autos to be found?
using the 68 - 95 - 99.7 rule, the central 95% of autos can be expected to be found in the interval from 12.37 to 37.25 mpg.
(type integers or decimals. do not round. use ascending order.)
c) about what percent of autos should get more than 31.03 mpg?
using the 68 - 95 - 99.7 rule, about % of autos should get more than 31.03 mpg.
(type an integer or a decimal. do not round.)

Explanation:

Step1: Recall the 68 - 95 - 99.7 Rule

The 68 - 95 - 99.7 Rule for a normal distribution states that about 68% of the data lies within 1 standard - deviation of the mean, 95% within 2 standard - deviations, and 99.7% within 3 standard - deviations.

Step2: Analyze part (b)

We are given that the central 95% of autos are in the interval from 12.37 to 37.25 mpg. This implies that the mean is $\frac{12.37 + 37.25}{2}=24.81$ mpg (since the normal distribution is symmetric about the mean). Also, the interval from the mean to one of the endpoints is 2 standard - deviations. So, if $\mu = 24.81$ and one endpoint is $x = 37.25$, then $37.25=24.81 + 2\sigma$, and $\sigma=\frac{37.25 - 24.81}{2}=6.22$.

Step3: Analyze part (c)

The value 31.03 is $31.03=24.81+1\sigma$ (since $\mu = 24.81$ and $\sigma = 6.22$). Since about 68% of the data lies within 1 standard - deviation of the mean ($\mu\pm\sigma$), the percentage of data outside of $\mu\pm\sigma$ is $100 - 68=32\%$. And since the normal distribution is symmetric, the percentage of data greater than $\mu+\sigma$ is $\frac{100 - 68}{2}=16\%$.

Answer:

b) 12.37, 37.25
c) 16