Question

fuel economy estimates for automobiles built one year predicted a mean of 25.8 mpg and a standard deviation of 6.2 mpg for highway driving. assume that a normal model can be applied. use the 68 - 95 - 99.7 rule to complete parts a) through e).
a) draw the model for auto fuel economy.
a.

b.

c.

b) in what interval would you expect the central 99.7% of autos to be found?
using the 68 - 95 - 99.7 rule, the central 99.7% of autos can be expected to be found in the interval from 7.2 to 44.4 mpg.
(do not round. type integers or decimals.)
c) about what percent of autos should get more than 32 mpg?
using the 68 - 95 - 99.7 rule, about □% of autos should get more than 32 mpg.
(do not round. type an integer or a decimal.)

Explanation:

Step1: Recall the 68 - 95 - 99.7 Rule

The 68 - 95 - 99.7 rule for a normal distribution states that:

• Approximately 68% of the data lies within \( \mu\pm\sigma\)
• Approximately 95% of the data lies within \( \mu\pm2\sigma\)
• Approximately 99.7% of the data lies within \( \mu\pm3\sigma\)

Here, \(\mu = 25.8\) and \(\sigma=6.2\)

Step2: Calculate the value for part c

We want to find the percentage of autos that get more than \(x = 32\) mpg.
First, find the \(z\) - score. The formula for the \(z\) - score is \(z=\frac{x-\mu}{\sigma}\)
Substitute \(x = 32\), \(\mu = 25.8\), and \(\sigma = 6.2\) into the formula:
\(z=\frac{32 - 25.8}{6.2}=\frac{6.2}{6.2}=1\)
Since the total area under the normal curve is 1. The area to the left of \(z = 1\) (using the 68 - 95 - 99.7 rule, the area within \(\mu\pm\sigma\) is 68%, so the area to the left of \(z = 1\) is \(\frac{1 + 0.68}{2}=0.84\))
The area to the right of \(z = 1\) is \(1-0.84 = 0.16\) or 16%

Answer:

16