Question
Fuel economy estimates for automobiles built in a certain year predicted a mean of 24.2 mpg and a standard deviation of 6.8 mpg for highway driving. Assume that a normal distribution can be applied. Within what range are \( 68 \% \) of the automobiles? The range for \( 68 \% \) of the automobiles is \( \square \) to \( \square \mathrm{mpg} \).
Ask by Todd Martin. in the United States
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The range for \( 68\% \) of the automobiles is **17.4 mpg** to **31.0 mpg**.
Solution
To determine the range within which \( 68\% \) of the automobiles fall, we can use the properties of the normal distribution. In a normal distribution:
- Approximately \( 68\% \) of the data lies within **one standard deviation** of the mean.
Given:
- **Mean (\( \mu \))** = 24.2 mpg
- **Standard Deviation (\( \sigma \))** = 6.8 mpg
**Calculation:**
- **Lower Bound**: \( \mu - \sigma = 24.2 - 6.8 = 17.4 \) mpg
- **Upper Bound**: \( \mu + \sigma = 24.2 + 6.8 = 31.0 \) mpg
**Conclusion:**
\( 68\% \) of the automobiles have a fuel economy between **17.4 mpg** and **31.0 mpg**.
**Answer:**
The range for \( 68 \% \) of the automobiles is 17.4 to 31.0 mpg.
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Simplify this solution Beyond the Answer
To find the range where \( 68\% \) of the automobiles fall, we can use the empirical rule, which states that about \( 68\% \) of observations in a normal distribution lie within one standard deviation of the mean. Using the given mean of \( 24.2 \) mpg and a standard deviation of \( 6.8 \) mpg, we calculate: - Lower range: \( 24.2 - 6.8 = 17.4 \) mpg - Upper range: \( 24.2 + 6.8 = 31.0 \) mpg Thus, the range for \( 68\% \) of the automobiles is \( 17.4 \) to \( 31.0 \mathrm{mpg} \).
