3.12 Speeding on the I-5, Part I: The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. (a) What percent of passenger vehicles travel slower than 80 miles/hour? to two decimal places) (b) What percent of passenger vehicles travel between 60 and 80 miles/hour? (please round to two decimal places) (c) How fast do the fastest 5\% of passenger vehicles travel? round to two decimal places) (d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5. \% (please round to two decimal places)

To tackle this problem, we can use the properties of the normal distribution and the Z-score formula. Here’s how to get the answers: (a) To find the percent of passenger vehicles traveling slower than 80 miles/hour, we first calculate the Z-score: \[ Z = \frac{(X - \mu)}{\sigma} = \frac{(80 - 72.6)}{4.78} \approx 1.54 \] Using a standard normal distribution table or calculator, the cumulative probability at Z = 1.54 is approximately 0.9382, or 93.82%. (b) For the percent of passenger vehicles traveling between 60 and 80 miles/hour, we calculate the Z-scores for both 60 and 80: \[ Z_{60} = \frac{(60 - 72.6)}{4.78} \approx -2.64 \] The cumulative probability for Z = -2.64 is approximately 0.0041, or 0.41%. From part (a), we already have the probability for 80 mph, which is 93.82%. So: \[ P(60 < X < 80) = 93.82\% - 0.41\% = 93.41\% \] (c) To find the speed at which the fastest 5% of vehicles travel, we need to find the Z-score that corresponds to 95% (the top 5%). The Z-score for 95% is approximately 1.645. Using the Z-score formula to find the speed: \[ X = \mu + Z \cdot \sigma = 72.6 + (1.645 \cdot 4.78) \approx 79.3421 \text{ mph} \] (d) To find the percentage of vehicles traveling above the speed limit of 70 mph: \[ Z_{70} = \frac{(70 - 72.6)}{4.78} \approx -0.55 \] The cumulative probability for Z = -0.55 is approximately 0.2910, or 29.10%. Thus, the percentage of vehicles traveling above the speed limit is: \[ 100\% - 29.10\% = 70.90\% \] Summarizing: (a) 93.82% (b) 93.41% (c) 79.34 mph (d) 70.90%