Final answer:
The distribution of passenger vehicle speeds on the I-5 Freeway is analyzed and various properties of the distribution are calculated based on given information. The results include comparing a given speed to a limit, calculating the percentage of vehicles that exceed the speed limit, determining the speed interval for the middle 70% of vehicles, finding the cutoff value for the middle 50% of speeds, and identifying the speed of the slowest 10% of vehicles.
Step-by-step explanation:
To answer part (A) of the question, we need to convert kilometers per hour to miles per hour. We know that 1 kilometer is equal to 0.62 miles. So, to convert 128 km/h to miles per hour, we can multiply 128 by 0.62 to get 79.36 miles per hour. Since the speed limit on the I-5 is 72.6 miles per hour, any speed above 72.6 miles per hour is considered faster than 128 km/h.
In part (B), we can calculate the percentage of passenger vehicles traveling above the speed limit by finding the area under the normal distribution curve above the speed limit. We find the z-score for the speed limit by subtracting the mean from the speed limit and dividing by the standard deviation. The z-score is (72.6 - 65) / 4.78 = 1.62. We then use a standard normal distribution table or calculator to find the area to the right of the z-score, which is approximately 0.0516. This means that about 5.16% of the passenger vehicles travel above the speed limit. Since this probability is less than 0.05, it is considered unusual.
For part (C), we need to find the speed interval that corresponds to the middle 70% of passenger vehicles. To do this, we find the z-scores for the lower and upper percentiles that bound the middle 70%. The lower percentile is 0.5 - (70/2) = 0.15, and the upper percentile is 0.5 + (70/2) = 0.85. We then use the z-score formula to find the corresponding speeds: lower speed = mean + (z-score * standard deviation) = 72.6 + (-1.04 * 4.78) = 67.84 miles per hour, and upper speed = mean + (z-score * standard deviation) = 72.6 + (1.04 * 4.78) = 77.36 miles per hour. Therefore, the authorities should expect the middle 70% of passenger vehicles to travel at speeds between 67.84 miles per hour and 77.36 miles per hour.
In part (D), we need to find the cutoff value for the middle 50% of passenger vehicle speeds, which represents the interquartile range (IQR). The lower quartile corresponds to the 25th percentile, and the upper quartile corresponds to the 75th percentile. We can find the z-scores for these percentiles by using the formula (percentile - 0.5) * 2. We then use the z-score formula to find the corresponding speeds: lower speed = mean + (z-score * standard deviation) = 72.6 + (-0.68 * 4.78) = 69.84 miles per hour, and upper speed = mean + (z-score * standard deviation) = 72.6 + (0.68 * 4.78) = 75.36 miles per hour. Therefore, the cutoff value for the middle 50% of passenger vehicle speeds is between 69.84 miles per hour and 75.36 miles per hour.
In part (E), we need to find the speed at which the slowest 10% of the passenger vehicles travel. This can be done by finding the z-score corresponding to the 10th percentile and using the z-score formula to find the corresponding speed: z-score = percentile - 0.5 = 0.1 - 0.5 = -0.4. The speed is then calculated as mean + (z-score * standard deviation) = 72.6 + (-0.4 * 4.78) = 70.69 miles per hour. Therefore, the slowest 10% of the passenger vehicles travel at speeds slower than 70.69 miles per hour.