Final answer:
The mean weight of a uniformly distributed vehicle weight is 3,279 pounds, the standard deviation is 928.9565 pounds, and the probabilities for the vehicle's weight being less than 2,356 pounds, more than 4,093 pounds, and between 2,356 and 4,093 pounds are 0.2131, 0.2471, and approximately 0.5398, respectively.
Step-by-step explanation:
The question deals with the concepts of uniform distribution, mean, standard deviation, and probabilities for continuous random variables. The mean weight of a randomly chosen vehicle, when the weight is a uniformly distributed random variable ranging from 1,670 pounds to 4,888 pounds, is calculated by taking the average of the lower and upper limits. The mean weight is therefore (1,670 + 4,888) / 2, which equals 3,279 pounds.
The standard deviation for a uniform distribution is calculated using the formula: standard deviation = (upper limit - lower limit) / sqrt(12). For this scenario, it is (4,888 - 1,670) / sqrt(12), which rounds to a standard deviation of 928.9565 pounds.
Probabilities for a uniform distribution are found by calculating the area under the uniform distribution curve within the specified range. To find the probability that a vehicle weighs less than 2,356 pounds, we calculate the length of the interval from the lower limit to 2,356 and divide it by the total possible range. This gives us a probability of (2,356 - 1,670) / (4,888 - 1,670), which equals 0.2131. Similarly, the probability that a vehicle weighs more than 4,093 pounds is (4,888 - 4,093) / (4,888 - 1,670), equaling 0.2471.
To find the probability that a vehicle weighs between 2,356 and 4,093 pounds, we subtract the sum of the probabilities of the vehicle weighing less than 2,356 pounds and the vehicle weighing more than 4,093 pounds from 1. That is 1 - 0.2131 - 0.2471, which is approximately 0.5398.