Question

Assume the weight of a randomly chosen American passenger car is a uniformly distributed random variable ranging from 1,612 pounds to 4,117 pounds.

a. what is the probability that a vehicle will weigh between 2,582 and 4,046 pounds?

Answer 1

The probability that a vehicle will weigh between 2,582 and 4,046 pounds is approximately 0.5855.

Step-by-step explanation:

To find the probability that a vehicle will weigh between 2,582 and 4,046 pounds, we need to calculate the cumulative probability at the upper weight limit and subtract the cumulative probability at the lower weight limit from it.

First, we calculate the cumulative probability at 4,046 pounds using the formula:

P(X ≤ x) = (x - a) / (b - a)

where x is the upper weight limit, a is the lower weight limit, and b is the upper weight limit of the distribution. Plugging in the values, we get:

P(X ≤ 4,046) = (4,046 - 1,612) / (4,117 - 1,612) = 2,434 / 2,505 ≈ 0.9727

Next, we calculate the cumulative probability at 2,582 pounds:

P(X ≤ x) = (x - a) / (b - a)

Using the formula with the given values:

P(X ≤ 2,582) = (2,582 - 1,612) / (4,117 - 1,612) = 970 / 2,505 ≈ 0.3872

Finally, we subtract the cumulative probability at the lower weight limit from the cumulative probability at the upper weight limit:

P(2,582 ≤ X ≤ 4,046) = P(X ≤ 4,046) - P(X ≤ 2,582) ≈ 0.9727 - 0.3872 ≈ 0.5855