An intersection of road with a four-way stop sign has no cars at the intersection 51% of the time, one car at the intersection 11% of the time, two cars at the intersection 21% of the time, three cars at the intersection 8% of the time, and 4 cars at the intersection during the remaining possible time. Calculate the expected value of the number of cars at this intersection at a randomly selected time. Use two decimal place accuracy.

To calculate the expected value (or mean) of the number of cars at the intersection, we can use the formula: Expected Value = ∑ (x * P(x)) Where: - x is the number of cars at the intersection (0, 1, 2, 3, 4). - P(x) is the probability of having x cars at the intersection. Given the probabilities you provided: P(0) = 51% = 0.51 P(1) = 11% = 0.11 P(2) = 21% = 0.21 P(3) = 8% = 0.08 The probability of having 4 cars at the intersection can be calculated as follows: Since the sum of all probabilities must equal 1: P(4) = 1 - (P(0) + P(1) + P(2) + P(3)) = 1 - (0.51 + 0.11 + 0.21 + 0.08) = 1 - 0.91 = 0.09 Now, let's calculate the expected value: Expected Value = (0 * 0.51) + (1 * 0.11) + (2 * 0.21) + (3 * 0.08) + (4 * 0.09) = 0 + 0.11 + 0.42 + 0.24 + 0.36 = 1.13 Therefore, the expected value of the number of cars at this intersection at a randomly selected time is 1.13 (rounded to two decimal places).