Question

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.

Answer 1

The expected value of the number of cars at the intersection is calculated by multiplying each number of cars by its probability of occurrence and summing these products, resulting in an expected value of 1.13 cars.

Step-by-step explanation:

To calculate the expected value of the number of cars at the intersection at a randomly selected time, we use the probabilities given for 0, 1, 2, 3, and 4 cars at the intersection. The calculation involves multiplying each number of cars by its corresponding probability and adding up all these products.

First, we calculate the probability of having 4 cars at the intersection since it is not directly provided. We know the probabilities must add up to 100%, so the missing percentage for 4 cars is:

100% - (51% + 11% + 21% + 8%) = 9%

Next, we calculate the expected number of cars (E[X]) as follows:

• 0 cars * 51% (0.51)
• 1 car * 11% (0.11)
• 2 cars * 21% (0.21)
• 3 cars * 8% (0.08)
• 4 cars * 9% (0.09)

E[X] = (0 * 0.51) + (1 * 0.11) + (2 * 0.21) + (3 * 0.08) + (4 * 0.09)

After performing the multiplication and addition:

E[X] = 0 + 0.11 + 0.42 + 0.24 + 0.36 = 1.13

Therefore, the expected value of the number of cars at the intersection is 1.13.

Answer 2

The expected value of the number of cars at the intersection at a randomly selected time, given the probabilities for different numbers of cars, is 1.13 cars.

Step-by-step explanation:

The question asks for the calculation of the expected value of the number of cars at an intersection at a randomly selected time. To find the expected value, you multiply each outcome by its probability and sum these products. Given that there are no cars 51% of the time, one car 11% of the time, two cars 21% of the time, three cars 8% of the time, and the remaining percentage for four cars, we can calculate as follows:

• 0 cars × 0.51 = 0
• 1 car × 0.11 = 0.11
• 2 cars × 0.21 = 0.42
• 3 cars × 0.08 = 0.24
• 4 cars for the remaining 9% (100% - 51% - 11% - 21% - 8%) = 4 cars × 0.09 = 0.36

Adding these together:

0 + 0.11 + 0.42 + 0.24 + 0.36 = 1.13

The expected value of the number of cars at the intersection is 1.13 cars at any randomly selected time.