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The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 10 seconds. (b) What is the probability that the arrival time between vehicles is 10 seconds or less? (Round your answer to four decimal places.) (c) What is the probability that the arrival time between vehicles is 6 seconds or less? (Round your answer to four decimal places.) (d) What is the probability of 32 or more seconds between vehicle arrivals? (Round your answer to four decimal places.)

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Final answer:

The probability that the time between vehicle arrivals is 10 seconds or less is approximately 63.21%, for 6 seconds or less it is approximately 45.12%, and for 32 or more seconds it is about 4.08%.

Step-by-step explanation:

The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 10 seconds. The exponential distribution is defined as P(X < x) = 1 - e^(-λx), where λ (lambda) is the rate of occurrence, and P(X < x) is the probability that the time between arrivals is less than x seconds.

To find the probability that the arrival time between vehicles is 10 seconds or less, you would use the following steps:

  1. Calculate the rate (λ) from the given mean (mean = 1/λ), which is 1/10 seconds.
  2. Plug in the value of x (10 seconds) into the exponential distribution formula to calculate P(X <= 10).
  3. Calculate 1 - e^(-1) = 1 - 1/e = 1 - 0.3679 = 0.6321. After rounding, the probability is approximately 0.6321, or 63.21%.

Similarly, for the probability that the arrival time between vehicles is 6 seconds or less, plug in x = 6:

  1. e^(-6/10) = e^(-0.6).
  2. Calculate 1 - e^(-0.6) = 0.4512. The probability is approximately 0.4512, or 45.12%.

Lastly, for the probability of 32 or more seconds between vehicle arrivals, the calculation is P(X >= 32) = e^(-32/10) = e^(-3.2).

After calculation, P(X >= 32) ≈ 0.0408. Thus, the probability is approximately 0.0408, or 4.08%.

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