Question

Data indicate that the number of traffic accidents in Berkeley on a rainy day is a Poisson random variable with mean 9 , whereas on a dry day it is a Poisson random variable with mean \(3 .\) Let \(X\) denote the number of traffic accidents tomorrow. If it will rain tomorrow with probability \(0.6\), find (a) \(E[X]\); (b) \(P[X=0\\}\) (c) \(\operatorname{Var}(X)\)

Short answer

The expected number of traffic accidents (E[X]) tomorrow is \(6.6\), the probability of no accidents (P[X=0]) is approximately \(0.0199\), and the variance of the number of traffic accidents (Var[X]) is approximately \(5.884\).

Step-by-step solution

Define random variables

Let R be a Bernoulli random variable with: - R = 1 when tomorrow is rainy (with probability \(0.6\)) - R = 0 when tomorrow is dry (with probability \(0.4\)) Let X denote the number of traffic accidents tomorrow.

Calculate E[X]

To find the expected number of traffic accidents, we use the conditional expectation formula: \(E[X] = E[X|R=1]\times P(R=1)+E[X|R=0]\times P(R=0)\) Since the traffic accidents follow a Poisson distribution, we know that the mean for rainy days (R=1) is 9 and for dry days (R=0) is 3. Now, we plug in the values: \(E[X] = (9\times 0.6)+(3\times 0.4)\)

Evaluate E[X]

Evaluate the expression for E[X]: \(E[X] = (5.4)+(1.2)=6.6\)

Calculate P[X=0]

To find the probability of no traffic accidents, we use the conditional probability formula: \(P[X=0] = P[X=0|R=1]\times P(R=1)+P[X=0|R=0]\times P(R=0)\) Since the traffic accidents follow a Poisson distribution: \(P[X=0|R=1] = e^{-9}\) \(P[X=0|R=0] = e^{-3}\) Now, we plug in the values: \(P[X=0] = (e^{-9}\times 0.6)+(e^{-3}\times 0.4)\)

Evaluate P[X=0]

Evaluate the expression for P[X=0]: \(P[X=0] = (0.6e^{-9})+(0.4e^{-3})\approx 0.0199\)

Calculate Var(X)

To find the variance of the number of traffic accidents, we use the conditional variance formula: \(\operatorname{Var}(X) = E[\operatorname{Var}(X|R)]+ \operatorname{Var}(E[X|R])\) Since the traffic accidents follow a Poisson distribution, we know that the variance for rainy days (R=1) is 9 and for dry days (R=0) is 3. Now, we plug in the values: \(\operatorname{Var}(X) = (9\times 0.6)+(3\times 0.4)+\operatorname{Var}(E[X|R])\) To find the \(\operatorname{Var}(E[X|R])\), we need to find the variance of 9 with probability 0.6 and 3 with probability 0.4. \(\operatorname{Var}(E[X|R]) = E[X^2]-E[X]^2\) \(= (9^2\times 0.6)+(3^2\times 0.4)-(6.6)^2\)

Evaluate Var(X)

Evaluate the expression for \(\operatorname{Var}(X)\): \(\operatorname{Var}(X) = 5.4+(1.2)+\operatorname{Var}(E[X|R])\) \(\operatorname{Var}(X) = 6.6 + ((81\times 0.6)+(9\times 0.4)-43.56)\) \(\operatorname{Var}(X) \approx 5.884\) So, the expected number of traffic accidents (E[X]) is 6.6, the probability of no accidents (P[X=0]) is approximately 0.0199, and the variance of the number of traffic accidents (Var[X]) is approximately 5.884.

Key concepts

Understanding Expected Value (E[X])

The expected value, denoted by E[X], represents the average or mean value of a random variable over a long number of trials. In the context of the exercise, it's the average number of traffic accidents we can expect on any given day in Berkeley, considering the probability of rain or dry conditions. To calculate it, we weigh each possible outcome (the number of accidents) by the probability of the conditions under which they occur (rainy or dry day).
By calculating the expected number of accidents for rainy and dry days separately, and then combining them based on the probability of each type of day occurring, we can predict what a typical day will look like. In the given problem, since rainy days have a higher mean of accidents, they contribute more to the overall expected value, which reflects the riskier conditions associated with rain.

Decoding the Poisson Distribution

The Poisson distribution is a probability distribution that expresses how many times an event is likely to occur over a fixed interval of time or space. It's particularly useful for modeling the number of events that happen at a specific rate, but completely random in time, such as traffic accidents in our case. A key characteristic of the Poisson distribution is that the mean and variance are equal. This means that the distribution is described entirely by its mean, which makes calculations more straightforward.

• Parameter λ (Lambda): This is both the mean and the variance of the distribution.
• Discrete Outcomes: The Poisson distribution only applies to non-negative integers (0, 1, 2, ...).
• Poisson Probability Formula: For a given value of λ, the probability of observing exactly k events is given by the formula: \(P(X=k) = \frac{e^{-λ}λ^k}{k!}\), where 'e' is the base of the natural logarithm, and 'k' is the number of events.

The scenario in the exercise is a classic example where knowing the mean number of accidents on rainy and dry days allows us to use the Poisson distribution to predict the likelihood of different numbers of accidents occurring.

Variance of a Random Variable (Var[X])

The variance of a random variable, symbolized as Var[X], quantifies the spread of the data points around the mean, or expected value. It shows how much the number of accidents can vary from one day to another. In the Poisson distribution, the mean and variance are equal, which simplifies the computation. However, when considering a mixed scenario like rain or shine, we need a method to incorporate both types of days and their probabilities.
In the textbook solution, the variance is calculated by taking into account each type of weather's variance and the variance of the expected number of accidents given each weather condition. This is done through the law of total variance, which combines the variance within each group (for rainy and dry days) and the variance between the groups (the weather's impact on accidents). The result gives us an idea of the variability and predictability of traffic accidents, a crucial piece of information for city planning and risk assessment.