Chapter 4, Problem 1UA

In Exercises 1–3, assume the fire department guidelines are correct and that they respond to an average of 15 emergency incidents per day. Use the graph of the Poisson distribution and technology to answer the questions. Explain your reasoning.

1. On a random day, what is more likely, 15 emergency incidents or at least 20 incidents?

To determine

The more likely event.

Answer to Problem 1UA

The probability of getting at least 20 emergency incidents is more likely to happen.

Explanation of Solution

Given info:

The mean number of emergency incidents per day is 15. On a random day, two events are considered: one is 15 emergency incidents and the other is at least 20 incidents.

Calculation:

Define the random variable x as the number of emergency incidents that happened on any given day. Here, the average number of emergency incidents per day is 15 incidents and each occurrence of emergency incidents is independent of the other. Hence, the random variable follows Poisson distribution.

The Poisson distribution formula is given below:

$P\left(x\right)=\frac{{\mu }^x{e}^{-\mu }}{x!}$

Here, x is the number occurrences of the event in a given interval, $\mu$ is the mean occurrences of the event, e is a constant and its value is 2.71828.

Consider x as 15, $\mu$ as 15 and e as 2.71828.

Software procedure:

Step-by-step procedure for calculating the Poisson probability is given below:

• Click on Calc, Probability distributions and select Poisson.
• Choose Probability.
• Under Mean enter 15.
• Under Input constant, enter 15.
• Click ok.

Output obtained from MINITAB is given below:

Thus, the probability of getting 15 emergency incidents per day is 0.102.

The probability of getting at least 20 emergency incidents per day is given below:

Software procedure:

Step-by-step procedure for calculating the Poisson probability is given below:

• Click on Calc, Probability distributions and select Poisson.
• Choose Cumulative Probability.
• Under Mean enter 15.
• Under Input constant, enter 19.
• Click ok.

Output obtained from MINITAB is given below:

The probability of getting at least 20 emergency incidents is as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Getting at least}\\ \text{20 emergency incidents}\end{array}\right)=P\left(x\ge 20\right)\\ =1-P\left(x<20\right)\\ =1-P\left(x\le 19\right)\\ =1-0.875\end{array}$

                                              =0.125

Thus, the probability of getting at least 20 emergency incidents is 0.125.

Conclusion:

The probability of getting at least 20 emergency incidents is more likely to happen when compared to 15 emergency incidents. This is because the probability of at least 20 emergency incidents is more than the probability of 15 emergency incidents.

Justification:

The events are unlikely to occur if the probability is less than 0.05.

From the MINITAB output, it can be seen that the probability of getting at least 20 emergency incidents per day is 0.125 and the probability of getting 15 emergency incidents is 0.102.

From the graph, it can be seen that the probability value corresponding to 15 incidents is 0.102 and the probability values corresponding to at least 20 incidents is observed as follows:

$\begin{array}{c}P\left(\text{At least 20 incidents}\right)=P\left(20\right)+P\left(22\right)+\mathrm{...}+P\left(26\right)+P\left(28\right)\\ =\left(\begin{array}{l}0.041+0.03+0.021+0.013+0.009+\\ 0.005+0.003+0.002+0.001\end{array}\right)\\ =0.125\end{array}$