Problem 9
Question
The number of accidents that occur at a certain intersection known as "Five Corners" on a Friday afternoon between the hours of 3 p.m. and 6 p.m., along with the corresponding probabilities, are shown in the following table. Find the expected number of accidents during the period in question. $$\begin{array}{lccccc}\hline \text { Accidents } & 0 & 1 & 2 & 3 & 4 \\ \hline \text { Probability } & .935 & .030 & .020 & .010 & .005 \\\\\hline\end{array}$$
Step-by-Step Solution
Verified Answer
The expected number of accidents on a Friday afternoon between 3 p.m. and 6 p.m. at the "Five Corners" intersection is \(0.120\).
1Step 1: Identify the corresponding probabilities for each accident
We have the following probabilities and their associated number of accidents in the table:
Accidents: 0 | 1 | 2 | 3 | 4
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Probabilities: 0.935 | 0.030 | 0.020 | 0.010 | 0.005
2Step 2: Calculate the product of accidents and their probabilities
Multiply each accident value by its corresponding probability:
0 × 0.935 = 0
1 × 0.030 = 0.030
2 × 0.020 = 0.040
3 × 0.010 = 0.030
4 × 0.005 = 0.020
3Step 3: Sum up the products to find the expected value
Add the products found in Step 2:
Expected value (E) = 0 + 0.030 + 0.040 + 0.030 + 0.020 = 0.120
The expected number of accidents on a Friday afternoon between 3 p.m. and 6 p.m. at the "Five Corners" intersection is 0.120.
Key Concepts
Probability DistributionDiscrete Random VariablesStatistical Analysis
Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. In our exercise, we are examining the number of accidents at a busy intersection during a specific time period.
For probability distributions with discrete outcomes, like accidents in our case, each potential outcome has an associated probability. This list of probabilities needs to sum up to 1, which indicates all possible outcomes together represent the whole scenario.
- In our example, the intersection can experience anywhere from 0 to 4 accidents during the specified time frame.
- Each number of accidents has a probability assigned to it, i.e., 0 accidents with a 0.935 probability, and so on up to 4 accidents with a 0.005 probability.
Discrete Random Variables
Discrete random variables are variables that can take on a countable number of distinct values. In our context, the random variable is the number of accidents during a specific time period at the intersection.
- These variables differ from continuous random variables, which can take on an infinite number of values within a given range.
- For discrete random variables, each specific value has a probability attached to it. This makes them easier to manage since we can directly calculate probabilities for specific outcomes.
Statistical Analysis
Statistical analysis involves methods for collecting, reviewing, and drawing conclusions from data. One application is finding the expected value, as shown in our exercise with accident occurrences.
- This calculation helps in understanding the data by summarizing it into a single metric that offers insights into what to generally expect.
- Beyond simple calculations, statistical analysis often involves interpreting results to guide decision-making or predict future events.
Other exercises in this chapter
Problem 8
If a sample of three batteries is selected from a lot of ten, of which two are defective, what is the expected number of defective batteries?
View solution Problem 8
Give the range of values that the random variable \(X\) may assume and classify the random variable as finite discrete, infinite discrete, or continuous. \(X=\)
View solution Problem 9
Give the range of values that the random variable \(X\) may assume and classify the random variable as finite discrete, infinite discrete, or continuous. \(X=\)
View solution Problem 9
If \(A\) and \(B\) are independent events, \(P(A)=.4\), and \(P(B)=.6\), find a. \(P(A \cap B)\) b. \(P(A \cup B)\)
View solution