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Chapter 7: Problem 18

The arrival times of the 8 a.m. Bostonbased commuter train as observed in the suburban town of Sharon over 120 weekdays is summarized below: $$ \begin{array}{lc} \hline & \begin{array}{c} \text { Frequency of } \\ \text { Arrival Time, } \boldsymbol{x} \end{array} & \text { Occurrence } \\ \hline 7: 56 \text { a.m. }

Short Answer

Expert verified
a. The sample space for the experiment is given by: \[S = \{x: 7:56 \; \text{a.m.} < x \leq 8:10 \; \text{a.m.}\}\] or in minutes: \[S = \{X: 476 < X \leq 490 \}\] b. The empirical probability distribution is: \[\begin{array}{cc} \hline \boldsymbol{X} & \boldsymbol{P(X)} \\ \hline 476 < X \leq 478 & \frac{4}{120} \\ \hline 478 < X \leq 480 & \frac{18}{120} \\ \hline 480 < X \leq 482 & \frac{50}{120} \\ \hline 482 < X \leq 484 & \frac{32}{120} \\ \hline 484 < X \leq 486 & \frac{9}{120} \\ \hline 486 < X \leq 488 & \frac{4}{120} \\ \hline 488 < X \leq 490 & \frac{3}{120} \\ \hline \end{array}\]

Step by step solution

01

a. Describe the sample space

Based on the given data, the commuter train arrives between 7:56 a.m. to 8:10 a.m. Therefore, an appropriate sample space would be the set of all possible arrival times (in minutes) within this range. Let \(X\) represent the arrival times in minutes. The sample space for the experiment is given by: \[S = \{x: 7:56 \; \text{a.m.} < x \leq 8:10 \; \text{a.m.}\}\] This includes all the arrival times within the intervals mentioned in the table. We can represent the intervals in minutes as: \[S = \{X: 476 < X \leq 490 \}\]
02

b. Finding the empirical probability distribution

The empirical probability distribution can be found by dividing the frequency of each interval by the total number of observations, which is 120 in this case. Let us represent the probability of the train arriving within a given time interval by \(P(X)\). 1. P(476 < X <= 478): \(P(476 < X \leq 478) = \frac{4}{120}\) 2. P(478 < X <= 480): \(P(478 < X \leq 480) = \frac{18}{120}\) 3. P(480 < X <= 482): \(P(480 < X \leq 482) = \frac{50}{120}\) 4. P(482 < X <= 484): \(P(482 < X \leq 484) = \frac{32}{120}\) 5. P(484 < X <= 486): \(P(484 < X \leq 486) = \frac{9}{120}\) 6. P(486 < X <= 488): \(P(486 < X \leq 488) = \frac{4}{120}\) 7. P(488 < X <= 490): \(P(488 < X \leq 490) = \frac{3}{120}\) Thus, the empirical probability distribution is given by the following table: \[\begin{array}{cc} \hline \boldsymbol{X} & \boldsymbol{P(X)} \\ \hline 476 < X \leq 478 & \frac{4}{120} \\ \hline 478 < X \leq 480 & \frac{18}{120} \\ \hline 480 < X \leq 482 & \frac{50}{120} \\ \hline 482 < X \leq 484 & \frac{32}{120} \\ \hline 484 < X \leq 486 & \frac{9}{120} \\ \hline 486 < X \leq 488 & \frac{4}{120} \\ \hline 488 < X \leq 490 & \frac{3}{120} \\ \hline \end{array}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Space
The sample space in probability is the set of all possible outcomes of a random experiment. In the case of the Boston-based commuter train arrivals, our sample space is the range of times the train can arrive at the suburban town of Sharon.

For this exercise, the train arrives between 7:56 a.m. and 8:10 a.m. We can express time in minutes for simplicity, with 7:56 a.m. corresponding to 476 minutes and 8:10 a.m. corresponding to 490 minutes from midnight.

Therefore, our sample space, denoted by \( S \), includes all possible arrival minutes:
  • \( S = \{ x: 476 < x \leq 490 \} \)
This represents every potential time interval in which the train may arrive based on our observations.
Frequency Distribution
Frequency distribution shows how often each outcome occurs in a dataset. In this experiment, we collected 120 observations of train arrival times.

Each frequency corresponds to the number of times the train arrived within each specified time interval.

Here is the frequency distribution for these arrival times:
  • Interval 476 - 478: Frequency 4
  • Interval 478 - 480: Frequency 18
  • Interval 480 - 482: Frequency 50
  • Interval 482 - 484: Frequency 32
  • Interval 484 - 486: Frequency 9
  • Interval 486 - 488: Frequency 4
  • Interval 488 - 490: Frequency 3
Understanding this frequency distribution helps us visualize which time intervals the arrivals typically fall into, thus providing a foundational understanding for further probability analysis.
Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random phenomena. It helps us predict how likely an event is to occur. In our train arrival example, we use the concept of empirical probability, which is based on observed data.

To find the probability of the train arriving within each specific interval, we divide the frequency of that interval by the total number of observations, which is 120 in our case.

These are the empirical probabilities of each interval:
  • \( P(476 < X \leq 478) = \frac{4}{120} \)
  • \( P(478 < X \leq 480) = \frac{18}{120} \)
  • \( P(480 < X \leq 482) = \frac{50}{120} \)
  • \( P(482 < X \leq 484) = \frac{32}{120} \)
  • \( P(484 < X \leq 486) = \frac{9}{120} \)
  • \( P(486 < X \leq 488) = \frac{4}{120} \)
  • \( P(488 < X \leq 490) = \frac{3}{120} \)
By examining these probabilities, we can understand which intervals have the highest chance for train arrival.
Random Variables
Random variables are functions that assign numerical values to the outcomes of random experiments. In our train arrival scenario, the random variable is the time of arrival, represented as \( X \).

Each possible arrival time is mapped to a specific measurement in minutes, creating a clear relationship between each possible time interval and its respective probability.

With random variables, we can calculate not just the probabilities for single intervals, but also for combinations of events. For instance, we might want to know the probability of the train arriving within the first 4 minutes of the target period, which would involve summing the probabilities of corresponding intervals.

Random variables are critical in transforming real-world scenarios into quantifiable data that can be analyzed using statistical methods, providing deep insights into the patterns of train arrivals.

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Most popular questions from this chapter

Let \(S\) be a sample space for an experiment, and let \(E\) and \(F\) be events of this experiment. Show that the events \(E \cup F\) and \(E^{c} \cap F^{c}\) are mutually exclusive. Hint: Use De Morgan's law.

According to Mediamark Research, 84 million out of 179 million adults in the United States correct their vision by using prescription eyeglasses, bifocals, or contact lenses. (Some respondents use more than one type.) What is the probability that an adult selected at random from the adult population uses corrective lenses?

In an attempt to study the leading causes of airline crashes, the following data were compiled from records of airline crashes from 1959 to 1994 (excluding sabotage and military action). $$ \begin{array}{lc} \hline \text { Primary Factor } & \text { Accidents } \\ \hline \text { Pilot } & 327 \\ \hline \text { Airplane } & 49 \\ \hline \text { Maintenance } & 14 \\ \hline \text { Weather } & 22 \\ \hline \text { Airport/air traffic control } & 19 \\ \hline \text { Miscellaneous/other } & 15 \\ \hline \end{array} $$ Assume that you have just learned of an airline crash and that the data give a generally good indication of the causes of airline crashes. Give an estimate of the probability that the primary cause of the crash was due to pilot error or bad weather.

The number of cars entering a tunnel leading to an airport in a major city over a period of 200 peak hours was observed, and the following data were obtained: $$ \begin{array}{rc} \hline \begin{array}{l} \text { Number of } \\ \text { Cars, } x \end{array} & \begin{array}{c} \text { Frequency of } \\ \text { Occurrence } \end{array} \\ \hline 01000 & 15 \\ \hline \end{array} $$ a. Describe an appropriate sample space for this experiment. b. Find the empirical probability distribution for this experiment.

In a survey conducted in the fall 2006, 800 homeowners were asked about their expectations regarding the value of their home in the next few years; the results of the survey are summarized below: $$ \begin{array}{lc} \hline \text { Expectations } & \text { Homeowners } \\ \hline \text { Decrease } & 48 \\ \hline \text { Stay the same } & 152 \\ \hline \text { Increase less than } 5 \% & 232 \\ \hline \text { Increase 5-10\% } & 240 \\ \hline \text { Increase more than 10\% } & 128 \\ \hline \end{array} $$ If a homeowner in the survey is chosen at random, what is the probability that he or she expected his or her home to a. Stay the same or decrease in value in the next few years? b. Increase \(5 \%\) or more in value in the next few years?
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