Model Answer
0 min readIntroduction
In the realm of probability theory, understanding the distribution of random events is fundamental. This question delves into a scenario where a train's arrival time is uniformly distributed over a given interval, a common model for events whose occurrence is equally likely at any point within a continuous range. By applying the principles of continuous probability distributions, we can quantify uncertainties related to arrival times, predict expected outcomes, and assess the variability of such events. This problem not only tests basic statistical definitions but also the application of these concepts in practical scenarios, such as scheduling and operational planning.
1. (a) Train Arrival Time Analysis
The problem describes a train arriving at a station at a random time between 09:00 AM and 09:30 AM, with the actual arrival time equally likely to be any moment within this interval. This indicates a continuous uniform distribution. Let 'T' be the random variable representing the arrival time of the train in minutes past 09:00 AM. The interval is from 09:00 AM to 09:30 AM, which is a duration of 30 minutes. So, the interval for T is [0, 30].(i) Define a suitable probability density function (PDF)
For a continuous uniform distribution over an interval [a, b], the probability density function (PDF) is given by:$f(x) = \begin{cases} \frac{1}{b-a} & \text{for } a \le x \le b \\ 0 & \text{otherwise} \end{cases}$
In this case, $a = 0$ (corresponding to 09:00 AM) and $b = 30$ (corresponding to 09:30 AM). Therefore, the suitable probability density function (PDF) is:$f(t) = \begin{cases} \frac{1}{30-0} = \frac{1}{30} & \text{for } 0 \le t \le 30 \\ 0 & \text{otherwise} \end{cases}$
Where 't' is the time in minutes past 09:00 AM.(ii) What is the probability that the train will arrive before 09:13 AM?
Arriving before 09:13 AM means the arrival time 't' is less than 13 minutes past 09:00 AM. We need to find $P(T < 13)$.$P(T < 13) = \int_{0}^{13} f(t) \,dt = \int_{0}^{13} \frac{1}{30} \,dt$
$P(T < 13) = \frac{1}{30} [t]_{0}^{13} = \frac{1}{30} (13 - 0) = \frac{13}{30}$
The probability that the train will arrive before 09:13 AM is $\frac{13}{30}$ or approximately 0.4333.(iii) Find the expected arrival time and the variance of the arrival time.
For a continuous uniform distribution over the interval [a, b]:- Expected Value (Mean): $E[X] = \frac{a+b}{2}$
- Variance: $Var[X] = \frac{(b-a)^2}{12}$
$E[T] = \frac{0+30}{2} = \frac{30}{2} = 15$ minutes
The expected arrival time is 15 minutes past 09:00 AM, which is 09:15 AM. Variance of the Arrival Time:$Var[T] = \frac{(30-0)^2}{12} = \frac{30^2}{12} = \frac{900}{12} = 75$ (minutes$^2$)
The variance of the arrival time is 75 minutes$^2$.(iv) Given that the train has not arrived by 09:11 AM, what is the expected arrival time?
This is a conditional expectation problem. We are given that $T > 11$. The new effective interval for the arrival time becomes (11, 30]. Within this new interval, the arrival time is still uniformly distributed. Let's define a new random variable $T'$ representing the arrival time given $T > 11$. The new interval is $[a', b'] = [11, 30]$. The expected value for a uniform distribution over $[a', b']$ is $\frac{a'+b'}{2}$.$E[T | T > 11] = \frac{11+30}{2} = \frac{41}{2} = 20.5$ minutes
The expected arrival time, given that the train has not arrived by 09:11 AM, is 20.5 minutes past 09:00 AM. This corresponds to 09:20 AM and 30 seconds.Conclusion
This comprehensive analysis demonstrates the application of continuous uniform distribution to model real-world scenarios like train arrival times. From defining the appropriate probability density function to calculating probabilities, expected values, and variances, each step provides valuable insights into the behavior of the random variable. The concept of conditional expectation further refines our predictions when partial information is available, highlighting the adaptability of statistical tools in dynamic situations. Such foundational understanding is crucial for operational efficiency, resource allocation, and risk management in various fields.
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