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Ann is expected at 7:00 \(\mathrm{pm}\) after an all-day drive. She may be as much as \(1 \mathrm{~h}\) early or as much as \(3 \mathrm{~h}\) late. Assuming that her arrival time \(X\) is uniformly distributed over that interval, find the pdf of \(|X-7|\), the unsigned difference between her actual and predicted arrival times.

Short Answer

Expert verified
The PDF of \(|X-7|\) is \( \frac{1}{3} \) for \( 0 \leq y \leq 3 \).

Step by step solution

01

Determine the Interval for X

Ann's expected arrival time follows a uniform distribution with parameters defined by being at most 1 hour early or 3 hours late. Thus the interval for her arrival time, \( X \), is between 6:00 pm and 10:00 pm. Therefore, \( X \sim U(6, 10) \).
02

Define the Transformation

We need to find the unsigned difference \(|X - 7|\). Since \( X \) is uniformly distributed between 6 and 10, it transforms the interval into \(|6-7|\) to \(|10-7|\), or 0 to 3. Thus, we define \( Y = |X - 7| \), with \( Y \) having the interval [0, 3].
03

Compute the Probability Density Function

For a uniform distribution, the probability density function (PDF) is constant across the interval of \( Y \). The total length of the interval [6, 10] is 4 hours. Since \( Y \) spans an interval of length 3 (from 0 to 3), we need to normalize the probability over this interval. The constant PDF should be \( \frac{1}{3} \) because the uniform PDF should integrate to 1 over the interval.
04

Express the PDF of Y

The PDF of \( Y = |X - 7| \) can be expressed as follows: \[ f_Y(y) = \begin{cases} \frac{1}{3}, & \text{for } 0 \leq y \leq 3 \ 0, & \text{otherwise} \end{cases} \] This distribution is uniform over the interval [0, 3], as derived from the original conditions.

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Key Concepts

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

Uniform Distribution
The uniform distribution is one of the simplest probability distributions. It is used when we know that all outcomes in a given interval are equally likely. In Ann's case, her arrival time falls within a specified time range, with no preference for any specific moment within that range. This means her arrival time can be described as uniformly distributed. The notation for a uniform distribution is given by \( X \sim U(a, b) \), where \( a \) and \( b \) are the lower and upper bounds of the interval, respectively. In this scenario, Ann might arrive anytime between 6:00 pm and 10:00 pm, making the interval \([6, 10]\). Hence, her arrival time \( X \) follows a uniform distribution between these bounds.
Transformation of Variables
The concept of transformation of variables is about converting one random variable into another through a mathematical function. In probability, this often involves changing the form in which we express our problem, while preserving the essential probabilistic characteristics. For Ann's scenario, we're interested in the unsigned difference between her actual arrival time and her expected arrival time of 7:00 pm. This is written as \(|X - 7|\). By doing this transformation, we're essentially looking at how far her actual arrival time deviates from the expected time, without considering direction. The transformation takes the initial interval [6, 10] to a new range [0, 3]. This range is derived from the absolute difference between each boundary of the original interval and 7.
Interval Normalization
Interval normalization is an important step when dealing with transformations. It ensures the probabilities remain consistent across a transformed interval. When Ann's arrival interval \([6, 10]\) is transformed to \(|X - 7|\), or \([0, 3]\), we need to maintain the uniform character over this new interval. The length of the original interval is 4 hours, spanning from 6:00 pm to 10:00 pm, and the transformation leads to a new interval of 3 hours. The uniform distribution must be normalized over this new interval, which means every subinterval is given equal probability sums to 1 over the entire interval [0, 3]. This involves adjusting the uniform density function across this interval appropriately.
PDF Calculation
Calculating the probability density function (PDF) for Ann's transformed arrival time involves determining a constant value that applies uniformly across the interval [0, 3]. Given that the transformation of Ann's arrival time to \(|X-7|\) results in a range of length 3, we need the PDF to integrate to 1 over this interval. The PDF for a uniformly distributed random variable \( Y \sim U(0, 3) \) is calculated by taking the reciprocal of the interval's length. Consequently, the value of \( f_Y(y) = \frac{1}{3} \) because the total probability (area under the curve) should sum up to 1 over [0, 3]. This ensures each point within the interval holds an equal likelihood of occurrence, maintaining the uniform distribution characteristic even after the transformation.

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