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The waiting time for service at a hospital emergency department (in hours) follows a distribution with probability density function \(f(x)=0.5 \exp (-0.5 x)\) for \(02)\) c. Value \(x\) (in hours) exceeded with probability 0.05 .

Short Answer

Expert verified
a. 0.2212, b. 0.3679, c. 5.9914 hours.

Step by step solution

01

Understanding the Probability Density Function

We are given a probability density function (pdf) for a waiting time at a hospital emergency department: \[ f(x) = 0.5 \exp(-0.5x) \text{ for } x > 0 \] This is an exponential distribution with the rate parameter \( \lambda = 0.5 \). For an exponential distribution, the cumulative distribution function (CDF) is given by \( F(x) = 1 - \exp(-\lambda x) \).
02

Calculating P(X

To find \( P(X<0.5) \), we use the cumulative distribution function:\[ P(X<0.5) = F(0.5) = 1 - \exp(-0.5 \times 0.5) \]Evaluating gives:\[ F(0.5) = 1 - \exp(-0.25) \approx 1 - 0.7788 = 0.2212 \]
03

Calculating P(X>2)

To find \( P(X>2) \), we use the complementary rule:\[ P(X>2) = 1 - P(X \leq 2) = 1 - F(2) \]Calculate \( F(2) \):\[ F(2) = 1 - \exp(-0.5 \times 2) = 1 - \exp(-1) \approx 1 - 0.3679 = 0.6321 \]Thus, \( P(X>2) = 1 - 0.6321 = 0.3679 \).
04

Value x Exceeded with Probability 0.05

We need to find the value \( x \) such that \( P(X > x) = 0.05 \). This is equivalent to \( F(x) = 0.95 \):\[ 1 - \exp(-0.5x) = 0.95 \]Solving for \( x \):\[ \exp(-0.5x) = 0.05 \]Take the natural log on both sides:\[ -0.5x = \ln(0.05) \]\[ x = \frac{\ln(0.05)}{-0.5} \approx \frac{-2.9957}{-0.5} = 5.9914 \]

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Key Concepts

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

Probability Density Function
The Probability Density Function (pdf) is a crucial concept in continuous probability distributions. In our context, it tells us how the waiting times at a hospital emergency department are distributed. In simple terms, the pdf, denoted as \( f(x) \), is a function that describes the likelihood of random variable \( X \) occurring at a particular value \( x \).
For the exponential distribution, relevant to our problem, the pdf is given by \( f(x) = 0.5 \exp(-0.5x) \) for \( x > 0 \). The parameter \( 0.5 \) in the function is the rate at which events happen. This pdf decreases exponentially, indicating that shorter waiting times are more probable than longer ones.
  • \( f(x) = 0.5 \exp(-0.5x) \) shows an exponential decline.
  • It works only for \( x > 0 \), suitable for positive waiting times.
  • The pdf integrates to 1 over its entire range, ensuring that the total probability is 1.
Understanding the pdf helps in making predictions about waiting times and identifying how probability is spread over different durations. It sets the stage for computing more detailed probabilities using tools like the cumulative distribution function.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is the next step in understanding the behavior of waiting times in our problem. The CDF, represented as \( F(x) \), is built from the pdf and offers a more intuitive perspective on probabilities.
While the pdf gives us the likelihood of waiting exactly a certain amount of time, the CDF tells us the probability that the waiting time \( X \) is less than or equal to a specific value \( x \). For an exponential distribution, the CDF is defined as:\[ F(x) = 1 - \exp(-\lambda x) \]
For our problem, with \( \lambda = 0.5 \), the CDF becomes \( F(x) = 1 - \exp(-0.5x) \). It is useful for computing probabilities of more general intervals.
  • \( F(x) \) offers the probability of waiting \( X \leq x \) hours.
  • As \( x \) increases, \( F(x) \) approaches 1, indicating higher cumulative probabilities for longer times.
  • We used it to find \( P(X<0.5) = 0.2212 \), indicating a 22.12% chance of waiting less than half an hour.
With the CDF, predicting waiting times becomes much easier, enabling us to answer diverse questions related to probabilities over time.
Complementary Rule
The Complementary Rule is a handy tool when dealing with probabilities of events that do not fall within the scope of the Cumulative Distribution Function (CDF) directly. Essentially, it uses the relation between probabilities of complementary events.
For instance, to find the probability that the waiting time \( X \) is more than a certain value \( x \), i.e., \( P(X > x) \), we use:\[ P(X > x) = 1 - F(x) \]
In our exercise, this rule helped calculate \( P(X>2) \). By applying the complementary rule:\[ P(X>2) = 1 - P(X \leq 2) = 1 - F(2) = 0.3679 \]
This value indicates approximately a 36.79% chance of waiting over 2 hours.
  • Complements what the CDF gives, broadening possible inquiries to outcomes exceeding a certain \( x \).
  • Important for finding probabilities of scenarios typically difficult to measure directly.
  • Allows for easy conversion between \( P(X \leq x) \) and \( P(X > x) \) values.
By employing this rule, tasks involving upper-tail probabilities in exponentially distributed events become more straightforward.
Rate Parameter
The Rate Parameter, often denoted by \( \lambda \), is a defining characteristic of the exponential distribution. It helps in explaining how fast the process occurs, putting a number to the frequency of events happening per unit time.
In our waiting time scenario, \( \lambda = 0.5 \) reflects how the probability density and cumulative distribution decrease with time.
The rate parameter directly affects both the pdf and CDF, shaping the distribution profile.
  • Higher \( \lambda \) values imply more frequent occurrences or shorter waiting periods.
  • It influences the steepness of the exponential curve, leading to quicker drops in probability with increasing \( x \).
  • Helps in adjusting the model to fit observed data or align with expected waiting times.
Understanding \( \lambda \) is crucial as it guides interpretation of the likelihood and timing aspects of the distribution. It allows for effective analysis and prognosis regarding time-dependent processes.

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