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Magazine Chapter 4: Problem 4
The diameter of a particle of contamination (in micrometers) is modeled with
the probability density function \(f(x)=2 / x^{3}\) for \(x>1 .\) Determine the
following:
(a) \(P(X<2)\)
(b) \(P(X>5)\)
(c) \(P(4
Short Answer
Expert verified
(a) 0.75, (b) 0.04, (c) 0.046875, (d) 0.953125, (e) 4.472.
Step by step solution
01
Identify CDF
The probability density function (PDF) is given by \( f(x) = \frac{2}{x^3} \) for \( x > 1 \). To find probabilities for continuous random variables, we need the cumulative distribution function (CDF), which is \( F(x) = \int_{1}^{x} f(t) \, dt \).
02
Calculate CDF
Perform the integration: \[ F(x) = \int_{1}^{x} \frac{2}{t^3} \, dt = \left[ -\frac{1}{t^2} \right]_{1}^{x} = -\frac{1}{x^2} + \frac{1}{1^2} = 1 - \frac{1}{x^2}.\] Thus, the CDF is \( F(x) = 1 - \frac{1}{x^2} \).
03
Solve (a) P(X
Substitute \( x = 2 \) in the CDF: \[ P(X<2) = F(2) = 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = 0.75. \] So, \( P(X < 2) = 0.75 \).
04
Solve (b) P(X>5)
Use the CDF to find \( P(X > 5) \): \[ P(X > 5) = 1 - P(X \leq 5) = 1 - F(5) = 1 - \left( 1 - \frac{1}{5^2} \right) = \frac{1}{25}. \] So, \( P(X > 5) = 0.04 \).
05
Solve (c) P(4
Calculate the probability: \[ P(4 < X < 8) = F(8) - F(4) = \left( 1 - \frac{1}{8^2} \right) - \left( 1 - \frac{1}{4^2} \right) = \frac{1}{16} - \frac{1}{64} = \frac{3}{64}. \] So, \( P(4 < X < 8) = 0.046875 \).
06
Solve (d) P(X8)
This uses the rule \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Since events are mutually exclusive, \( P(A \cap B) = 0 \): \[ P(X < 4 \text{ or } X > 8) = P(X < 4) + P(X > 8). \]Calculate individually:\[ P(X < 4) = F(4) = 1 - \frac{1}{16} = \frac{15}{16}, \]\[ P(X > 8) = 1 - F(8) = \frac{1}{64}. \]Therefore, \[ P(X < 4 \text{ or } X > 8) = \frac{15}{16} + \frac{1}{64} = \frac{4 \times 15 + 1}{64} = \frac{61}{64}. \] So, \( P(X < 4 \text{ or } X > 8) = 0.953125. \)
07
Solve (e) x such that P(X
We need to solve for \( x \) such that \( F(x) = 0.95 \):\[ 1 - \frac{1}{x^2} = 0.95 \rightarrow \frac{1}{x^2} = 0.05 \rightarrow x^2 = 20 \rightarrow x = \sqrt{20} = 2\sqrt{5}. \] Thus, \( x \approx 4.472. \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cumulative Distribution Function
When we talk about the cumulative distribution function (CDF), it refers to a function that maps a value to its probability of occurrence in a given distribution. The CDF gives us the probability that the random variable is less than or equal to a certain value. This is crucial for continuous random variables. The key difference between a CDF and a probability density function (PDF) is that the PDF gives the probability density at a point, while the CDF gives the cumulative probability.
To calculate the CDF from a PDF, you would integrate the PDF over the range of interest. As seen in the example, if the PDF is given by \( f(x) = \frac{2}{x^3} \), then the CDF \( F(x) \) is the integral of \( f(x) \) evaluated from a starting point to \( x \). For \( x > 1 \), the CDF becomes \( F(x) = 1 - \frac{1}{x^2} \). With CDF, you get the cumulative probability up to any particular value of \( x \).
To calculate the CDF from a PDF, you would integrate the PDF over the range of interest. As seen in the example, if the PDF is given by \( f(x) = \frac{2}{x^3} \), then the CDF \( F(x) \) is the integral of \( f(x) \) evaluated from a starting point to \( x \). For \( x > 1 \), the CDF becomes \( F(x) = 1 - \frac{1}{x^2} \). With CDF, you get the cumulative probability up to any particular value of \( x \).
Continuous Random Variables
Continuous random variables are those that can take any value within a given range. Unlike discrete random variables, that take specific values, continuous variables are represented with intervals. For instance, the diameter of a particle, as mentioned in the problem, is a continuous random variable because it could be any value greater than 1 micrometer.
The probability of a continuous random variable taking on an exact value is practically zero. Instead, we talk about the probability between ranges. This is expressed using a probability density function (PDF). By integrating the PDF over a certain interval, you can determine the probability that the random variable falls within that interval. Understanding continuous random variables is crucial in fields that model data over a range, like physics and engineering.
The probability of a continuous random variable taking on an exact value is practically zero. Instead, we talk about the probability between ranges. This is expressed using a probability density function (PDF). By integrating the PDF over a certain interval, you can determine the probability that the random variable falls within that interval. Understanding continuous random variables is crucial in fields that model data over a range, like physics and engineering.
Integration in Probability
Integration plays an essential role in probability for continuous random variables. It is through integration that the cumulative probabilities are calculated from the probability density function (PDF). When you integrate a PDF over an interval, it provides the total probability that the random variable will fall within that interval.
The definite integral of the PDF from a lower limit to an upper limit gives the cumulative probability for that range. For example, if you want to know \( P(4 < X < 8) \), you would subtract \( F(4) \) from \( F(8) \), where \( F(x) \) represents the CDF. This calculation gives insights into the likelihood of the variable being within that specific range, which is crucial when analyzing data that follow a continuous distribution.
The definite integral of the PDF from a lower limit to an upper limit gives the cumulative probability for that range. For example, if you want to know \( P(4 < X < 8) \), you would subtract \( F(4) \) from \( F(8) \), where \( F(x) \) represents the CDF. This calculation gives insights into the likelihood of the variable being within that specific range, which is crucial when analyzing data that follow a continuous distribution.
Mutually Exclusive Events
Mutually exclusive events are those that cannot happen at the same time. In probability theory, if one event occurs, the other cannot. Consider the event of drawing a card; drawing a King and a Queen simultaneously is impossible. For continuous variables, mutually exclusive refers to intervals that do not overlap.
In the solution, calculating the probability that either \( X < 4 \) or \( X > 8 \) involves using the principle that these events are mutually exclusive. The probabilities are calculated separately for \( X < 4 \) and \( X > 8 \), and then simply summed, without worrying about overlap. This leverages the property that for mutually exclusive events, \( P(A \cup B) = P(A) + P(B) \), thus simplifying calculations for non-overlapping events.
In the solution, calculating the probability that either \( X < 4 \) or \( X > 8 \) involves using the principle that these events are mutually exclusive. The probabilities are calculated separately for \( X < 4 \) and \( X > 8 \), and then simply summed, without worrying about overlap. This leverages the property that for mutually exclusive events, \( P(A \cup B) = P(A) + P(B) \), thus simplifying calculations for non-overlapping events.
