Continuous Probability Models: Videos & Practice Problems

Understanding probability density functions (PDFs) is crucial for determining the likelihood of continuous random variables within specific intervals. A PDF must be non-negative and integrate to one over its entire range. The cumulative distribution function (CDF), derived from the PDF, provides the probability that a variable is less than or equal to a certain value. Key properties of the CDF include being bounded between zero and one and being non-decreasing. For example, to find the probability of a variable falling between two values, one can use the CDF by calculating the difference between its values at those points.

Which of the following is a discrete random variable?

• A. The amount of milk in a jug (in gallons)
• B. The number of times a die shows 4 in 10 rolls
• C. The height (in cm) of a student
• D. The distance a car travels before running out of fuel  

Answer: (B)

Which of the following is a continuous random variable?

• A. The number of goals scored in a soccer match
• B. The number of cars passing through a toll booth in an hour
• C. The number of students present in class
• D. The amount (in inches) of rainfall in a month

Answer: (D)

Probability Density Function (PDF)

A continuous random variable can take on any real number value, and often we need to determine the probability of this variable falling within a specific interval. For instance, we might want to find the probability that the daily average temperature in April is between 75 and 81 degrees Fahrenheit. To calculate such probabilities, we utilize a probability density function (PDF), denoted as \( f(x) \), which describes the likelihood of a continuous random variable, represented by \( X \), taking on a value within a given interval.

Since continuous random variables have an infinite number of possible outcomes, the probability of a specific value occurring is represented as an infinite sum, which can be calculated using definite integrals. A function qualifies as a PDF if it meets two essential criteria: first, \( f(x) \) must be greater than or equal to zero for all \( x \), ensuring that the function remains positive. Second, the integral of \( f(x) \) from negative infinity to positive infinity must equal one, indicating that the total area under the curve of the PDF is one, which corresponds to the sum of probabilities of all possible outcomes.

For example, if we know that the daily average temperature in April ranges from 0 to 100 degrees Fahrenheit, we typically seek the probability of the temperature falling within a narrower range, such as between 75 and 81 degrees. To find this probability, we integrate the PDF \( f(x) \) over the interval from 75 to 81 degrees, expressed mathematically as:

\[P(c < X < d) = \int_{c}^{d} f(x) \, dx\]

In a practical scenario, consider a piecewise function defined as \( f(x) = \frac{3}{2}x^2 + 2x^3 \) for \( x \) in the interval [0, 1], and \( f(x) = 0 \) otherwise. To verify that this function is a valid PDF, we first check that \( f(x) \geq 0 \) for all \( x \). Since the function is non-negative within the specified interval and zero elsewhere, it satisfies the first condition. Next, we compute the integral of \( f(x) \) from 0 to 1:

\[\int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{1} \left( \frac{3}{2}x^2 + 2x^3 \right) \, dx\]

Applying the power rule for integration, we find:

\[\int_{0}^{1} \left( \frac{3}{2}x^2 + 2x^3 \right) \, dx = \left[ \frac{1}{2}x^3 + \frac{1}{2}x^4 \right]_{0}^{1} = \frac{1}{2} + \frac{1}{2} = 1\]

This confirms that the function meets the second condition for being a PDF. Now, to find the probability that \( X \) lies between 0.1 and 0.3, we set up the integral:

\[P(0.1 < X < 0.3) = \int_{0.1}^{0.3} f(x) \, dx = \int_{0.1}^{0.3} \left( \frac{3}{2}x^2 + 2x^3 \right) \, dx\]

Using the power rule again, we evaluate this integral from 0.1 to 0.3:

\[= \left[ \frac{1}{2}x^3 + \frac{1}{2}x^4 \right]_{0.1}^{0.3}\]

Calculating the upper and lower bounds gives us:

\[= \left( \frac{1}{2}(0.3)^3 + \frac{1}{2}(0.3)^4 \right) - \left( \frac{1}{2}(0.1)^3 + \frac{1}{2}(0.1)^4 \right)\]

After performing the arithmetic, we find that the probability is approximately 0.017. This process illustrates how to use a probability density function to compute probabilities for continuous random variables effectively.

Probability Density Function (PDF) Example 1

In this analysis of the probability density function \( f(x) = \frac{3}{2}x^2 + 2x^3 \) defined on the interval \([0, 1]\), we first confirm that it is indeed a valid probability density function. This is established by verifying that the function is non-negative and that the integral over its entire range equals one.

To find the probability that \( x \) equals 0.5, we recognize that for continuous random variables, the probability of \( x \) taking on any specific value is zero. This is because the probability density function provides probabilities over intervals, not discrete points. Therefore, when we set up the integral with both upper and lower bounds at 0.5, we find:

\[P(X = 0.5) = \int_{0.5}^{0.5} f(x) \, dx = 0\]

Next, we calculate the probability that \( x \) is less than or equal to 0.6. Since the function is defined only on the interval from 0 to 1, we can express this probability as:

\[P(X \leq 0.6) = \int_{0}^{0.6} f(x) \, dx\]

Substituting the function into the integral, we have:

\[P(X \leq 0.6) = \int_{0}^{0.6} \left( \frac{3}{2}x^2 + 2x^3 \right) \, dx\]

Applying the power rule of integration, we find the antiderivative:

\[\int f(x) \, dx = \frac{1}{2}x^3 + \frac{1}{2}x^4\]

Evaluating this from 0 to 0.6 gives:

\[P(X \leq 0.6) = \left[ \frac{1}{2}(0.6)^3 + \frac{1}{2}(0.6)^4 \right] - \left[ \frac{1}{2}(0)^3 + \frac{1}{2}(0)^4 \right]\]

Calculating the upper bound:

\[= \frac{1}{2}(0.216) + \frac{1}{2}(0.1296) = 0.108 + 0.0648 = 0.1728\]

Thus, the probability that \( x \) is less than or equal to 0.6 is \( 0.1728 \). This illustrates how to work with continuous probability density functions and reinforces the understanding that probabilities for specific values are zero, while probabilities over intervals can be calculated through integration.

Probability Density Function (PDF) Example 2

In this analysis, we explore the lifetime of a rechargeable battery modeled by the probability density function (PDF) given by \( f(x) = 2(1 - x) \) for \( 0 \leq x \leq 1 \) and \( f(x) = 0 \) otherwise. To confirm that this function is indeed a probability density function, we must verify two key properties.

First, we check that \( f(x) \geq 0 \) for all \( x \). Within the interval from 0 to 1, the expression \( 2(1 - x) \) remains non-negative since \( 1 - x \) is positive or zero. Outside this interval, \( f(x) = 0 \), which is also non-negative. Thus, the first property is satisfied.

Next, we need to ensure that the integral of \( f(x) \) over its entire range equals 1. This is calculated as follows:

\[\int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{1} 2(1 - x) \, dx\]

Calculating this integral, we can factor out the 2:

\[= 2 \int_{0}^{1} (1 - x) \, dx = 2 \left[ x - \frac{1}{2} x^2 \right]_{0}^{1} = 2 \left( 1 - \frac{1}{2} \right) = 2 \cdot \frac{1}{2} = 1\]

Since both properties are satisfied, \( f(x) \) is confirmed as a valid probability density function.

Next, we determine the probability that the battery lasts at least 0.75 years. This is represented as:

\[P(X \geq 0.75) = \int_{0.75}^{1} f(x) \, dx = \int_{0.75}^{1} 2(1 - x) \, dx\]

Using the previously integrated function, we evaluate:

\[= \left[ 2x - x^2 \right]_{0.75}^{1} = \left( 2(1) - (1)^2 \right) - \left( 2(0.75) - (0.75)^2 \right)\]

Calculating this gives:

\[= (2 - 1) - (1.5 - 0.5625) = 1 - 0.9375 = 0.0625\]

Thus, the probability that the battery lasts at least 0.75 years is 0.0625.

Finally, we find the probability that the battery lasts between 0.2 and 0.6 years:

\[P(0.2 < X < 0.6) = \int_{0.2}^{0.6} f(x) \, dx = \int_{0.2}^{0.6} 2(1 - x) \, dx\]

Using the same integrated function, we evaluate:

\[= \left[ 2x - x^2 \right]_{0.2}^{0.6} = \left( 2(0.6) - (0.6)^2 \right) - \left( 2(0.2) - (0.2)^2 \right)\]

Calculating this gives:

\[= (1.2 - 0.36) - (0.4 - 0.04) = 0.84 - 0.36 = 0.48\]

Therefore, the probability that the battery lasts between 0.2 and 0.6 years is 0.48.

Probability Density Function (PDF) Example 3

To determine the value of \( k \) that makes the function \( f(x) = kx^3 \) a probability density function (PDF) on the interval [1, 3], we need to ensure two key properties are satisfied. First, the function must be non-negative for all \( x \) in the specified interval, and second, the integral of the function over the interval must equal 1.

For the first property, since \( x^3 \) is positive for \( x \) in [1, 3], \( k \) must also be a positive value to keep \( f(x) \) non-negative. Thus, we conclude that \( k \geq 0 \).

Next, we focus on the second property, which requires us to evaluate the integral:

\[\int_{1}^{3} kx^3 \, dx = 1\]

We can factor out the constant \( k \) from the integral:

\[k \int_{1}^{3} x^3 \, dx = 1\]

The integral of \( x^3 \) is calculated as follows:

\[\int x^3 \, dx = \frac{1}{4} x^4\]

Now, we evaluate this from 1 to 3:

\[\left[ \frac{1}{4} x^4 \right]_{1}^{3} = \frac{1}{4} (3^4) - \frac{1}{4} (1^4) = \frac{1}{4} (81 - 1) = \frac{80}{4} = 20\]

Substituting this back into our equation gives:

\[k \cdot 20 = 1\]

To solve for \( k \), we divide both sides by 20:

\[k = \frac{1}{20}\]

Thus, the value of \( k \) that makes \( f(x) = kx^3 \) a valid probability density function on the interval [1, 3] is \( \frac{1}{20} \).

Cumulative Distribution Function (CDF)

In the study of probability, we often encounter the concept of probability density functions (PDFs), which describe the likelihood of a continuous random variable taking on a specific value within an interval. To enhance our understanding of probabilities associated with continuous random variables, we introduce the cumulative distribution function (CDF), denoted as \( F(x) \). The CDF is defined as the integral of the PDF from negative infinity to a specific value \( x \), mathematically expressed as:

\[ F(x) = \int_{-\infty}^{x} f(t) \, dt \]

Here, \( f(t) \) represents the PDF, and \( t \) is a dummy variable used for integration. The CDF provides the probability that the random variable \( X \) is less than or equal to \( x \), contrasting with the PDF, which gives the probability of \( X \) falling within a specific interval.

To find the probability that \( X \) lies between two values \( c \) and \( d \), we can utilize the CDF by calculating the difference:

\[ P(c < X < d) = F(d) - F(c) \]

This relationship stems from the fundamental theorem of calculus, linking the CDF and PDF. Notably, the CDF has several important properties: it is always non-decreasing, bounded between 0 and 1, and the derivative of the CDF returns the PDF wherever the PDF is continuous.

To illustrate the application of the CDF, consider a specific PDF given by:

\[ f(x) = 3x^2 e^{-x^3} \quad \text{for } x \geq 0 \quad \text{and } 0 \text{ otherwise} \]

To derive the CDF, we integrate the PDF from 0 to \( x \) (since the PDF is zero for negative values):

\[ F(x) = \int_{0}^{x} 3t^2 e^{-t^3} \, dt \]

Using a substitution where \( u = -t^3 \) leads to \( du = -3t^2 dt \), we can rewrite the integral as:

\[ F(x) = -\int e^{u} \, du = -e^{-t^3} \bigg|_{0}^{x} = -e^{-x^3} + 1 \]

Thus, the CDF is:

\[ F(x) = 1 - e^{-x^3} \quad \text{for } x \geq 0 \]

Next, to find the probability that \( X \) is between 1 and 3, we compute:

\[ P(1 \leq X \leq 3) = F(3) - F(1) \]

Calculating these values gives:

\[ F(3) = 1 - e^{-27} \quad \text{and} \quad F(1) = 1 - e^{-1} \]

Thus, the probability becomes:

\[ P(1 \leq X \leq 3) = (1 - e^{-27}) - (1 - e^{-1}) = e^{-1} - e^{-27} \approx 0.368 \]

This result indicates the probability that the continuous random variable \( X \) falls between 1 and 3. Understanding the CDF and its properties is crucial for effectively calculating probabilities in continuous distributions.

Cumulative Distribution Function (CDF) Example 4

In this scenario, we are examining the time students require to complete a standardized test, which is represented by a probability density function (PDF) defined as \( f(t) = 2t \) for \( 0 \leq t < 1 \). To find the cumulative distribution function (CDF), denoted as \( F(x) \), we need to compute the integral of the PDF from negative infinity to \( x \). However, since the PDF is only defined for \( t \geq 0 \), we can simplify our limits of integration to \( 0 \) to \( x \).

The CDF is calculated as follows:

\[F(x) = \int_{0}^{x} f(t) \, dt = \int_{0}^{x} 2t \, dt\]

Using the power rule for integration, we find:

\[F(x) = \left[ t^2 \right]_{0}^{x} = x^2\]

Thus, the cumulative distribution function is \( F(x) = x^2 \) for \( 0 \leq x < 1 \).

Next, we can use this CDF to determine the probability that a randomly selected student completes the test in 45 minutes or less. Since time is measured in hours, we convert 45 minutes to hours, which is \( 0.75 \) hours. We are interested in finding \( P(X \leq 0.75) \), which can be expressed using the CDF:

\[P(X \leq 0.75) = F(0.75) - F(0)\]

Substituting the values into the CDF:

\[P(X \leq 0.75) = F(0.75) - F(0) = (0.75)^2 - (0)^2 = 0.5625\]

Therefore, the probability that a randomly chosen student finishes the test in 45 minutes or less is \( 0.5625 \), or 56.25%.

Cumulative Distribution Function (CDF) Example 5

To derive the probability density function (PDF), denoted as \( f(x) \), from the cumulative distribution function (CDF), denoted as \( F(x) \), we start with the given piecewise function for \( F(x) \). The CDF is defined as follows:

• \( F(x) = 0 \) for \( x < 0 \)
• \( F(x) = \frac{x^2}{4} \) for \( 0 \leq x \leq 2 \)
• \( F(x) = 1 \) for \( x > 2 \)

To find the PDF, we differentiate the CDF with respect to \( x \). The differentiation process is as follows:

1. For \( x < 0 \):

Since \( F(x) = 0 \), the derivative is:

\[f(x) = \frac{d}{dx}(0) = 0\]

2. For \( 0 \leq x \leq 2 \):

We differentiate \( F(x) = \frac{x^2}{4} \) using the power rule:

\[f(x) = \frac{d}{dx}\left(\frac{x^2}{4}\right) = \frac{2x}{4} = \frac{1}{2}x\]

3. For \( x > 2 \):

Since \( F(x) = 1 \), the derivative is:

\[f(x) = \frac{d}{dx}(1) = 0\]

Combining these results, we can express the PDF in a concise piecewise format:

• \( f(x) = \frac{1}{2}x \) for \( 0 \leq x \leq 2 \)
• \( f(x) = 0 \) otherwise

Thus, the probability density function \( f(x) \) derived from the cumulative distribution function \( F(x) \) is:

\[f(x) = \begin{cases} \frac{1}{2}x & \text{if } 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}\

Find the cumulative distribution function for the probability density function $f\left(x\right)=\frac{3}{8}{(x-1)}^2$ on the interval $[1,3]$. Use the CDF to find the probability that $X$ is between $2$ and $2.5$.

• A. $P(2\le X\le 2.5)=0.2969$
• B. $P(2\le X\le 2.5)=0.0625$
• C. $P(2\le X\le 2.5)=0.8906$
• D. $P(2\le X\le 2.5)=0.0234$

Answer: (A)

Here’s what students ask on this topic:

What is a probability density function (PDF) and what are its key properties?

A probability density function (PDF) is a mathematical function that describes the likelihood of a continuous random variable taking on a value within a specific interval. The key properties of a PDF are:

• Non-Negativity: The function must be greater than or equal to zero for all values of the random variable, ensuring probabilities are never negative.
• Normalization: The integral of the PDF over its entire range (from negative infinity to positive infinity) must equal 1, representing the total probability of all possible outcomes.

These properties ensure the PDF accurately models probabilities for continuous random variables. To calculate the probability of the variable falling within a specific interval, you compute the definite integral of the PDF over that interval.

How do you calculate probabilities using a cumulative distribution function (CDF)?

To calculate probabilities using a cumulative distribution function (CDF), you use the relationship between the CDF and the probability density function (PDF). The CDF, denoted as F(x), gives the probability that a random variable X is less than or equal to a specific value x. To find the probability that X lies within an interval [c, d], you compute:

$F_d-F_c$

This is equivalent to finding the difference between the CDF values at d and c. The CDF is derived by integrating the PDF from negative infinity to x, and its properties ensure it is always between 0 and 1 and non-decreasing.

What is the relationship between a PDF and a CDF?

The relationship between a probability density function (PDF) and a cumulative distribution function (CDF) is fundamental in probability theory. The CDF, denoted as F(x), is the integral of the PDF, f(x), from negative infinity to x:

$F\left(x\right)={\int }_{-}^{\infty }f\left(t\right)dt$

The CDF represents the probability that the random variable X is less than or equal to x. Conversely, the PDF is the derivative of the CDF wherever the PDF is continuous. This relationship allows you to move between the two functions depending on whether you need probabilities for intervals (PDF) or cumulative probabilities (CDF).

How do you verify if a function is a valid probability density function?

To verify if a function is a valid probability density function (PDF), you must check two key conditions:

• Non-Negativity: Ensure the function is greater than or equal to zero for all values of the random variable. This guarantees probabilities are not negative.
• Normalization: Compute the integral of the function over its entire range (typically from negative infinity to positive infinity). The result must equal 1, representing the total probability of all possible outcomes:

${\int }_{-}^{\infty }f\left(x\right)dx=1$

If both conditions are satisfied, the function is a valid PDF.

What are the key properties of a cumulative distribution function (CDF)?

The cumulative distribution function (CDF) has several key properties:

• Range: The CDF is always between 0 and 1 for all values of the random variable, as it represents cumulative probabilities.
• Non-Decreasing: The CDF never decreases because probabilities accumulate as the variable increases.
• Relationship to PDF: The derivative of the CDF is the PDF wherever the PDF is continuous. Conversely, the CDF is the integral of the PDF from negative infinity to x.
• Limits: As x approaches negative infinity, the CDF approaches 0, and as x approaches positive infinity, the CDF approaches 1.

These properties make the CDF a useful tool for calculating probabilities and understanding the distribution of a random variable.