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4.1.4 Solved Problems: Continuous Random Variables
- Find the constant $c$.
- Find $EX$ and Var$(X)$.
- Find $P(X \geq \frac{1}{2})$.
- To find $P(X \geq \frac{1}{2})$, we can write $$P(X \geq \frac{1}{2})=\frac{3}{2} \int_{\frac{1}{2}}^{1} x^2dx=\frac{7}{16}.$$
Let $X$ be a continuous random variable with PDF given by $$f_X(x)=\frac{1}{2}e^{-|x|}, \hspace{20pt} \textrm{for all }x \in \mathbb{R}.$$ If $Y=X^2$, find the CDF of $Y$.
Let $X$ be a continuous random variable with PDF \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} 4x^3 & \quad 0 \frac{1}{3})$.
Let $X$ be a continuous random variable with PDF \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} x^2\left(2x+\frac{3}{2}\right) & \quad 0 < x \leq 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} If $Y=\frac{2}{X}+3$, find Var$(Y)$.
First, note that $$\textrm{Var}(Y)=\textrm{Var}\left(\frac{2}{X}+3\right)=4\textrm{Var}\left(\frac{1}{X}\right), \hspace{15pt} \textrm{using Equation 4.4}$$ Thus, it suffices to find Var$(\frac{1}{X})=E[\frac{1}{X^2}]-(E[\frac{1}{X}])^2$. Using LOTUS, we have $$E\left[\frac{1}{X}\right]=\int_{0}^{1} x\left(2x+\frac{3}{2}\right) dx =\frac{17}{12}$$ $$E\left[\frac{1}{X^2}\right]=\int_{0}^{1} \left(2x+\frac{3}{2}\right) dx =\frac{5}{2}.$$ Thus, Var$\left(\frac{1}{X}\right)=E[\frac{1}{X^2}]-(E[\frac{1}{X}])^2=\frac{71}{144}$. So, we obtain $$\textrm{Var}(Y)=4\textrm{Var}\left(\frac{1}{X}\right)=\frac{71}{36}.$$
Let $X$ be a positive continuous random variable. Prove that $EX=\int_{0}^{\infty} P(X \geq x) dx$.
Probability Density Function
Continuous distributions are constructed from continuous random variables which take values at every point over a given interval and are usually generated from experiments in which things are “measured” as opposed to “counted”.
With continuous distributions, probabilities of outcomes occurring between particular points are determined by calculating the area under the probability density function (pdf) curve between those points. In addition, the entire area under the whole curve is equal to 1.
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14.1 - probability density functions.
A continuous random variable takes on an uncountably infinite number of possible values. For a discrete random variable \(X\) that takes on a finite or countably infinite number of possible values, we determined \(P(X=x)\) for all of the possible values of \(X\), and called it the probability mass function ("p.m.f."). For continuous random variables, as we shall soon see, the probability that \(X\) takes on any particular value \(x\) is 0. That is, finding \(P(X=x)\) for a continuous random variable \(X\) is not going to work. Instead, we'll need to find the probability that \(X\) falls in some interval \((a, b)\), that is, we'll need to find \(P(a<X<b)\). We'll do that using a probability density function ("p.d.f."). We'll first motivate a p.d.f. with an example, and then we'll formally define it.
Example 14-1 Section
Even though a fast-food chain might advertise a hamburger as weighing a quarter-pound, you can well imagine that it is not exactly 0.25 pounds. One randomly selected hamburger might weigh 0.23 pounds while another might weigh 0.27 pounds. What is the probability that a randomly selected hamburger weighs between 0.20 and 0.30 pounds? That is, if we let \(X\) denote the weight of a randomly selected quarter-pound hamburger in pounds, what is \(P(0.20<X<0.30)\)?
In reality, I'm not particularly interested in using this example just so that you'll know whether or not you've been ripped off the next time you order a hamburger! Instead, I'm interested in using the example to illustrate the idea behind a probability density function.
Now, you could imagine randomly selecting, let's say, 100 hamburgers advertised to weigh a quarter-pound. If you weighed the 100 hamburgers, and created a density histogram of the resulting weights, perhaps the histogram might look something like this:
In this case, the histogram illustrates that most of the sampled hamburgers do indeed weigh close to 0.25 pounds, but some are a bit more and some a bit less. Now, what if we decreased the length of the class interval on that density histogram? Then, the density histogram would look something like this:
Now, what if we pushed this further and decreased the intervals even more? You can imagine that the intervals would eventually get so small that we could represent the probability distribution of \(X\), not as a density histogram, but rather as a curve (by connecting the "dots" at the tops of the tiny tiny tiny rectangles) that, in this case, might look like this:
Such a curve is denoted \(f(x)\) and is called a (continuous) probability density function .
Now, you might recall that a density histogram is defined so that the area of each rectangle equals the relative frequency of the corresponding class, and the area of the entire histogram equals 1. That suggests then that finding the probability that a continuous random variable \(X\) falls in some interval of values involves finding the area under the curve \(f(x)\) sandwiched by the endpoints of the interval. In the case of this example, the probability that a randomly selected hamburger weighs between 0.20 and 0.30 pounds is then this area:
Now that we've motivated the idea behind a probability density function for a continuous random variable, let's now go and formally define it.
The probability density function (" p.d.f. ") of a continuous random variable \(X\) with support \(S\) is an integrable function \(f(x)\) satisfying the following:
\(f(x)\) is positive everywhere in the support \(S\), that is, \(f(x)>0\), for all \(x\) in \(S\)
The area under the curve \(f(x)\) in the support \(S\) is 1, that is:
\(\int_S f(x)dx=1\)
If \(f(x)\) is the p.d.f. of \(x\), then the probability that \(x\) belongs to \(A\), where \(A\) is some interval, is given by the integral of \(f(x)\) over that interval, that is:
\(P(X \in A)=\int_A f(x)dx\)
As you can see, the definition for the p.d.f. of a continuous random variable differs from the definition for the p.m.f. of a discrete random variable by simply changing the summations that appeared in the discrete case to integrals in the continuous case. Let's test this definition out on an example.
Example 14-2 Section
Let \(X\) be a continuous random variable whose probability density function is:
\(f(x)=3x^2, \qquad 0<x<1\)
First, note again that \(f(x)\ne P(X=x)\). For example, \(f(0.9)=3(0.9)^2=2.43\), which is clearly not a probability! In the continuous case, \(f(x)\) is instead the height of the curve at \(X=x\), so that the total area under the curve is 1. In the continuous case, it is areas under the curve that define the probabilities.
Now, let's first start by verifying that \(f(x)\) is a valid probability density function.
What is the probability that \(X\) falls between \(\frac{1}{2}\) and 1? That is, what is \(P\left(\frac{1}{2}<X<1\right)\)?
What is \(P\left(X=\frac{1}{2}\right)\)?
It is a straightforward integration to see that the probability is 0:
\(\int^{1/2}_{1/2} 3x^2dx=\left[x^3\right]^{x=1/2}_{x=1/2}=\dfrac{1}{8}-\dfrac{1}{8}=0\)
In fact, in general, if \(X\) is continuous, the probability that \(X\) takes on any specific value \(x\) is 0. That is, when \(X\) is continuous, \(P(X=x)=0\) for all \(x\) in the support.
An implication of the fact that \(P(X=x)=0\) for all \(x\) when \(X\) is continuous is that you can be careless about the endpoints of intervals when finding probabilities of continuous random variables. That is:
\(P(a\le X\le b)=P(a<X\le b)=P(a\le X<b)=P(a<x<b)\)
for any constants \(a\) and \(b\).
Example 14-3 Section
\(f(x)=\dfrac{x^3}{4}\)
for an interval \(0<x<c\). What is the value of the constant \(c\) that makes \(f(x)\) a valid probability density function?
Probability Density Function
Probability density function is a function that provides the likelihood that the value of a random variable will fall between a certain range of values. We use the probability density function in the case of continuous random variables. For discrete random variables, we use the probability mass function which is analogous to the probability density function.
The graph of a probability density function is in the form of a bell curve. The area that lies between any two specified values gives the probability of the outcome of the designated observation. We solve the integral of this function to determine the probabilities associated with a continuous random variable. In this article, we will do a detailed analysis of the probability density function and take a look at the various aspects related to it.
What is Probability Density Function?
Probability density function and cumulative distribution function are used to define the distribution of continuous random variables. If we differentiate the cumulative distribution function of a continuous random variable it results in the probability density function. Conversely, on integrating the probability density function we get the cumulative distribution function.
Probability Density Function Definition
Probability density function defines the density of the probability that a continuous random variable will lie within a particular range of values. To determine this probability, we integrate the probability density function between two specified points.
Probability Density Function Example
Say we have a continuous random variable whose probability density function is given by f(x) = x + 2, when 0 < x ≤ 2. We want to find P(0.5 < X < 1). Then we integrate x + 2 within the limits 0.5 and 1. This gives us 1.375. Thus, the probability that the continuous random variable lies between 0.5 and 1 is 1.375.
Probability Density Function Formula
The probability density function of a continuous random variable is analogous to the probability mass function of a discrete random variable. Discrete random variables can be evaluated at a particular point while continuous random variables have to be evaluated between a certain interval . This is because the probability that a continuous random variable will take an exact value is 0. Given below are the various probability density function formulas.
Probability Density Function of Continuous Random Variable
Suppose we have a continuous random variable, X. Let F(x) be the cumulative distribution function of X. Then the formula for the probability density function, f(x), is given as follows:
f(x) = \(\frac{\mathrm{d} F(x)}{\mathrm{d} x}\) = F'(x)
If we want to find the probability that X lies between lower limit 'a' and upper limit 'b' then using the probability density function this can be given as:
P(a < X ≤ b) = F(b) - F(a) = \(\int_{a}^{b}f(x)dx\)
Here, F(b) and F(a) represent the cumulative distribution function at b and a respectively.
Probability Density Function of Normal Distribution
Normal distribution is the most widely used type of continuous probability distribution. The notation for normal distribution is given as \(X \sim N(\mu ,\sigma ^{2})\). The probability density function of a normal distribution is given below.
f(x) = \(\frac{1}{\sigma \sqrt{2\Pi}}e^{\frac{-1}{2}\left ( \frac{x - \mu }{\sigma } \right )^{2}}\)
Here, \(\mu\) is the mean and \(\sigma\) is the standard deviation while \(\sigma\) 2 is the variance.
Probability Density Function Graph
If X is a continuous random variable then the probability distribution of this variable is given by the probability density function. The graph given below depicts the probability that X will lie between two points a and b.
Mean of Probability Density Function
In the case of a probability density function, the mean is the expected value or the average value of the random variable. If f(x) is the probability density function of the random variable X, then mean is given by the following formula:
E[X] = \(\mu = \int_{-\infty }^{\infty}xf(x)dx\)
Median of Probability Density Function
The median is the value that splits the probability density function curve into two equal halves. Suppose x = m is the value of the median. The area under the curve from \(-\infty\) to m will be equal to the area under the curve from m to \(\infty\). This implies that the value of the median is 1 / 2. Thus, the median of the probability density function is given as follows:
\(\int_{-\infty }^{m}f(x)dx = \int_{m}^{\infty }f(x)dx\) = 1/2
Variance of Probability Density Function
The expected value of the squared deviation from the mean is the variance of a random variable. Expressing this definition mathematically we get,
Var(X) = \(E[(X - \mu )^{2}]\)
To represent this variance with the help of the probability density function, the formula is given as:
Var(X) = \(\sigma ^{2} = \int_{-\infty }^{\infty }(x - \mu )^{2}f(x)dx\)
Properties of Probability Density Function
The properties of the probability density function help to solve questions faster. If f(x) is the probability distribution of a continuous random variable, X, then some of the useful properties are listed below:
- f(x) ≥ 0. This implies that the probability density function for all real numbers can be either equal to or greater than 0. But it can never be negative or lesser than 0.
- \(\int_{-\infty }^{\infty}f(x)dx\) = 1. Thus, the total area under the probability density curve will be equal to 1.
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Important Notes on Probability Density Function
- Probability density function determines the probability that a continuous random variable will fall between a range of specified values.
- On differentiating the cumulative distribution function, we obtain the probability density function.
- The mean of the probability density function can be give as E[X] = \(\mu = \int_{-\infty }^{\infty}xf(x)dx\).
- As the median divides the probability density function curve into 2 equal halves, its value will be equal to 1 / 2.
- The variance of a probability density function is given by Var(X) = \(\sigma ^{2} = \int_{-\infty }^{\infty }(x - \mu )^{2}f(x)dx\)
Examples on Probability Density Function
- Example 1: If the probability density function is given as: f(x) = \(\left\{\begin{matrix} x(x-1) &0\leq x < 3 \\ x& x\geq3 \end{matrix}\right.\) Find P(1< X < 2). Solution: Integrating the function, \(\int_{1}^{2}x(x-1)dx\) = \(\int_{1}^{2}(x^{2}-x)dx\) = \([\frac{x^{3}}{3} - \frac{x^{2}}{2}]_{1}^{2}\) = 5 / 6 Answer: P(1< X < 2) = 5 / 6
- Example 2: If X is a continuous random variable with the probability density function given as: f(x) = \(\left\{\begin{matrix} \frac{be^{-x}}{2} & x\geq 0\\ 0 & \text{otherwise} \end{matrix}\right.\) Find the value of b. Solution: We know from properties of probability density function that \(\int_{-\infty }^{\infty}f(x)dx\) = 1 \(\int_{0}^{\infty }\frac{be^{-x}}{2}dx\) = 1 On integrating, \(\frac{b}{2}\left [ -e^{-x} \right ]_{0}^{\infty }\) = 1 b / 2 = 1 b = 2 Answer: b = 2
- Example 3: Find the expected value of X if the probability density function is defined as: f(x) = \(\left\{\begin{matrix} \frac{3}{2}x^{2} & 0\leq x\leq 2\\ 0& \text{otherwise} \end{matrix}\right.\) Solution: We know that, E[X] = \(\int_{-\infty }^{\infty}xf(x)dx\) = \(\int_{-\infty }^{0}x(0)dx + \int_{0}^{2}x (\frac{3x^{2}}{2})dx + \int_{2}^{\infty }x(0)dx\) = \(\frac{3}{2}[\frac{x^{4}}{4}]_{0}^{2}\) = (3/8) [16] = 6 Answer: Mean = 6
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Practice Questions on Probability Density Function
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FAQs on Probability Density Function
What is meant by the probability density function.
Probability density function is a function that is used to give the probability that a continuous random variable will fall within a specified interval. The integral of the probability density function is used to give this probability.
What is the Probability Density Function Formula?
We can differentiate the cumulative distribution function (cdf) to get the probability density function (pdf). This can be given by the formula f(x) = \(\frac{\mathrm{d} F(x)}{\mathrm{d} x}\) = F'(x). Here, f(x) is the pdf and F'(x) is the cdf.
How to Calculate the Probability Density Function?
To calculate the probability density function we differentiate the cumulative distribution function. If we integrate the probability density function, we get the probability that a continuous random variable lies within a certain interval.
How to Find the Mean of Probability Density Function?
The expected value is also known as the mean. The mean of the probability density function is given by the formula \(\mu = \int_{-\infty }^{\infty}xf(x)dx\).
Is Probability Density Function Always Positive?
The value of the integral of a probability density function will always be positive. This is because probability can never be negative hence, as a consequence, the probability density function can also never be negative.
How Do You Find the Probability of a Probability Density Function?
To find the probability that a continuous random variable X, falls between an interval a and b we use the probability density function, f(x). This formula is given as P(a < X < b) = \(\int_{a}^{b}f(x)dx\)
What are the Features of Probability Density Function?
The features of the probability density function are given below:
- The probability density function will always be a positive value.
- The total area under the probability density function curve will always be equal to 1.
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Probability Density Function
What is Probability Density Function?
The probability density function gives the output indicating the density of a continuous random variable lying between a specific range of values. If a given scenario is calculated based on numbers and values, the function computes the density corresponding to the specified range.
The function, when solved, defines the relationship between the random variable and its probability. The variable’s probability is matched when the function is solved. The function is denoted by f(x), and the graphical portrayal gives a bell curve. The area between any two particular values gives the outcome probability of the designated observation.
Table of contents
- #1 – Discrete variable
- #2 – Continuous Variable
Examples with Calculation
Applications, frequently asked questions (faqs), recommended articles, key takeaways.
- The probability density function (PDF) gives the output indicating the density of a continuous random variable lying between a specific range of values.
- There are imperatively two types of variables: discrete and continuous. The PDF turns into the probability mass function when dealing with discrete variables.
- It is used in statistical calculations and graphically represented as a bell curve forming a relationship between the variable and its probability.
- Its application is significant in machine learning algorithms, analytics, probability theory, neural networks, etc.
Probability Density Function Explained
The probability density function (PDF) is associated with a continuous random variable by finding the probability that falls in a specific interval. A continuous random variable can take an uncountably infinite number of possible values. The probability mass function replaces the PDF for a discrete random variable that takes on finite or countable possible values.
PDFs have a wide range of applications. For example, it is used in modeling and predictions related to chemically reactive turbulent flows and analysis of stock price returns. For each application, the corresponding curve is depicted on a graph, and analyzing bell curve features like symmetry and sides like left or right gives important information.
Another important concept significant in understanding the PDF is the cumulative distribution function. It is also used to explain the distribution of random variables, primarily continuous random variables. The differentiating cumulative distribution function of a continuous random variable will give the value of PDF, and integrating the PDF gives the value of the cumulative distribution function.
Probability density function formula :
To calculate the PDF online probability density function calculator or formula based on cumulative distribution function is used, we differentiate the cumulative distribution function:
f(x) is the PDF and F(x) is the CDF
- X lies between lower limit ‘a’ and upper limit ‘b’
- F(b): Cumulative distribution function at a
- F(a): Cumulative distribution function at b
To better understand the formula and its application, consider the following PDF example:
The PDF is:
Find P(1<x<2)
The formula is:
Applying the upper and lower limits:
- It is used in machine learning algorithms, analytics, probability theory, neural networks, etc.
- Calculates probabilities associated with continuous random variables.
- The PDF of stock price returns is used for studying various market scenarios.
- It is used to model various processes and derive solutions to the problem. For example, it is applied to model chemically reacting turbulent flows and derive appropriate resolution to the closure problems.
The function, joint pdf, denotes the probability distribution of two or more continuous random variables, which together form a continuous random vector. If two random variables have a joint PDF, they are jointly continuous. Its calculation involves the application of multiple integrals.
Consider an example with PDF, f(x) = x + 3, when 1 < x ≤ 3. We have to find P(2 < X < 3). Integrating x + 3 within the limits 2 and 3 gives the answer 5.5.
This is a Guide to What is Probability Density Function (PDF) and its definition. We explain formulas, calculations, applications, examples & joint PDF. You can learn more from the following articles –
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If you are flipping a coin, it is pretty easy to see that the probability of getting a head is \(0.5\). But what if you want to find the probability of someone being exactly \(2\) metres tall? Height is a continuous variable, not a discrete one, so you can't use the basic probability rules you might already know. Instead, you will need a probability density function . So don't be dense, read on to find out about continuous random variables and probability density functions!
Probability mass function and probability density function
You might think that because the names 'probability mass function' and 'probability density function' are so close that they really describe the same thing. Both of them do describe probabilities, and both are functions. The big difference is in what kind of random variable they are used with:
If \(X\) is a discrete random variable, then use a probability mass function, which is a summation.
If \(X\) is a continuous random variable, then use a probability density function, which is an integral.
Going forward you will see information and examples involving the probability density function for a continuous random variable \(X\). If you are interested in probability mass functions, check out the article Discrete Probability Distributions or the article on the Poisson Distribution.
Probability density function graph
First of all, what is a probability density function?
The probability density function , or PDF, of a continuous random variable \(X\) is an integrable function \(f_X(x)\) satisfying the following:
- \(f_X(x) \ge 0\) for all \(x\) in \(X\); and
\(\displaystyle \int_X f_X(x) \, \mathrm{d} x = 1\).
Then the probability that \(X\) is in the interval \([a,b]\) is \[ P(a<X<b) = \int_a^b f_X(x) \, \mathrm{d} x .\]
That looks more complicated than it actually is. Let's relate it to the graph of a function.
Take the function
\[ f_X(x) = \begin{cases} 0.1 & 1 \le x \le 11 \\ 0 & \text{otherwise} \end{cases} \]
as seen in the graph below.
Let's check it for the properties of a probability density function. It is certainly at least always zero. The area under the curve is \(1\) since that area is just a rectangle with height \(0.1\) and width \(10\). And lastly, you can represent the probability as an area. For example, if you wanted to find \(P(5<X<7)\) you could do so by finding the area of the rectangle in the graph below, getting that \(P(5<X<7) = 0.2\).
So \(f_X(x)\) is a probability density function. If you were to graph the probability curve, you would need to integrate it, giving you
\[ P(a<X<b) = \begin{cases} 0 & a \text{ and } b \le 1 \\ 0.2(b-1) & a<1<b \le 11 \\ 0.2(b-a) & 1 \le a \le b \le 11 \\ 0.2(11-a) & 1 <a <11 < b \\ 1 & 11 \le a < b \end{cases} \]
That certainly seems like a lot of cases, but you can see it much more easily by looking at the graph below.
Notice that the minimum height of the graph above is \(0\), and the maximum height of the graph is \(1\). This makes sense because probabilities are always at least zero and at most one.
It turns out that the integral of the probability density function is quite useful, and it is called the Cumulative Distribution Function.
Probability density function properties
Using the definition of the probability density function, you can see an important property of them:
\[P(X=a) = 0.\]
It also doesn't matter if you use strict inequalities with continuous density functions:
\[ P(X<a) = P(X\le a).\]
Both of those properties come from the fact that
\[ P(a<X<b) = \int_a^b f_X(x) \, \mathrm{d} x .\]
You might ask if the probability density function can be greater than \(1\). Sure it can! The integral of the function still needs to be equal to \(1\), but the probability density function can take on values larger than that as long as it is also at least zero. One example of this is the probability density function
\[ f_X(x) = \begin{cases} 2 & 0 \le x \le \dfrac{1}{2} \\ 0 & \text{otherwise} \end{cases} .\]
This function is always at least zero, it is integrable, and the integral is \(1\), so it could be a probability density function for a continuous random variable \(X\). Don't confuse the probability density function for actual probabilities!
Probability density function of normal distribution
One of the probability density functions you will see often is the normal distribution. You can see the graph of the standard normal distribution probability density function below.
Just like with other probability density functions, the area under the curve of the standard normal distribution is \(1\).
Probability density function example
Let's look at some examples.
Suppose that someone tells you that
\[ f_X(x) = \begin{cases} 2x & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}\]
is the probability density function for the length of time, in hours, you will spend waiting in the doctor's office.
(a) Check to be sure this is a probability density function.
(b) Find the probability you will wait less than half an hour to see the doctor.
(c) Find the probability you will wait more than half an hour to see the doctor.
(a) First note that \(X\) is in fact a continuous random variable. In addition, \(f_X(x)\) is always at least zero. It is also integrable, so now it just remains to check that the integral is one. Doing the integration,
\[\begin{align} \int_X f_X(x) \, \mathrm{d} x &= \int_0^1 2x \, \mathrm{d} x \\ &= \left. 2\left(\frac{1}{2}\right)x^2\right|_0^1 \\ &= 1^2-0^2 \\ &= 1.\end{align}\]
So this is, in fact, a probability density function.
(b) You want to know the probability that you will wait less than half an hour. In other words, you need to find \(P(X<0.5)\). Then
\[\begin{align} P(X<0.5) &= \int_0^{0.5} 2x \, \mathrm{d} x \\ &= \left.\phantom{\frac{1}{2}} x^2 \right|_0^{0.5} \\ &= (0.5)^2 - 0^2 \\ &=0.25. \end{align}\]
So the probability that you will wait less than half an hour is \(0.25\). So \(25\%\) of the time, you will wait less than half an hour to see the doctor.
(c) Now you want to find the probability that you will wait more than half an hour to see the doctor. Remember that the area under the probability density function is \(1\), so
\[ P(X > 0.5) = 1 - P(X<0.5).\]
Then using the previous part of the problem, \(P(X> 0.5) =0.75\). This means that \(75\%\) of the time, you will wait at least half an hour to see the doctor!
Probability Density Function - Key takeaways
- A probability mass function is used with discrete random variables, and a probability density function is used with continuous random variables.
The probability density function, or PDF, of a continuous random variable \(X\) is an integrable function \(f_X(x)\) satisfying the following:
The probability that a continuous random variable \(X\) is in the interval \([a,b]\) is \[ P(a<X<b) = \int_a^b f_X(x) \, \mathrm{d} x .\]
For a continuous random variable \(X\), \(P(X=a) = 0\), and it doesn't matter if you use strict inequalities: \( P(X<a) = P(x \le a)\).
Frequently Asked Questions about Probability Density Function
--> what is pdf and cdf .
The PDF is the probability density function and the CDF is the cumulative distribution function.
--> What is probability density function give example?
An example of a probability density function for a continuous random variable would be the standard normal distribution.
--> Can probability density function be negative?
No. They are always at least zero.
--> What does the probability density function tell us?
A probability density function can tell you the probability of a continuous random variable being within a certain range.
--> Can a probability density function be greater than 1?
Yes. Remember that the area under the probability density function is 1. As long as that is satisfied, and the function is at least zero, it can take on values larger than 1.
Final Probability Density Function Quiz
Suppose you are rolling a \(20\) sided die in a game. You could use a ____ to model the probability of rolling \(13\) twice in a row.
Show answer
Probability mass function.
Show question
Suppose you want to model the probability that someone will spend at least \(2\) minutes paying attention to an advertisement. To do this you would use a ____.
Probability density function.
Probability density functions are used with ____ random variables.
Continuous.
What three things do you need to check to be sure that \(f_X(x)\) is a probability density function for the continuous random variable \(X\)?
\(f_X(x)\) is always at least zero.
\(f_X(x)\) is integrable.
Which of the following is true about a probability density function for a continuous random variable?
The area under the curve is equal to \(1\).
True or False: The probability density function for a continuous random variable can't be larger than one.
True or False: The probability density function for a continuous random variable must be larger than zero.
True or False: For a continuous random variable \(X\), \( P(a\le X \le b) = P(a<X <b)\).
The probability that a continuous random variable \(X\) is in the interval \([a,b]\) is found using the formula ____.
\( P(a<X<b) = \displaystyle\int_a^b f_X(x) \, \mathrm{d} x\).
Which of the following are properties of the probability density function \(f_X(x)\) of a continuous random variable \(X\)?
\[ f(x) = \begin{cases} 1 & 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}\]
a probability density function for a continuous random variable \(X\)?
No. The area under the curve of \(f(x)\) is not equal to \(1\).
\[ f(x) = \begin{cases} 2 & 0 \le x \le \frac{1}{2} \\ 0 & \text{otherwise} \end{cases}\]
Yes. It satisfies all three properties of a probability density function.
\[ f(x) = \begin{cases} \frac{1}{2} & 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}\]
For a probability density function, what is \(P(X=a)\)?
For a probability mass function, what is \(P(X=a)\)?
Not enough information to tell.
- Mechanics Maths
- Probability and Statistics
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Section 8.5 : Probability
- Show that \(f\left( x \right)\) is a probability density function.
- Find \(P\left( {X \le 7} \right)\).
- Find \(P\left( {X \ge 7} \right)\).
- Find \(P\left( {3 \le X \le 14} \right)\).
- Determine the mean value of \(X\).
- Verify that \(f\left( t \right)\) is a probability density function.
- What is the probability that a light bulb will have a life span less than 8 months?
- What is the probability that a light bulb will have a life span more than 20 months?
- What is the probability that a light bulb will have a life span between 14 and 30 months?
- Determine the mean value of the life span of the light bulbs.
- Determine the value of \(c\) for which the function below will be a probability density function. \[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{c\left( {8{x^3} - {x^4}} \right)}&{{\mbox{if }}0 \le x \le 8}\\0&{{\mbox{otherwise}}}\end{array}} \right.\] Solution
IMAGES
VIDEO
COMMENTS
Problem. Let X be a continuous random variable with PDF given by fX(x)=12e−|x|,for all x∈R. If Y=X2, find the CDF of Y. Solution.
a) Verify that f is a probability density function. b) What is the probability that x is greater than 4. c) What is the probability that x is between 1 and 3
Solution. In reality, I'm not particularly interested in using this example just so that you'll know whether or not you've been ripped
In this video lecture you will know the relationship between probability and probability density function (PDF). This problem on probability
Say we have a continuous random variable whose probability density function is given by f(x) = x + 2, when 0 < x ≤ 2. We want to find P(0.5 < X < 1). Then we
Let the joint probability density function for (X, Y) be f(x, y) = 2 yx. +. , x > 0, y > 0,. 3x + y < 3, zero otherwise. a). Find the probability P(X < Y).
Consider an example with PDF, f(x) = x + 3, when 1 < x ≤ 3. We have to find P(2 < X < 3). Integrating x + 3 within the limits 2 and 3 gives the answer 5.5.
An example of a probability density function for a continuous random variable would be the standard normal distribution. Can probability density function be
Determine the mean value of X X . Solution; For a brand of light bulb the probability density function of the life span of the light bulb is
The lifetime of a certain brand of battery, in tens of hours, is modelled by the continuous random variable X with probability density function ( ).