What is the difference and when do u use a binomial pdf vs cdf?
binompdf is used to calculate the probability of obtaining a specific value in a binomial distribution. For example, finding the probability that somebody's height is 168 using a range of data. binomcdf is used to find the probability of getting a value between the lowest possible value (negative infinity) and the value that you go up to. This is used, for example, for finding the probability that somebody's height is LESS than 168 Easy way to remember is that cdf = cumulative distribution frequency
What is the technique for determining the distribution of a dataset?
Great Question!!This question lies at the heart of the arithmetic of uncertainty. Different practitioners will advocate for different approaches. Here is one way to think of it, however if this is an academic question, ignore all this and do what your professor says.Step 1 - do you really need a named distribution? Can you just sample from your dataset? If so, you might want to use this - Home (Probability Management - free download for Excel - SIPMath tools).Step 2 - if you must use something named for a dead mathematician, do you know about the source of the data? If it comes from adding up uniform distributions (like the game of craps), then the distribution will be something from a triangle to a Gaussian. If it comes from a set of true/false choices, it might be a binomial.Step 3 - If you got this far, you have two choices.3a - If you don’t care about classic names, and you just want a good fit, then the Metalog (also found at Probability Management) is a great alternative. Metalog is the most flexible distribution because it approximates any data set. Another name for this is the Keelan distribution for Tom Keelan (a VERY smart guy).3b - if you need to use a classic name, and you got this far, then start with how lopsided the distribution is - look at a graph. There are many names for being lopsided, but what ever you call it, that eliminates many choices either way. From there it boils down to how well you hope to fit the dataset.Again, there are many ways to do this, and your toolset may make some other approach better. Ignore anyone who says there is one “best” way.
Statistics, determine the CDF?
This is an instructive question. Part b) first: The CDF - Cumulative Distribution Function is the integal of the PDF as follows. x>0. Fx(x) = P (X
Use the cdf technique to find the pdf of U=(X-.5)^2 when f(x)=4x^3?
P(U <= u) = P((x-.5)^2 <= u) = P( -√u < (x-5) < √u) = P( 0.5 - √u < x < 0.5 + √u) = 4∫ x^3 from (0.5 -√u) to (0.5 + √u) = x^4 from (0.5 -√u) to (0.5 + √u) F(u) = (0.5 + √u)^4 - (0.5 -√u)^4 F'(u) = 4(0.5 + √u)^3 (1/2√u) - 4(0.5 - √u)^3 (-1/2√u) F'(u) = 4(0.5 + √u)^3 (1/2√u) + 4(0.5 - √u)^3 (1/2√u) F'(u) = (2/√u) [ (0.5 + √u)^3 + (0.5 - √u)^3] , 0 < u < 0.25 f(u) = (2/√u) [ (0.5 + √u)^3 + (0.5 - √u)^3] , 0 < u < 0.25 --------------------------------------... As x varies from 0 to 1, (x-0.5)^2 varies from 0 to 0.25
In statistics, when do you know whether to use binomial cdf or pdf?
You use the binomial pdf {probability density function} if you are trying to find the probability of exactly x occurances of random variable X given n trials. You use the binomial cdf {cummulative distribution function} if you are trying to find the probability of x or FEWER occurances of an event given n trials. so pdf Pr{X= x} cdf Pr{X≤x} --------------------------------------... You asked how would you find Pr {X> x} All you have to do is use the properties of probability Pr {X> x} = 1 - Pr{X≤x} and you get the quantity after the subtraction sign , as we have said by using the binomial cumulatitive probability distribution function cdf
What is the difference between a CDF and a PDF?
When a random variable (r.v.) can take any value over a range (finite or infinite), then its distribution is modelled using its Probability Density Function (PDF). An example of such a r.v. would be the height of students in a class; the height of a particular student could take any value between some reasonable interval.Because the r.v. can take any value over a range, effectively an infinite number of possibilities, the probability that it will take a particular value is 0. Therefore, for a continous r.v. we define the notion of infinitisimal[1] size intervals. Take the example of heights of students in a class. Lets say that the height of all the students is between 160 cm and 170 cm. It makes no sense to ask the probability that the height of the student is 165 cm; that probability is zero (for reasons explained above). We can, however, define the probability that the height of the student lies in the infinitisimal interval around 165 cm. The function that gives this probability density is referred to as the probability density function or PDF.A cumulative probability function or CDF is defined over any interval where the PDF is defined. Suppose a PDF is defined over the interval [a,b] and let [math]a
How does one integrate the gauss distribution formula and manually solve for a p-value without the z-table?
If you mean manually, it involves a LOT of work, because you would have to use numerical methods such as Riemann sums, and you would have to use series representation to calculate the function being integrated. This means you would be making thousands of repetitive calculations to get the value. This is why the tables were designed in the first place.Of course, with computers, thousands of repetitive calculations can be made in a second. In addition to the error function, there is some computer software that can perform high precision numerical integration for you.There is a free computer software calculator I like to use that performs numerical integration called Precise Calculator - programmable scientific calculator. On that calculator, I could calculate the CDF of the standard normal distribution by entering this command:z=1;0.5+integral(x,0,z,15,exp(-x^2/2)/sqrt(2*pi))The value z=1 at the start could be changed to whatever z-score you want to compute the CDF for. The number 15 inside the integral function is how many digits of precision you want, numerical integration can get very slow for large numbers of digits like 25, but if you are willing to wait for it, you can do so.My computer performs that calculation in 0.155 seconds, giving a result of [math]\Phi(1) = 0.841344746068543[/math]Another computer option would be using Microsoft Excel. There is a function in excel that will give you a very good estimate of the standard normal CDF. Using the example of z=1 again, you would enter this:=NORM.S.DIST(1,TRUE)Which would return the value 0.841344746068543Notice that is the same answer given by the numerical integration above. However, the advantage of the numerical integration is you can get more precision if you are willing to wait for it.
How do you get the CDF of the Laplace distribution from the PDF?
The cumulative distribution function is the integral of the probability density function. Since the pdf of the Laplace distribution is f(x) := (1/(2b)) exp(- |x - μ| / b), the cdf is F(x) = integral_{t ≤ x} (1/(2b)) exp(- |t - μ| / b). If x ≤ μ, then |t - μ| = μ - t over the range of integration, so F(x) = integral_{t ≤ x} (1/(2b)) exp((t - μ) / b) = (1/2) exp((x - μ)/b). (*) If x ≥ μ, then F(x) can be computed by breaking the range of integration into two intervals, (-∞, μ] and [μ, x]. From (*), we see that the integral over (-∞, μ] is 1/2. Over the second interval, |t - μ| = t - μ, so F(x) = 1/2 + integral_{μ ≤ t ≤ x} (1/(2b)) exp((μ - t) / b) = 1/2 + (-(1/2) exp((μ - x) / b) + 1/2) = 1 - (1/2) exp((μ - x) / b). Therefore, F(x) = (1/2) exp((x - μ) / b), if x ≤ μ, F(x) = 1 - (1/2) exp((μ - x) / b), if x ≥ μ.
How do I find the CDF of Gaussian distribution?
The probability density function of a Gaussian with mean [math]\mu[/math] and standard deviation [math]\sigma[/math] is:[math]f(x \; | \; \mu, \sigma^2) = \frac{1}{\sigma\sqrt{2\pi} } \; e^{ -\frac{(x-\mu)^2}{2\sigma^2} }[/math]The CDF [math]F(x \; | \; \mu, \sigma^2)[/math] is the integral of [math]f(x \; | \; \mu, \sigma^2)[/math] :[math]F(x \; | \; \mu, \sigma^2) = \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^x e^{ -\frac{(\chi-\mu)^2}{2\sigma^2}} d\chi [/math]Unfortunately this integral doesn’t have a closed form solution and is only solved by numerical methods. It is written often by using the error function [math]\erf()[/math] which is calculated with numerical approximations.So one usually consult tables of a normalized Gaussian distribution with mean 0 and unitary standard deviation, [math]F(x\; | \; 0,1)[/math], after using a variable transformation from the generic [math]\mu[/math] and [math]\sigma[/math]. This is taught at Normal distribution - WikipediaThere is a section at that same Wikipedia article where some numerical approximations of [math]f(x)[/math] and [math]F(x)[/math] are given which guarantee a very small error along an interval of [math]\pm 9 \sigma[/math] of the Gaussian.