how to find lambda in exponential distributionhow to find lambda in exponential distribution

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Understanding the average [], The mask calculator is a tool that helps you decide whether or not you need to wear a mask. The "Rule of 70" refers to the totaltime it takes to double a quantity or value. WebFind the maximum likelihood estimator of \lambda of the exponential distribution, f(x) = \lambda e^{-\lambda x). This estimator can be considered as good. E(\hat\lambda) = & E\left(\frac{1}{\bar X}\right) = E\left(\frac{n}{\sum X_i}\right)= E\left(\frac{n}{y}\right)\\ is the scale parameter which is the reciprocal of the mean value. I have seven steps to conclude a dualist reality. x : the value (s) of the variable and, rate : rate parameter of exponential distribution. rev2023.4.5.43379. . 3.146 c. 3.149 d. Let X be an exponentially distributed random variable with parameter lambda = 1 / 2. Now, calculate the probability function at different values of x to derive the distribution curve. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Now consider 40 random variables, independent and identically distributed to Y. What is the standard deviation of X? $$, $$ Define Y = [X + 1] (viz., the integ . Even though it is almost the same as exponential distribution, we usually called negative due to the negative sign of the exponent. Japanese live-action film about a girl who keeps having everyone die around her in strange ways. . 3. Arcu felis bibendum ut tristique et egestas quis: Suppose \(X\), following an (approximate) Poisson process, equals the number of customers arriving at a bank in an interval of length 1. , Xn form a random sample from a Normal distribution for which mean μ is known, but the variance σ^2 is unknown. Hint: consider the t. Let Xi, i = 1, 2, . \implies \hat\lambda =& \frac{n}{\sum x} = \frac{1}{\bar x}\end{aligned} Thus, the density of X is: f (x,)=ex for 0x,=0.25. is what R calls

Given that Y = y, the random variable X has a Poisson distribution with mean y, Let X_n, n = 1,2 ,N be a set of N uncorrelated normal (Gaussian) random variables with mean mu_n and variance sigma ^2_n Let Y = summation _n=1 ^N a_n X_n, where a_n are real constants. curl --insecure option) expose client to MITM. Let X be an exponential random variable with parameter \lambda =2. The probabilit. It only takes a minute to sign up. . The mean of \(X\) is \(\displaystyle{\text{E}[X]= \frac{1}{\lambda}}\). Find the unconditional mean and variance of Y. We thus aim to obtain a parameter which will maximize the likelihood. Suppose X_1, . random variables from an exponential (lambda) distribution for i = 1, . Mean Squared Error (MSE) is a common metric for measuring the average squared difference between the predicted and actual values in a regression problem. If \(X\sim\text{exponential}(\lambda)\), then the following hold. Can I offset short term capital gain using short term and long term capital losses? Learn more about Stack Overflow the company, and our products.

We define it as the reciprocal of the scale parameter and indicate how quickly decay of the exponential function occurs. ,Xn of size n is taken from a Poisson distribution with a mean of \lambda, 0 is less than \lambda is less than \infty. for \(0c__DisplayClass228_0.b__1]()", "4.2:_Expected_Value_and_Variance_of_Continuous_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.3:_Uniform_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.4:_Normal_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.5:_Exponential_and_Gamma_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.6:_Weibull_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.7:_Chi-Squared_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.8:_Beta_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1:_What_is_Probability" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "2:_Computing_Probabilities" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "3:_Discrete_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4:_Continuous_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "5:_Probability_Distributions_for_Combinations_of_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "showtoc:yes", "authorname:kkuter" ], https://stats.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fstats.libretexts.org%2FCourses%2FSaint_Mary's_College_Notre_Dame%2FMATH_345__-_Probability_(Kuter)%2F4%253A_Continuous_Random_Variables%2F4.5%253A_Exponential_and_Gamma_Distributions, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\).

Let Y_1, Y_2, . \begin{aligned} Ahmed is a hard-working Software engineering student, and has a passion toward mobile app and system development. I'm trying to solve a problem but I'm not sure where to go with it: Y is an exponential random variable with parameter . Lastly, you would like to look at the MSE of your estimator. Determine the probability distribution function of the random variable Y = X / 2. Variance is one of the properties of an exponential distribution. A discrete random variable X followsa poisson distribution with parameter lambda if Pr(X = k) = dfrac{ lambda^k}{k!}. Let N have a Poisson distribution with mean \lambda and let the conditional distribution of X given N = n be binomial with n trials and success probability p; that is, P(N = n) = e^{-\lambda} \lambd. Suppose that, conditioned on N=n, the random variable X is binomially distributed with parameters N=n and p. Set Y=N-X. Also, x is a continuous random variable.

It determines the wait time for the occurrence, success, or failure of an event. Suppose X and Y are independent. The term how to find a good estimator is quite broad. Plotting your data on log-log paper , Xn).

Suppose X_1, , X_n is a random sample from a normal distribution with mean theta and variance theta where theta greater than 0 is an unknown parameter. Suppose X and Y are independent random variables where X has exponential distribution with parameter lambda = 4 and Y has gamma distribution with parameters Alpha = 2 and lambda = 4. a. For example, suppose the mean number of customers to arrive at a bank in a 1-hour interval is 10. \frac{g^{\prime}(\lambda)^{2}}{n I(\lambda)}=\frac{1 / \lambda^{4}}{n \lambda^{2}}=\frac{1}{n \lambda^{2}} But what exactly do we consider as a good estimator? All that is left is to input these values to our calculator or the given formula. Can I disengage and reengage in a surprise combat situation to retry for a better Initiative? describes the time between events in [] a process in which events Show that Y is, (a) Prove that the variance of the Poisson distribution is Var[X] = lambda (derive that equation) Var[X] = E[(X - E [X])^2] = lambda. lambda is just the inverse of your mean, in is case, 1/5. x_n! (a) Let X be a Poisson random variable with variance lambda.

This memoryless random distribution facilitates the estimation of an events occurrence, success, or failure. Thanks for contributing an answer to Cross Validated! 0 & \text{otherwise,} If mu = lambda, what, Find by MLE method the estimator of the parameter of the following distribution based on a random sample of size n, when f(x, \lambda) = \lambda exp(-\lambda x), \lambda is greater than 0, and 0 \leq, Suppose that the random variable X has a Weibull distribution with parameters alpha = 23 and lambda = 1.7 (where lambda = frac{1}{ beta} ). The median formula in statistics is used to determinethe middle number in a data set that is arranged in ascending order. Previously, our focus would have been on the discrete random variable \(X\), the number of customers arriving. $$\int^{\infty}_0 \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x} dx = \int^{\infty}_0 \frac{\lambda \lambda^{\alpha-1}}{\Gamma(\alpha)} x^{\alpha-1}e^{-\lambda x} dx = \frac{1}{\Gamma(\alpha)}\int^{\infty}_0 u^{\alpha-1}e^{-u} du = \frac{1}{\Gamma(\alpha)}\Gamma(\alpha) = 1.

The skewness of the exponential distribution does not rely upon the value of the parameter A. Suppose X is a Poisson distributed random variable with parameter lambda. a. But, if the selected time interval is roughly constant, then the exponential distribution can be used as a good approximate model. The cumulative distribution function for Y is: F(t, Let X_1, . ) is the digamma function.

(a) What are E[X] and E[X2 ]? exponential order statistics, Sum of two independent exponential random variables, complementary cumulative distribution function, the only memoryless probability distributions, Learn how and when to remove this template message, bias-corrected maximum likelihood estimator, Relationships among probability distributions, "Calculating CVaR and bPOE for common probability distributions with application to portfolio optimization and density estimation", "Maximum entropy autoregressive conditional heteroskedasticity model", "The expectation of the maximum of exponentials", NIST/SEMATECH e-Handbook of Statistical Methods, "A Bayesian Look at Classical Estimation: The Exponential Distribution", "Power Law Distribution: Method of Multi-scale Inferential Statistics", "Cumfreq, a free computer program for cumulative frequency analysis", "Frequentist predictions intervals and predictive distributions", Universal Models for the Exponential Distribution, Online calculator of Exponential Distribution, https://en.wikipedia.org/w/index.php?title=Exponential_distribution&oldid=1147097347, Infinitely divisible probability distributions, Articles with unsourced statements from September 2017, Articles lacking in-text citations from March 2011, Creative Commons Attribution-ShareAlike License 3.0, The exponential distribution is a limit of a scaled, Exponential distribution is a special case of type 3, The exponential distribution is a limit of the, Exponential distribution is a limit of the, The time it takes before your next telephone call, The time until default (on payment to company debt holders) in reduced-form credit risk modeling, a profile predictive likelihood, obtained by eliminating the parameter, an objective Bayesian predictive posterior distribution, obtained using the non-informative. Book where Earth is invaded by a future, parallel-universe Earth. I can't seem to find an explanation anywhere. Let M be the median of X . Step 3 - Click on Calculate button to calculate exponential probability. Inprobabilitytheory andstatistics, it describes the time between events happening in a process that occurs independently and at a constant average rate, in other words,the Poisson process. The continuous random variable \(X\) follows an exponential distribution if its probability density function is: Because there are an infinite number of possible constants \(\theta\), there are an infinite number of possible exponential distributions. A random variable (Y) has the exponential distribution, so its density function is: f(y) = \lambda \exp (-y) , for y > 0, and f (y) = 0, elsewhere. . Aprobabilitydistribution, such as exponential distribution, is uniquely determined by its mtf. Well this might confuse you. Whenever there is a case of 'At most' take all the outcomes which are either equal to the given and less than that. Sa . Find a. Definition ofmeanprobability and statistics is that it is an average of a dataset, and we express it with a symbol . b. Exponential distribution probability function For x=0 will be, Similarly, calculate exponential distribution probability function for x=1 to x=30. . The events should occur continuously and should be independent of each other. Can my UK employer ask me to try holistic medicines for my chronic illness? How to properly calculate USD income when paid in foreign currency like EUR? Split a CSV file based on second column value. The variance of \(X\) is \(\displaystyle{\text{Var}(X)= \frac{\alpha}{\lambda^2}}\). You'll find the area is 1/lambda. You are free to use this image on your website, templates, etc., Please provide us with an attribution linkHow to Provide Attribution?Article Link to be HyperlinkedFor eg:Source: Exponential Distribution (wallstreetmojo.com). WebThe syntax to compute the probability density function for Exponential distribution using R is. For instance, it can be used to determine the approximate time it will take for a consumer to make a purchase. The mean and variance of Y . Lambda provides us with an indication of the strength of the relationship between independent and dependent variables.As an asymmetrical measure of association, lambdas value may vary (b)Is \bar{X_n} =T/n a minimum variance unbias. What is the variance of this estimator? Learn the meaning and definition of the mean squared error (MSE). Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. To learn more, see our tips on writing great answers. Memorylessness property of exponential random variable, Conditional Value at Risk (Expected Shortfall), Buffered Probability of Exceedance (bPOE), Distribution of the minimum of exponential random variables, Joint moments of i.i.d. You'll find the area is 1/lambda. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. Show: \(\displaystyle{\int^{\infty}_0 \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x} dx = 1}\), In the integral, we can make the substitution: \(u = \lambda x \rightarrow du = \lambda dx\). Using the CLT, compute the probability that the average of these 40 r.v. Suppose X1, . Recall that the mean and variance of Poisson( \lambda) are both \lambda. Given a Poisson distribution with The result is that there is 48% chance that you will wait less than 2 minutes for the next customer. dexp (x,rate=1) where. Compute a. Identification of the dagger/mini sword which has been in my family for as long as I can remember (and I am 80 years old). Determine the (a) mean and (b) variance of X. I need formulas to calculate it. Let (bar)X_n denote the sample mean. \ln f(x \mid \lambda)=\ln \lambda-\lambda x, \quad \frac{\partial^{2} f(x \mid \lambda)}{\partial \lambda^{2}}=-\frac{1}{\lambda^{2}} \implies& E\left(\frac{n-1}{n}\hat\lambda\right) = \lambda What is an exponential probability distribution? Show the mean and variance are given by the fol. As the picture suggests, however, we could alternatively be interested in the continuous random variable \(W\), the waiting time until the first customer arrives.

A random sample X1, X2, , Xn of size n is taken from a Poisson distribution with a mean of lambda greaterthan 0. It is calculated using integration by parts, and the formula is \frac{1}{\Lambda} . In standard tuning, does guitar string 6 produce E3 or E2? Improving the copy in the close modal and post notices - 2023 edition, statistics: probability, normal distribution, Probability and Statistics Binomials distribution, Computing expected cost for exponential random variable, Statistics: "hybrid" of Poisson and Binomial distribution. a) Find the method of, A random sample X1, X2, , Xn of size n is taken from a Poisson distribution with a mean of lambda, 0 lessthan lambda lessthan infinity. .

Let X have exponential distribution with rate parameter lambda and Y have exponential distribution with rate parameter mu. Ordinarily, we say that the random variable $X$ has exponential distribution with parameter $\lambda$ if $X$ has density function $\lambda e^{-\lambda x}$ (for positive $x$). Let W be an exponential random variable with parameter lambda. , Xn form a random sample with Bernoulli distribution with parameter p unknown . If Y is a Poisson rv with λ = 16, what are E(Y) and V(Y)?

For a censo, I want to know how to show if the estimator is a consistent estimator for parameter theta? Other such examples would be: The fundamental formulas for exponential distribution analysis allow you to determine whether the time between two occurrences is less than or more than X, the target time interval between events: Our calculator also includes more values: mean \; = \frac{1}{a}. a number of cars that will pass in a minute. Should we always use 100 samples for an equivalence test given the KS test size problems? Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. \implies& E\left(\frac{n-1}{n}\hat\lambda\right) = \lambda \ Lambda is sometimes also called the rate perimeter, and it determines the constant average rate at which events should happen. WebIf lambda is an exponential value of a random variable having a Gamma distribution with alpha = 2 and beta = 1, compute the P (X = 1). What must the constant c equal if c(2Y1 + 3Y2) is an unbiased estimator of 1/theta? As its name suggests, we use the moment generating function (mgf) to compute themomentsof adistribution. \end{array}\right.

\begin{aligned} It is always better to understand the theory of the probability The assumption of a regular rate is very rarely satisfied in real-world scenarios. Where is the additional lambda while solving problems for an Exponential distribution? a. 1. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. 98.765). The exponential family of distributions is a very rich family from which you can select distributions with nearly any shape you want. They have som P(x X) = 1 - exp(-ax) => P(x 2) = 1 - exp(-0.33 \cdot 2) = 0.48. Find the probability P [Z less than or equal to 1]. c) P(0.5 less th, Let X_1, , X_n be a random sample from the Poisson distribution with parameter lambda = 1. It has great practical importance, mainly because we can use it to derive moments; itsderivativesat 0 are identical to the moments of a random variable. . Let Z = max(X,Y). In calculating the conditional probability, the exponential distribution "forgets" about the condition or the time already spent waiting and you can just calculate the unconditional probability that you have to wait longer. We prove Properties #1 & #3, the others are left as an exercise. E(\hat\lambda) = & E\left(\frac{1}{\bar X}\right) = E\left(\frac{n}{\sum X_i}\right)= E\left(\frac{n}{y}\right)\\ We express it as Var(x)=\frac{1} {\Lambda^2}. First, decide whether the event under consideration is continuous and independent. (b) Find the probability mass, Let X and Y be two independent Gaussian random variables with means 0 and variances 1, each. f(y) = theta e^-theta y, y greaterthan 0). b. $$f(x) = \left\{\begin{array}{l l} Suppose that X1, . $$ It means you have a variable that ranges from [math]0[/math] to [math]\infty[/math] and the probability of observing a value greater than [math]c[/ Adjust S to form an, Random variables X and Y are independent exponential random variables with expected values E(X) = 1/lambda and E(Y) = 1/mu. Required fields are marked *. . Does disabling TLS server certificate verification (E.g. Let N have a Poisson distribution with parameter lambda > 0. (b) Let X be an exponential random variable with mean 1. Random variables X1,, Xn are independent binary (0,1) variables with parameter p such that p = Pr(x. The continuous probability distribution is used for time modeling, reliability modeling, and service time modeling. Why is my multimeter not measuring current?

No quarrel with that. In exponential distribution, it is the same asthe mean. Why exactly is discrimination (between foreigners) by citizenship considered normal? Suppose that X_1, X_2, are IID standard normal random variables, and Z_i = Xi^2 for each i - therefore, each Zi is a chi1 squared with mean 1 and variance 2. Is there a connector for 0.1in pitch linear hole patterns? \begin{aligned} Why does the right seem to rely on "communism" as a snarl word more so than the left? Except where otherwise noted, content on this site is licensed under a CC BY-NC 4.0 license. Identify the type of this distribution. This article has been a guide to Exponential Distribution. You could also look at Consistency, Asymptotic Normality and even Robustness. Figure 1: Graph of pdf for exponential(\(\lambda=5\)) distribution. . , n be nidentical independent exponentially distributed random variables with parameter λ.

Is this a fallacy: "A woman is an adult who identifies as female in gender"? Find the likelihood ratio test for testing H_0 : lambda = lambda_0 vs H_a : lambda no. The expected value of an exponential distribution, Moment generating function of exponential distribution. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. , Xn form a random sample with Normal distribution with mean μ and variance σ^2, both parameters are unknown. Recall:\quad& \sum X_i = y \sim \Gamma(\alpha=n, \beta = \lambda) \text{ where } \beta\text{ is the rate parameter}\\ If you have already been waiting 5 minutes at the bus stop, the probability that you have to wait 4 more minutes (so more than 9 minutes total) is equal to the probability that you only had to wait more than 4 minutes once arriving at the bus stop. Often we assume an underlying distribution and put forth the claim that data follows the given distribution. Our estimator above is biased. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Now, $\bar{X}$ is an unbiased estimator for $h(\lambda)=1 / \lambda$ with variance How many unique sounds would a verbally-communicating species need to develop a language? Find the UMVUE (uniformly minimum-variance unbiased estimator) for p. Let X have a Poisson distribution with parameter lambda. It is clear that the CNML predictive distribution is strictly superior to the maximum likelihood plug-in distribution in terms of average KullbackLeibler divergence for all sample sizes n > 0. Probability distribution could be defined as the table or equations showing respective probabilities of different possible outcomes of a defined event or scenario. How to use the exponential distribution calculator?

Find the distribution of each of them. b) Variance of X can be calculated using the formula: Var (X) = 1/^2. I CBA to get my calculator as I am in bed, but I'll tell you how to do it. So consider the first student taking a pen. The probability of them taki This article will provide information about the concept of the exponential distribution,its formula,examples, and how to use it inreallife. What is Lambda in exponential distribution? Define U = X + Y and V = X - Y . First, if \(x<0\), then the pdf is constant and equal to 0, which gives the following for the cdf: Let X, Y be independent and identically distributed exponential random variables with mean lambda. Statistics is the science behind identifying, collecting, organizing and summarizing, analyzing, interpreting, and finally, presenting such data, either qualitative or quantitative, which helps make better and effective decisions with relevance. A sample of size two, Y1, Y2, is taken from a distribution with pdf.

Lambda . The standard formula for it is ^2 = \frac{1}{a^2}. How do you find lambda exponential distribution? We have by the definition of a median : Pr ( X < M) = 1 0 M e x d x = 1 2. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. There are two parametrizations of the exponential distribution. The most common is [math]\begin{align} \begin{cases} f(x) &= \lambda e^{-\lambda x} The mean of \(X\) is \(\displaystyle{\text{E}[X]= \frac{\alpha}{\lambda}}\). WebInstructions: Compute exponential distribution probabilities using the form below. You can learn more about statistical modeling from the articles below , Your email address will not be published. , Y_n constitute a random sample from a Poisson distribution with mean lambda. Its designed for people who work []. For all pdf's, \(\lambda=5\). X is also an exponential random variable and independent of Y with .. Find the PDF , where .

In statisticsStatisticsStatistics is the science behind identifying, collecting, organizing and summarizing, analyzing, interpreting, and finally, presenting such data, either qualitative or quantitative, which helps make better and effective decisions with relevance.read more, the exponential distribution function determines the constant rate of time-lapse between the occurrence of two independent and continuous events.