Gamma & Exponential Process are frequently found in the Engineering and Management sciences, particularly in Queuing Models and Simulation of Working of various Electrical and Mechanical components for their life (Reliability Analysis) in case of Exponential distribution.
So First we go with a standard Definition of these two Distributions from Wikipeida:
In probability theory and statistics, the Exponential distribution (a.k.a. negative exponential distribution) is a family of continuous probability distributions. It describes the time between events in a Poisson process, i.e. a process in which events occur continuously and independently at a constant average rate. It is the continuous analogue of thegeometric distributionand...
In probability theory and statistics, the Gamma distribution is a two-parameter family of continuous probability distributions. There are three different parameterizations in common use:since i am not a statistician, so i will stop here leaving further theoretical part to your end! and move on to how we can solve the problems pertaining to these two distributions with MS Excel 2007.
1. With a shape parameter k and a scale parameter θ.
2. With a shape parameter α = k and an inverse scale parameter β = 1/θ, called a rate parameter.
3. With a shape parameter k and a mean parameter μ = k/β.
Syntex of Gamma & Exponential Functions in Excel 2007:
= GAMMA.DIST(x,alpha,beta,cumulative)
X: Is the value at which you want to evaluate the distribution.
Alpha: A parameter to the distribution.
Beta: If beta = 1, GAMMA.DIST returns the standard gamma distribution.
Cumulative: A logical value that determines the form of the function.
= GAMMA.INV(probability,alpha,beta)
Probability: The probability associated with the gamma distribution.
Alpha: A parameter to the distribution.
Beta: If beta = 1, GAMMA.INV returns the standard gamma distribution.
= EXPON.DIST(x,lambda,cumulative)
X: The value of the function.
Lambda: The parameter value.
Cumulative: A logical value that indicates which form of the function to provide.
Example: An
engineer examines the edges of steel girders for hairline fractures. The
girders are 10 m long, and it is discovered that they have an average of 42
fractures each. If a girder has 42 fractures, then there are 43 “gaps” between
fractures or between the ends of the girder and the adjacent fractures. Find
the probability that (a) The probability that a gap is less than 10 cm long. (b) The probability that a gap is longer than 30 cm.
Solution:
According
to this model, the length of a gap between any two adjacent fractures has an
exponential distribution with λ = 4.3, the
probability that a gap is less than 10 cm long is:
P(X ≤ 0.10) = F(0.10) = 1 − e−4.3×0.10 = 0.35
The
probability that a gap is longer than 30 cm is:
P(X ≥ 0.30) = 1 − F(0.30) = e−4.3×0.30 = 0.28
Using
Excel’s EXPONDIST() function, the solution can be reduced to:
1. =EXPONDIST(0.1,4.3,1)
= 0.35
2. =1-EXPONDIST(0.3,4.3,1)
= 0.28
Example: Suppose
that the random variable X measures the length between one end of a girder and the fifth fracture along the girder, as shown in Figure
4.15. If the fracture locations are modeled by a Poisson process as discussed
previously, X has a gamma
distribution with parameters k = 5 and λ = 4.3. Find:
1.
Probability that fifth fracture is at least 46 cm away from the end of the
girder.
2.
Probability that the fifth fracture is within 1 m of the end of the girder
Using Excel's GAMMADIST() function we can easily calculate this:
1. =GAMMADIST(1,5,1/4.3,1) = 0.430
2. =GAMMADIST(0.458,5,1/4.3,1) = 0.050
You can download this file to play with it, its simple!! :)
You can use the GAMMAINV() to reverse the process. Hope you enjoy this post, please give me some feedback on contents of the Material.
Thanks & Take Care
