#set document(title: "6.3 Exponential Distribution", author: "Rachel Webb")
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== 6.3#h(0.6em)Exponential Distribution


    An #strong[exponential distribution] models a continuous random variable over time, area or space where the rate of occurrences decreases as X gets larger.


    The probability density function (PDF) for an exponential curve is


    #math.equation(block: true, alt: "λ e to the power minus x λ , for x greater than or equal to 0; 0 , elsewhere")[$λ e^(− x λ) , "for " x ≥ 0 \ 0 , "elsewhere"$]


    The value lambda λ is the fixed rate of occurrence and is equal to one divided by the mean, #math.equation(block: false, alt: "the fraction 1 over μ")[$frac(1, μ)$].


    If the mean is given in the problem then you write the PDF as f(x)= #math.equation(block: false, alt: "the fraction 1 over μ e to the power open parenthesis minus the fraction x over μ close parenthesis")[$frac(1, μ) e^(( − frac(x, μ) ))$], where e is a mathematical constant approximately equal to 2.71828, x ≥ 0 and x is the value you are trying to find the probability for, μ is the mean number of a successes over an interval of time, space, volume, etc. The distribution is denoted as X~Exp(λ).


    Figure 6-8 gives example graphs for a mean of 5, 10 and 20. Note the curve hits the y-axis at 1/μ and keeps going forever to the right with an asymptote at y = 0.


    #figure(figph[Three exponential density curves labeled f(x) = (1/5)e^(-x/5) in blue, f(x) = (1/10)e^(-x/10) in green, and f(x) = (1/20)e^(-x/20) in magenta; they start at 0.2, 0.1, and 0.05 on the y-axis respectively and decay toward zero as x runs to 19.], alt: "Three exponential density curves labeled f(x) = (1/5)e^(-x/5) in blue, f(x) = (1/10)e^(-x/10) in green, and f(x) = (1/20)e^(-x/20) in magenta; they start at 0.2, 0.1, and 0.05 on the y-axis respectively and decay toward zero as x runs to 19.", caption: none)


    Figure 6-8


    You would need integral calculus skills to find the area under this curve. To get around having the calculus requirement, we have three scenarios that we can use to find probability for an exponential distribution where we will not have to use the PDF.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
To find the probability (area) under the exponential curve, use the following formulas.


      - #math.equation(block: false, alt: "P open parenthesis X greater than or equal to x close parenthesis equals P open parenthesis X greater than x close parenthesis equals e to the power minus x / μ")[$P ( X ≥ x ) = P ( X > x ) = e^(− x / μ)$]
- #math.equation(block: false, alt: "P open parenthesis X less than or equal to x close parenthesis equals P open parenthesis X less than x close parenthesis equals 1 minus e to the power minus x / μ")[$P ( X ≤ x ) = P ( X < x ) = 1 − e^(− x / μ)$]
- #math.equation(block: false, alt: "P open parenthesis x sub 1 less than or equal to X less than or equal to x sub 2 close parenthesis equals P open parenthesis x sub 1 less than X less than x sub 2 close parenthesis equals e to the power open parenthesis minus x sub 1 / μ close parenthesis minus e to the power open parenthesis minus x sub 2 / μ close parenthesis")[$P ( x_(1) ≤ X ≤ x_(2) ) = P ( x_(1) < X < x_(2) ) = e^(( − x_(1) / μ )) − e^(( − x_(2) / μ ))$]
]


    #examplebox("Example 1")[][
The time it takes to help a customer at the customer service desk is exponentially distributed with an average help time of 45 seconds. Find the probability that a customer waits less than two minutes.


        \# Help time ~ Exponential, mean = 45 s  (rate = 1/mean)
pexp(120, rate = 1/45)   \# P(X \< 120 s) -\> 0.9305
      
      #solutionbox[
We need to have the same units as the mean in the question so instead of finding P(X \< 2 minutes) we will use P(X \< 120 seconds). Also note that \< and ≤ find the same probabilities so use the equation \\(\\mathrm{P}(X


        #math.equation(block: true, alt: "P open parenthesis X less than 120 close parenthesis equals 1 minus e to the power minus 120 / 45 equals 0.9305.")[$P ( X < 120 ) = 1 − e^(− 120 / 45) = 0.9305 .$]


        In Excel use =EXPON.DIST(x,λ,TRUE) =EXPON.DIST(120,1/45,TRUE) = 0.9305.


        #figure(figph[Exponential distribution web calculator titled X ~ exp(lambda) with rate lambda = 1/45 and x = 120; the output reads P(X \< x) = 0.93052, and the density curve below shades the area to the left of 120 in red over an x-axis from 0 to 300.], alt: "Exponential distribution web calculator titled X ~ exp(lambda) with rate lambda = 1/45 and x = 120; the output reads P(X < x) = 0.93052, and the density curve below shades the area to the left of 120 in red over an x-axis from 0 to 300.", caption: none)


        Figure 6-9


        Alternatively, as shown in Figure 6-9, the following website will calculate the exponential probability: #link("https://homepage.divms.uiowa.edu/~mbognar/applets/exp.html")[https://homepage.divms.uiowa.edu/~mbognar/applets/exp.html].
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