#set document(title: "6.2 Uniform Distribution", author: "Rachel Webb")
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== 6.2#h(0.6em)Uniform Distribution


    The continuous uniform distribution models the probability that is the same on an interval from a to b. We use the following probability density function (PDF) to graph a straight line.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
f(x)= #math.equation(block: false, alt: "the fraction 1 over b minus a , for a less than or equal to x less than or equal to b; 0 , elsewhere")[$frac(1, b − a) , " for " a ≤ x ≤ b \ 0 , " elsewhere "$]
]


    The probability is found by taking the area between two points within the rectangle formed from the x-axis, between the endpoints a and b, the length, and f(x) = 1/(b-a), the height. When working with continuous distributions it is helpful to draw a picture of the distribution, then shade in the area of the probability that you are trying to find. See Figure 6-5.


    #figure(figph[Uniform density function drawn on axes labeled f(x) and x: a rectangle of height 1/(b-a) sits above the x-axis between the points a and b.], alt: "Uniform density function drawn on axes labeled f(x) and x: a rectangle of height 1/(b-a) sits above the x-axis between the points a and b.", caption: none)


    Figure 6-5


    If a continuous random variable X has a uniform distribution with starting point a and ending point b then the distribution is denoted as X~U(a,b).


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Area of a Rectangle = length\*height
]


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To find the probability (area) under the uniform distribution, 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 open parenthesis the fraction 1 over b minus a close parenthesis times open parenthesis b minus x close parenthesis")[$P ( X ≥ x ) = P ( X > x ) = ( frac(1, b − a) ) · ( b − x )$]
- #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 open parenthesis the fraction 1 over b minus a close parenthesis times open parenthesis x minus a close parenthesis")[$P ( X ≤ x ) = P ( X < x ) = ( frac(1, b − a) ) · ( x − a )$]
- #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 open parenthesis the fraction 1 over b minus a close parenthesis times open parenthesis x sub 2 minus x sub 1 close parenthesis")[$P ( x_(1) ≤ X ≤ x_(2) ) = P ( x_(1) < X < x_(2) ) = ( frac(1, b − a) ) · ( x_(2) − x_(1) )$]
]


    #examplebox("Example 1")[][
The arrival time between trains at a train stop is uniformly distributed between 0 and 15 minutes. A student does not check the schedule and has arrived at the train stop.


        \# Train arrival ~ Uniform(0, 15) minutes
punif(10, 0, 15, lower.tail = FALSE)   \# P(X \> 10)  -\> 0.3333
punif(8, 0, 15) - punif(2, 0, 15)      \# P(2 \<= X \<= 8) -\> 0.4
        + Compute the probability they wait more than 10 minutes.
+ Compute the probability of waiting between 2 and 8 minutes.


      
      #solutionbox[
a) First plug in the endpoints a = 0 and b = 15 into the PDF to get the height of the rectangle. The height is f(x)= #math.equation(block: false, alt: "the fraction 1 over 15 minus 0 equals the fraction 1 over 15")[$frac(1, 15 − 0) = frac(1, 15)$]. Draw and label the distribution with the a, b and the height as in Figure 6- 6. The probability is the area of the shaded rectangle P(X \> 10). Draw a vertical line at x = 10. We want x values that are greater than 10, so shade the area to the right of 10, stopping at b = 15. To find the area of the shaded rectangle in Figure 6-6, we can take the length times the height. The length would be b – a = 15 – 10 = 5 and the height is f(x) = 1/15.


        #figure(figph[Uniform density rectangle at height f(x) = 1/15; the region between x = 10 and x = 15 is shaded blue, representing P(X \> 10).], alt: "Uniform density rectangle at height f(x) = 1/15; the region between x = 10 and x = 15 is shaded blue, representing P(X > 10).", caption: none)


        Figure 6-6


        The area of the shaded rectangle is 5 (#math.equation(block: false, alt: "the fraction 1 over 15")[$frac(1, 15)$]) = #math.equation(block: false, alt: "the fraction 1 over 3")[$frac(1, 3)$] = 0.3333 or P(X \> 10) = 0.3333, which is the probability of waiting more than 10 minutes. Note that this would be the same if we asked P(X ≥ 10) = 0.3333 since there is no area at the line X = 10.


        b) The area will be length times height. Draw the picture and shade the rectangle between 2 and 8, see Figure 6-7. The length is b – a = 8 – 2 = 6 and the height is still f(x) = 1/15. P(2 ≤ X ≤ 8) = 6(#math.equation(block: false, alt: "the fraction 1 over 15")[$frac(1, 15)$]) = 0.4


        #figure(figph[Uniform density rectangle at height f(x) = 1/15 extending to 15; the region between x = 2 and x = 8 is shaded blue, representing P(2 \<= X \<= 8).], alt: "Uniform density rectangle at height f(x) = 1/15 extending to 15; the region between x = 2 and x = 8 is shaded blue, representing P(2 <= X <= 8).", caption: none)


        Figure 6-7
]
]