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9.3: The Uniform Distribution

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    261820

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    The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints.

    The mathematical statement of the uniform distribution is

    \[f(x)=\dfrac{1}{b-a} \text { for } a \leq x \leq b\]

    where \(a=\) the lowest value of \(x\) and \(b=\) the highest value of \(x\).
    Formulas for the theoretical mean and standard deviation are

    \[\mu=\dfrac{a+b}{2} \text { and } \sigma=\sqrt{\dfrac{(b-a)^2}{12}}\]

    Exercise \(\PageIndex{1}\)

    The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.

    a. What is the probability that a person waits fewer than 12.5 minutes?

    Answer

    a. Let \(X=\) the number of minutes a person must wait for a bus. \(a=0\) and \(b=15 . X \sim U(0,15)\). Write the probability density function. \(f(x)=\dfrac{1}{15-0}=\dfrac{1}{15}\) for \(0 \leq x \leq 15\).

    Find \(P(x<12.5)\). Draw a graph.

    \[P(x<k)=(\text { base })(\text { height })=(12.5-0)\left(\dfrac{1}{15}\right)=0.8333\]

    The probability a person waits less than 12.5 minutes is 0.8333 .

    This shows the graph of the function f(x) = 1/15. A horiztonal line ranges from the point (0, 1/15) to the point (15, 1/15). A vertical line extends from the x-axis to the end of the line at point (15, 1/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = 12.5. Figure \(\PageIndex{1}\):
    Exercise \(\PageIndex{2}\)

    b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ.

    Answer

    b. \(\mu=\dfrac{a+b}{2}=\dfrac{15+0}{2}=7.5\). On the average, a person must wait 7.5 minutes. \(\sigma=\sqrt{\dfrac{(b-a)^2}{12}}=\sqrt{\dfrac{(15-0)^2}{12}}=\) 4.3. The Standard deviation is 4.3 minutes.

    Exercise \(\PageIndex{3}\)

    c. Ninety percent of the time, the time a person must wait falls below what value?

    NOTE

    This asks for the 90th percentile.

    Answer

    c. Find the \(90^{\text {th }}\) percentile. Draw a graph. Let \(k=\) the \(90^{\text {th }}\) percentile.

    \[\begin{array}{l}
    P(x<k)=(\text { base })(\text { height })=(k-0)\left(\dfrac{1}{15}\right) \\
    0.90=(k)\left(\dfrac{1}{15}\right) \\
    k=(0.90)(15)=13.5
    \end{array}\]

    The \(90^{\text {th }}\) percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.

    f(X)=1/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1/15 on the y-axis, a vertical upward line from an arbitrary point on the x-axis, and the x and y-axes. A shaded region from points 0-k occurs within this area. The area of this probability region is equal to 0.90. Figure \(\PageIndex{2}\):
    Try It \(\PageIndex{1}\)

    The total duration of baseball games in the major league in a typical season is uniformly distributed between 447 hours and 521 hours inclusive.

    1. Find a and b and describe what they represent.
    2. Write the distribution.
    3. Find the mean and the standard deviation.
    4. What is the probability that the duration of games for a team in a single season is between 480 and 500 hours?

    This page titled 9.3: The Uniform Distribution was last modified on Wed, 18 Jun 2025 17:46:25 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.