9.3: The Uniform Distribution
- Page ID
- 261820
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
\[\mu=\dfrac{a+b}{2} \text { and } \sigma=\sqrt{\dfrac{(b-a)^2}{12}}\]
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
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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 .
Figure \(\PageIndex{1}\):
b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ.
- Answer
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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.
c. Ninety percent of the time, the time a person must wait falls below what value?
NOTE
This asks for the 90th percentile.
- Answer
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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.
Figure \(\PageIndex{2}\):
The total duration of baseball games in the major league in a typical season is uniformly distributed between 447 hours and 521 hours inclusive.
- Find a and b and describe what they represent.
- Write the distribution.
- Find the mean and the standard deviation.
- What is the probability that the duration of games for a team in a single season is between 480 and 500 hours?