In the realm of statistical distributions, the Uniform Distribution stands out for its simplicity and applicability in various business scenarios. When we say a random variable $X$ follows a uniform distribution, denoted as $X sim U(a, b)$, we mean that every value within the interval $[a, b]$ is equally likely to occur. This distribution is defined by two parameters: $a$, the minimum value, and $b$, the maximum value, with the condition that $b > a$.
A key characteristic of the uniform distribution is its constant probability density across its range. This implies that if you were to pick any two intervals of equal length within $[a, b]$, the probability of $X$ falling into either interval is the same.
The probability density function (PDF) for a uniform distribution is given by:
$$ f(x) = dfrac{1}{b-a} text{ for } a le x le b, text{ and } 0 text{ otherwise} $$
The cumulative distribution function (CDF), which gives the probability that $X$ is less than or equal to a certain value $x$, is:
$$ P(X leq x) = begin{cases} 0 & text{for } x < a dfrac{x-a}{b-a} & text{for } a le x le b 1 & text{for } x > b end{cases} $$
To understand the central tendency and spread of a uniform distribution, we look at its expectation (mean) and variance.
Expectation and Variance of Uniform Distribution
The expected value, or mean, $E[X]$, of a uniform distribution represents the average value we would expect to observe over many trials. For a uniform distribution $U(a, b)$, the expectation is simply the midpoint of the interval:
$$ E[X] = dfrac{a+b}{2} $$
Variance, denoted as $Var(X)$, on the other hand, measures the dispersion or spread of the distribution around its mean. A higher variance indicates that the values are more spread out, while a lower variance suggests they are clustered closer to the mean. For a uniform distribution, the variance is given by the formula:
$$ Var(X) = dfrac{(b-a)^2}{12} $$
This formula reveals that the variance of a uniform distribution depends on the square of the range $(b-a)$. A larger range leads to a significantly larger variance, which is intuitive as a wider interval means more variability in the possible values of $X$. The division by 12 is a constant factor derived from the mathematical properties of the uniform distribution.
Understanding the variance in uniform distribution is crucial in business contexts. For example, consider a scenario where the waiting time for a customer service representative is uniformly distributed between 0 and 10 minutes. The expectation tells us the average waiting time is 5 minutes. However, the variance helps us understand the potential fluctuation around this average. A higher variance would mean waiting times can vary significantly, leading to potential customer dissatisfaction, while a lower variance would indicate more consistent waiting times.
Visualizing Uniform Distribution and Probability
The probability density function of a uniform distribution is graphically represented as a rectangle, where the height is constant ($1/(b-a)$) over the interval $[a, b]$ and zero elsewhere. The total area under this rectangle is always 1, representing the total probability.
The following plot illustrates the PDF for a $U(3, 16)$ distribution. The equal areas for $P(5 leq x leq 7)$ and $P(10 leq x leq 12)$ visually demonstrate the constant probability characteristic of the uniform distribution.
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Worked Example: Post Box Location
Let’s consider a practical example. A local authority needs to install a post box along a 4km stretch of road. They assume that the location is uniformly distributed along this stretch. Let $Y$ be the distance from the east end of the road to the post box location, so $Y sim U(0, 4)$.
(A) Distribution, Expectation, and Variance
The random variable $Y$ follows a uniform distribution $U(0, 4)$.
The expected location is:
$$ E[Y] = dfrac{0+4}{2} = 2 text{ km} $$
The variance of the location is:
$$ Var(Y) = dfrac{(4-0)^2}{12} = dfrac{16}{12} = dfrac{4}{3} approx 1.33 text{ km}^2 $$
(B) Probability Calculation
What is the probability that the post box is located between 1 km and 1.5 km from the east end, where the pavement is narrow?
$$ P(1 leq Y leq 1.5) = P(Y leq 1.5) – P(Y leq 1) = dfrac{1.5-0}{4-0} – dfrac{1-0}{4-0} = dfrac{1.5}{4} – dfrac{1}{4} = dfrac{0.5}{4} = 0.125 $$
There is a 12.5% chance the post box will be placed in the section with narrow pavements.
Conclusion
The uniform distribution is a fundamental concept in probability and statistics, particularly useful for modeling situations where all outcomes within a given range are equally likely. Understanding not only the expectation but also the variance of the uniform distribution is essential for assessing the spread and variability in such scenarios, enabling better decision-making in business and other fields. The variance provides a measure of risk and uncertainty associated with uniformly distributed variables, complementing the average value given by the expectation.
