Uniform Percentiles and Quantiles

Why Uniform Quantiles Are Useful

A percentile gives a cutoff on the scale of the variable. For example, a 90th percentile is a value at or below which 90% of the modeled outcomes fall. In a uniform model, outcomes are spread evenly across a fixed interval, so finding the cutoff requires inverting a simple cumulative distribution function (CDF). The key is to use the probability for the correct tail and to keep the cutoff within the model's support.

Quantiles can summarize the location of a distribution or set a threshold for decisions. In a hypothetical model of processing times, for instance, a high percentile describes a time by which most modeled cases are complete. The result depends on the assumed distribution and its endpoints; it is not a guarantee about actual outcomes.

Definition and Quantile Function

Assume $X$ has a continuous uniform distribution on the interval $[a,b]$, where $a<b$. The model assigns equal probability to intervals of equal length within its support. Its CDF is zero below $a$, increases linearly from zero to one between $a$ and $b$, and is one at and above $b$.

Definition: For $0<p<1$, the $p$-quantile of a continuous distribution is the value $Q(p)$ satisfying $F_X(Q(p))=p$, when the CDF is strictly increasing at that value. The $p$-quantile is also called the $100p$th percentile. For a uniform distribution on $[a,b]$, define the endpoint quantiles by $Q(0)=a$ and $Q(1)=b$.

For $X\sim\operatorname{Uniform}(a,b)$, the CDF within the support is $F_X(x)=(x-a)/(b-a)$. Solving $F_X(x)=p$ for $x$ gives the quantile function:

Q(p)=a+p(b-a), \qquad 0\le p\le 1.

Here $p$ is a probability between zero and one, not a number between zero and one hundred. The quantity $b-a$ is the width of the interval. Thus the $p$-quantile is the lower endpoint plus a fraction $p$ of that width. This construction also shows that the quantile has the same units as $X$.

Why the Formula Holds

The quantile formula follows directly from the linear CDF. For an interior probability $0<p<1$, the quantile lies between $a$ and $b$, where the CDF is $(x-a)/(b-a)$. Setting the CDF equal to $p$ and solving gives:

\begin{aligned} \frac{x-a}{b-a}&=p,\\ x-a&=p(b-a),\\ x&=a+p(b-a). \end{aligned}

Because $b-a>0$, the CDF increases strictly over the support, so this solution is unique. At $p=0$, the formula gives $a$; at $p=1$, it gives $b$. The formula therefore covers the endpoints as well as the interior.

Theorem: If $X\sim\operatorname{Uniform}(a,b)$, then $P(X\le Q(p))=p$, where $Q(p)=a+p(b-a)$ for $0\le p\le1$. Since this distribution is continuous, $P(X<Q(p))=p$ as well.

The equality of the strict and non-strict probabilities holds because a continuous random variable has probability zero at any single point. For a uniform variable, $P(X=c)=0$ for every fixed $c$. The quantile formula describes a model cutoff, not a value that must occur in a finite sample.

Lower Tails, Upper Tails, and Central Ranges

The percentile specified in a problem may refer to the lower tail or to the upper tail. A $p$-quantile has probability $p$ below it and probability $1-p$ above it. Consequently, a cutoff with a fraction $r$ of outcomes above it is the $(1-r)$-quantile.

Key relationship: The cutoff leaving an upper-tail probability $r$ is $Q(1-r)=a+(1-r)(b-a)$. The probability between two quantiles is the difference between their percentile probabilities: for $0\le p_1\le p_2\le1$, $P(Q(p_1)\le X\le Q(p_2))=p_2-p_1$.

This relationship is especially convenient for central ranges. For example, the interval from the 10th to the 90th percentile contains 80% of the probability under the model. Both endpoints come from the same quantile function; no separate formula is needed.

Worked Example: Finding Two Percentiles and the Probability Between Them

A hypothetical processing-time model assumes $X\sim\operatorname{Uniform}(4,16)$, with time measured in minutes. Find the 35th and 90th percentiles, then calculate the probability that a processing time lies between them.

Method. Use $Q(p)=a+p(b-a)$, with $a=4$, $b=16$, and interval width $b-a=16-4=12$ minutes.

For the 35th percentile, $p=0.35$:

Q(0.35)=4+0.35(12)=4+4.2=8.2\text{ minutes}.

For the 90th percentile, $p=0.90$:

Q(0.90)=4+0.90(12)=4+10.8=14.8\text{ minutes}.

The probability between these quantiles is the difference in their percentile probabilities:

P(8.2\le X\le14.8)=0.90-0.35=0.55.

Check. The cutoffs are both within the support $[4,16]$, and the interval from the 35th to the 90th percentile contains $55\%$ of the modeled probability. The endpoints are in minutes, while the probability is dimensionless.

Common mistake avoided: The width is 12 minutes, not 16 minutes. The quantile is measured from the lower endpoint, so the calculation starts at 4 and adds a fraction of the width.

Worked Example: A Cutoff for the Highest-Cost Outcomes

In a hypothetical repair-cost model, $X\sim\operatorname{Uniform}(600,1800)$, where costs are measured in dollars. Find the cutoff that separates the highest 12% of modeled costs from the other 88%, and state how to interpret it.

Method. The highest 12% is an upper-tail probability of $r=0.12$. Therefore the probability below the cutoff is $1-0.12=0.88$, so the required cutoff is the 88th percentile. The interval width is $1800-600=1200$ dollars.

\begin{aligned} Q(0.88)&=600+0.88(1200)\\ &=600+1056\\ &=1656. \end{aligned}

The cutoff is $1,656.00. Under the stated continuous uniform model, 88% of costs are at or below this value and 12% are above it. As a check, the distance from the cutoff to the upper endpoint is $1800-1656=144$, and $144/1200=0.12$.

Common mistake avoided: Using $Q(0.12)$ would give $600+0.12(1200)=744$, the 12th percentile. That leaves 12% below the cutoff, not above it. For a highest-12% threshold, use $1-0.12=0.88$.

Worked Example: Quantiles and a Percentile Rank

A hypothetical sensor reading $X$ is modeled by $X\sim\operatorname{Uniform}(-3,9)$, in calibrated units. Find the 20th and 70th percentiles, the probability that $X$ falls between them, and the percentile rank of a reading of 3.

Method. The support width is $9-(-3)=12$. Apply the quantile formula to each percentile:

\begin{aligned} Q(0.20)&=-3+0.20(12)=-3+2.4=-0.6,\\ Q(0.70)&=-3+0.70(12)=-3+8.4=5.4. \end{aligned}

The probability between these cutoffs is:

P(-0.6\le X\le5.4)=0.70-0.20=0.50.

The percentile rank of a value $x$ is the probability $P(X\le x)$, expressed as a percentage. Since 3 lies inside the support, evaluate the CDF:

F_X(3)=\frac{3-(-3)}{9-(-3)}=\frac{6}{12}=0.50.

Thus a reading of 3 is at the 50th percentile. Check. The cutoffs $-0.6$ and $5.4$ are inside $[-3,9]$, and the central probability between them is 50%. The reading 3 is halfway through the support, consistent with a percentile rank of 50%.

Common mistake avoided: The lower endpoint is $-3$, so the distance from the endpoint to 3 is $3-(-3)=6$, not $3-3$ or $3$. Subtracting the lower endpoint correctly is essential when the support includes negative values.

A Reliable Calculation Process

A short sequence of checks prevents most errors in uniform quantile problems:

State the model and support.
Identify $a$, $b$, and the units, and confirm that the assumed distribution is uniform on $[a,b]$ with $a<b$.

Translate the wording into a probability.
A $k$th percentile uses $p=k/100$. For a cutoff with $r$ above it, use $p=1-r$.

Apply the quantile formula.
Calculate $Q(p)=a+p(b-a)$, keeping $p$ as a decimal probability.

Check the interpretation.
Verify that the cutoff is between $a$ and $b$, has the variable's units, and leaves the stated probability on the correct side.

Common Exam Traps

Using a percentage instead of a probability. For the 80th percentile, use $p=0.80$, not $p=80$. A probability must be between zero and one.
Reversing the tail. “The highest 5%” means 5% above the cutoff, so the cutoff is $Q(0.95)$, not $Q(0.05)$.
Multiplying by an endpoint instead of the width. The formula uses $b-a$, then adds the resulting distance to $a$. It is not generally $p b$.
Ignoring the stated model. A quantile computed from a uniform assumption is valid for that assumed distribution. It does not establish that observed data are actually uniform.
Confusing a quantile with a percentile rank. A quantile starts with a probability and returns a value. A percentile rank starts with a value and returns a probability.

Quantiles are useful because they translate a probability requirement into a value on the original scale. For the uniform model, that translation is linear: a probability fraction $p$ corresponds to the same fraction $p$ of the distance from $a$ to $b$. Always state whether the probability is below or above the cutoff; that interpretation is as important as the arithmetic.

Check Your Understanding

For each problem, identify the relevant percentile probability before calculating. Assume the stated uniform model applies.

A hypothetical waiting time $X\sim\operatorname{Uniform}(10,30)$ is measured in minutes. Find its 25th percentile. (Answer: 15 minutes)
A hypothetical equipment cost $X\sim\operatorname{Uniform}(100,700)$ is measured in dollars. Find the cutoff with 15% of costs above it. (Answer: $610.00)
A calibrated quantity $X\sim\operatorname{Uniform}(-8,16)$ is measured in units. Find the 35th percentile. (Answer: 0.4 units)
A hypothetical duration $X\sim\operatorname{Uniform}(2,22)$ is measured in hours. Find the 5th and 95th percentiles and the probability between them. (Answer: 3 hours and 21 hours; probability 0.9000)
A hypothetical measurement $X\sim\operatorname{Uniform}(0,50)$ is observed to equal 42. Find its percentile rank. (Answer: 84th percentile)

What You'll Learn