Uniform Probability Problems with Unknown Cutoffs

From a Probability Statement to an Unknown Cutoff

Uniform models make many probability questions questions about lengths. If a variable is equally likely to fall anywhere in an interval, the probability of landing in a subinterval is the subinterval’s length divided by the full interval’s length. When a cutoff is unknown, the main task is to translate the event carefully before solving the resulting equation.

This tutorial focuses on finding an unknown cutoff or support endpoint from a probability statement. The methods build on the earlier material on uniform densities, cumulative distribution functions, interval probabilities, and quantiles. The same approach applies to hypothetical measurements, waiting times, or amounts, provided the uniform model is appropriate. This skill is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let $X\sim\operatorname{Uniform}(a,b)$, where $a<b$ are known or partly unknown support endpoints. Its density is $f_X(x)=1/(b-a)$ for $a\le x\le b$, and zero outside that interval. For $a\le r\le s\le b$, $$ P(r\le X\le s)=\frac{s-r}{b-a}. $$ Thus, when a cutoff lies inside the support, the probability of an event is the length of the event’s interval divided by the full support width. Because $X$ is continuous, including or excluding an individual endpoint does not change the probability.

Translate the Event Before Solving

The denominator is the width of the full support, $b-a$. The numerator is the length of the part of the support that satisfies the event. For an unknown cutoff, write the event as an interval first, identify that length, and only then form the equation. This order helps distinguish lower-tail probabilities from upper-tail probabilities.

Probability statement	Length equation when the cutoff is within the support
$P(X\le c)=p$	$(c-a)/(b-a)=p$
$P(X>c)=p$	$(b-c)/(b-a)=p$
$P(r<X\le c)=p$, with $a\le r\le c\le b$	$(c-r)/(b-a)=p$
$P(c<X\le s)=p$, with $a\le c\le s\le b$	$(s-c)/(b-a)=p$

These equations assume the stated cutoffs lie within the support, so the indicated interval lengths are valid. A solution must also satisfy the support restrictions. For example, a lower-tail cutoff must lie between $a$ and $b$; a value outside that range cannot be accepted merely because it solves an algebraic equation derived without checking the model.

Worked Example: Finding an Upper-Tail Cutoff

A hypothetical quality-control measurement $X$, in millimeters, follows $X\sim\operatorname{Uniform}(20,80)$. Find the cutoff $c$ such that the probability a measurement exceeds $c$ is $0.25$. No interest or mortality assumptions are involved.

Method. The event is $X>c$, so the favorable interval runs from $c$ to the upper support endpoint, 80. Its length is $80-c$, and the full support width is $80-20=60$.

Set the upper-tail probability equal to the given value:

\frac{80-c}{60}=0.25

Multiply by 60 to obtain $80-c=15$. Therefore,

$$ c=80-15=65. $$

Check. The cutoff satisfies $20\le65\le80$. The interval above it has length $80-65=15$, and $15/60=0.25$, as required. The cutoff is $65$ millimeters.

Common mistake avoided. The upper-tail interval is not the interval from 20 to $c$. That is the lower-tail interval. The event here starts at $c$ and ends at 80.

Why the Length Equation Works

The probability calculation follows directly from the uniform density: every point in the support has the same density, so integrating that constant density over an interval multiplies the density by the interval’s width. This gives the length ratio rather than requiring a separate probability formula for each cutoff.

Justification: For $a\le r\le s\le b$, the density of $X$ is $1/(b-a)$ throughout $[r,s]$. Therefore, $$ P(r\le X\le s) =\int_r^s \frac{1}{b-a}\,dx =\frac{s-r}{b-a}. $$ The probability of the entire support is $\int_a^b 1/(b-a)\,dx=1$, which confirms that the density is normalized. When an unknown cutoff is an endpoint of the event interval, it appears in that interval’s length; equating the length ratio to the stated probability produces the required algebraic equation.

An unknown support endpoint can appear in the denominator as well as in the numerator. In that case, use the probability statement to form an equation involving the unknown width, solve it, and then verify that the resulting support contains the event cutoff. A probability statement cannot be used with a cutoff outside the assumed support interval as though the interval-length formula still applied.

Worked Example: Recovering an Unknown Support Endpoint

A hypothetical sensor’s turnaround time $T$, in minutes, is modeled as $T\sim\operatorname{Uniform}(0,B)$, where $B$ is an unknown positive endpoint. The probability that turnaround time is at most 12 minutes is $0.30$. Find $B$. No interest or mortality assumptions are involved.

Method. The event $T\le12$ covers an interval of length 12, provided $12\le B$. The full support width is $B$. Translate the probability statement into

\frac{12}{B}=0.30.

Multiply both sides by $B$, then divide by $0.30$:

12=0.30B, \qquad B=\frac{12}{0.30}=40.

Check. The resulting endpoint satisfies $40\ge12$, as required for the event interval to be inside the support. Substituting back gives $P(T\le12)=12/40=0.30$. Thus the support is $[0,40]$ minutes.

Common mistake avoided. Do not treat 12 as the full support width. It is the length of the event interval; $B$ is the full support width and must be found from the stated probability.

Conditional Probabilities and Unknown Cutoffs

A conditional probability changes the interval over which probability is being measured. For events $A$ and $B$ with $P(B)>0$, the definition is $P(A\mid B)=P(A\cap B)/P(B)$. With a uniform variable, both the numerator and denominator can be calculated from interval lengths. The denominator is the length associated with the conditioning event, not necessarily the width of the original support.

Worked Example: Solving a Conditional Cutoff

A hypothetical calibration offset $Z$, measured in units, follows $Z\sim\operatorname{Uniform}(10,50)$. Find $c$ such that $P(Z\le c\mid Z>18)=0.60$. No interest or mortality assumptions are involved.

Method. The conditioning event $Z>18$ corresponds, within the support, to an interval of length $50-18=32$. For the conditional event $Z\le c$ to occur while $Z>18$, the offset must lie between 18 and $c$. We therefore require $18<c\le50$, and the numerator interval has length $c-18$.

By the definition of conditional probability, the required conditional probability is the ratio of these two interval lengths:

P(Z\le c\mid Z>18) =\frac{P(18<Z\le c)}{P(Z>18)} =\frac{(c-18)/40}{(50-18)/40} =\frac{c-18}{32}.

Set this equal to $0.60$ and solve:

\frac{c-18}{32}=0.60, \qquad c-18=19.2, \qquad c=37.2.

Check. The cutoff satisfies $18<37.2\le50$. The conditional interval length is $37.2-18=19.2$, and $19.2/32=0.60$. Thus $c=37.2$ units.

Common mistake avoided. Dividing $c-18$ by the original support width of 40 would give an unconditional probability, not the required conditional probability. Conditioning restricts the relevant total interval to $(18,50]$, whose width is 32.

A Reliable Solution Process

A short sequence of checks makes these problems less error-prone, especially when the wording describes a tail or a condition rather than directly naming an interval.

State the model and support.
Write $X\sim\operatorname{Uniform}(a,b)$ and identify which endpoints are known and which are unknown.

Translate the event.
Identify the portion of the support satisfying the event. For a conditional probability, first restrict attention to the conditioning event.

Form the length ratio.
Use event length divided by full support width for an unconditional probability. For a conditional probability, use the intersection length divided by the conditioning-event length.

Solve and check.
Rearrange the equation, verify the solution lies in the required support interval, and substitute it back into the original probability statement.

Common Traps and Why the Check Matters

The most frequent error is reversing the interval for a tail event. If $X\sim\operatorname{Uniform}(20,80)$ and $P(X>c)=0.25$, the incorrect equation $c/60=0.25$ uses a lower endpoint of zero, which is not part of the model, and treats the cutoff as though it described a lower-tail length. It gives $c=15$, outside the support. The correct upper-tail equation is

\frac{80-c}{80-20}=0.25, \qquad c=65.

Substitution confirms the correction: $P(X>65)=(80-65)/60=0.25$. The support check exposes the incorrect value, while the interval translation identifies the underlying error. In a uniform model, a computed value should not be accepted until both the event direction and the support restrictions have been checked.

These problems matter because probability statements often conceal an unknown threshold inside a practical decision rule: a measurement may be flagged above a cutoff, a service time may need to fall below a target, or a conditional probability may describe a subgroup already selected by a restriction. In each case, the interpretation of the event determines the correct interval. The algebra is usually straightforward once that interval is identified.

Check Your Understanding

For each problem, translate the event into an interval-length equation, solve for the unknown, and check that the result is consistent with the stated support.

A hypothetical inspection reading $X$ follows $X\sim\operatorname{Uniform}(4,28)$. Find $c$ if $P(X>c)=0.35$. (Answer: $c=19.6$.)
A hypothetical processing time $T$ follows $T\sim\operatorname{Uniform}(0,B)$, where $B$ is unknown. If $P(T<9)=0.45$, find $B$. (Answer: $B=20$.)
A hypothetical component rating $R$ follows $R\sim\operatorname{Uniform}(10,70)$. Find $c$ if $P(25<R<c)=0.50$. (Answer: $c=55$.)
A hypothetical measurement $W$ follows $W\sim\operatorname{Uniform}(0,40)$. Find $c$ if $P(W\le c\mid W>8)=0.75$. (Answer: $c=32$.)
A hypothetical score $Y$ follows $Y\sim\operatorname{Uniform}(a,50)$, with unknown $a<50$. If $P(Y>30)=0.40$, find $a$ and verify that 30 is in the support. (Answer: $a=0$, with $0\le30\le50$.)

Probability statement	Length equation when the cutoff is within the support
\(P(X\le c)=p\)	\((c-a)/(b-a)=p\)
\(P(X>c)=p\)	\((b-c)/(b-a)=p\)
\(P(r<X\le c)=p\), with \(a\le r\le c\le b\)	\((c-r)/(b-a)=p\)
\(P(c<X\le s)=p\), with \(a\le c\le s\le b\)	\((s-c)/(b-a)=p\)

What You'll Learn