Continuous families and their mechanisms

Try

A uniform travel model spreads probability evenly over a time interval. An exponential model describes a memoryless waiting time. A normal model is symmetric around a center and extends over the entire real line. Before comparing their formulas, identify which support and mechanism fit the question.

Predict

Predict whether a lifetime that becomes more failure-prone with age can be exactly memoryless. Then predict whether a normal model with substantial negative-time probability would be appropriate for a nonnegative waiting time. Matching a mean alone does not establish a suitable distribution family.

Experiment

For X uniform on (1,5), an interval from 2 to 4 has probability 2/4=1/2. For an exponential T with rate 3 per hour, survival is P(T>t)=e^(−3t) for t≥0, so waiting beyond one hour has probability e^(−3).

For a normal measurement with mean 30 and SD 5, the interval [25,40] standardizes to [−1,2]. Using supplied Φ(2)=0.9772 and Φ(−1)=0.1587 gives probability 0.8185. The calculation uses a normal assumption; those two moments would not determine the probability for an arbitrary law.

Understand

Uniform(a,b), with a<b, has density 1/(b−a) inside the interval and zero outside. Probabilities are interval-length ratios after intersecting with the support. Its mean is the midpoint and its variance is (b−a)²/12. Uniformity depends on coordinates: a nonlinear transformation of a uniform variable is not generally uniform.

The exponential density with rate λ>0 is λe^(−λt) for t>0. Its CDF is 1−e^(−λt), mean 1/λ and variance 1/λ². Rate has inverse-time units; replacing a rate with a mean in the exponent would be dimensionally wrong. Some notation uses a scale parameter instead, so identify the convention explicitly.

Memorylessness is the survival ratio P(T>s+t|T>s)=e^(−λ(s+t))/e^(−λs)=e^(−λt). This applies for nonnegative s,t with positive conditioning probability. It describes the remaining waiting time after survival, not the statement that all events in a real system are independent. Physical aging or changing rates can invalidate the model.

A normal law with mean μ and positive variance σ² has density exp(−(x−μ)²/(2σ²))/(σ√(2π)). Standardizing Z=(X−μ)/σ produces the standard normal law. Its CDF generally needs numerical evaluation or a supplied table; no elementary antiderivative is assumed. Use SD σ rather than variance σ² in the denominator.

Normal linear combinations have additional structure when variables are jointly normal; merely having normal marginals is not enough to assume every dependence property. The present one-variable calculations rely only on the stated marginal normal model. Later lessons address joint laws and independence.

All three families are mathematical models. Distributional support, tails, independence and parameter stability should be checked against the intended application rather than inferred from convenient formulas.

Apply

Translate each practice statement into support, parameter convention and units. Derive an exponential median by solving a survival equation rather than confusing median with mean. For a normal tail, write standardized endpoints before consulting the supplied table, and keep the final probability labeled as conditional on that model.

Continue

Next compute expectations of transformed values directly from the original law. This often avoids deriving a new distribution, but integrability must remain explicit.