Central Quantiles and Tail Quantiles Across Families

Why Compare Quantiles?

A median describes the halfway point of a distribution, and the interquartile range (IQR) describes the width of its middle half. A high quantile, such as the 95th percentile, describes a point below which most observations lie. Together, these summaries distinguish the center and central spread from the upper tail. They are especially useful when comparing distributions whose tails behave differently.

The formulas below use the normal, logistic, and Laplace location-scale parameterizations, the Weibull shape-scale parameterization, and the Pareto Type I parameterization. The goal is not only to calculate a quantile, but also to see which parts of its formula respond to location, scale, or shape.

Quantiles and the Central Summaries

All five families considered here have continuous CDFs that are strictly increasing on their supports where they rise, so their quantiles for \(0<p<1\) are unique. For the Weibull and Pareto distributions, the CDF is zero below the lower endpoint of the support; it is not strictly increasing on the whole real line. This does not prevent their interior quantiles from being uniquely determined.

An upper-tail probability \(\varepsilon\) corresponds to the quantile \(q_{1-\varepsilon}\): for these continuous distributions, \(P(X>q_{1-\varepsilon})=\varepsilon\). Thus the 95th percentile is a threshold with 5% of the probability above it.

Quantile Formulas Across the Five Families

Write \(\Phi\) for the standard normal CDF and \(\Phi^{-1}\) for its inverse. For \(X\sim N(\mu,\sigma^2)\), with \(\sigma>0\), the quantile is \(q_p=\mu+\sigma\Phi^{-1}(p)\). Symmetry gives a median of \(\mu\); since \(\Phi^{-1}(0.75)\approx0.67449\), the IQR is approximately \(1.34898\sigma\). The 95th percentile is \(\mu+1.64485\sigma\).

For a logistic distribution with location \(\mu\) and scale \(s>0\), inversion of its CDF gives \(q_p=\mu+s\log\{p/(1-p)\}\). Its median is \(\mu\), its IQR is \(2s\log 3\), and its 95th percentile is \(\mu+s\log 19\).

For a Laplace distribution with location \(\mu\) and scale \(b>0\), the quantile formula is piecewise: \(q_p=\mu+b\log(2p)\) for \(0<p\le 0.5\), and \(q_p=\mu-b\log\{2(1-p)\}\) for \(0.5\le p<1\). The two expressions agree at \(p=0.5\). Consequently, its median is \(\mu\), its IQR is \(2b\log 2\), and its 95th percentile is \(\mu+b\log 10\).

For a Weibull distribution with shape \(k>0\) and scale \(\lambda>0\), supported on \([0,\infty)\), the CDF is \(F(x)=1-\exp\{-(x/\lambda)^k\}\) for \(x\ge0\). Solving \(F(q_p)=p\) gives \(q_p=\lambda[-\log(1-p)]^{1/k}\). Thus its median is \(\lambda(\log 2)^{1/k}\), its IQR is \(\lambda\{(\log 4)^{1/k}-[\log(4/3)]^{1/k}\}\), and its 95th percentile is \(\lambda(\log 20)^{1/k}\).

For a Pareto Type I distribution with minimum \(x_m>0\) and shape \(\alpha>0\), supported on \([x_m,\infty)\), the CDF on its support is \(F(x)=1-(x_m/x)^\alpha\). Inverting gives \(q_p=x_m(1-p)^{-1/\alpha}\). Its median is \(x_m2^{1/\alpha}\), its IQR is \(x_m\{4^{1/\alpha}-(4/3)^{1/\alpha}\}\), and its 95th percentile is \(x_m20^{1/\alpha}\).

Why the Parameter Effects Differ

Proof. The CDF of \(Y\) is \(F_Y(y)=P(a+cX\le y)=F_X((y-a)/c)\), because \(c>0\). Since \(c>0\), \(\{y:F_Y(y)\ge p\}=\{a+cx:F_X(x)\ge p\}\), so taking infima gives \(q_p(Y)=a+cq_p(X)\). Subtracting the transformed first quartile from the transformed third quartile gives \((a+cq_{0.75}(X))-(a+cq_{0.25}(X))=c\operatorname{IQR}(X)\). \(\square\)

This result explains why \(\mu\) shifts normal, logistic, and Laplace quantiles without changing their IQRs, while their positive scale parameters multiply both the quantiles’ distances from the center and the IQR. It also explains why the Weibull scale \(\lambda\) and Pareto minimum \(x_m\) multiply their quantiles and IQRs when their shape parameters are held fixed.

For the Weibull distribution, put \(A=-\log(1-p)>0\). Its quantile can be written \(q_p=\lambda A^{1/k}\), so \(\log q_p=\log\lambda+(\log A)/k\). Differentiation with respect to \(k\) gives \(\partial\log q_p/\partial k=-(\log A)/k^2\). Since \(A<1\) when \(p<1-e^{-1}\), the quantile increases with \(k\) at those probabilities. It decreases with \(k\) when \(p>1-e^{-1}\), and is unchanged when \(p=1-e^{-1}\). In particular, the median decreases as \(k\) increases because \(\log2<1\); the 95th percentile increases because \(\log20>1\).

For Pareto, \(q_p=x_m\exp\{\log(1/(1-p))/\alpha\}\). At fixed \(p\in(0,1)\), the exponent is positive, so increasing \(\alpha\) decreases \(q_p\), while increasing \(x_m\) scales it linearly. The IQR also decreases with \(\alpha\): writing \(t=1/\alpha\), its factor apart from \(x_m\) is \(4^t-(4/3)^t\), which increases with \(t>0\). Indeed, its derivative is \(4^t\log4-(4/3)^t\log(4/3)>0\), since \(4^t>(4/3)^t\) and \(\log4>\log(4/3)>0\).

Common Mistake and a Practical Check

• Using the wrong normal parameter. In \(N(\mu,\sigma^2)\), the second argument is the variance; the standard deviation used in quantile calculations is its square root.
• Confusing a high quantile with an upper-tail probability. The 95th percentile has 5% above it, not 95%.
• Assuming a shape parameter has one effect at every quantile. For the Weibull family, increasing \(k\) lowers the median but raises the 95th percentile when \(\lambda\) is fixed.
• Claiming strict increase on the whole real line for every CDF here. Weibull and Pareto CDFs are flat below their support. Their interior quantiles are nevertheless unique because each CDF is continuous and strictly increasing where it rises on its support.

A useful check after calculating an upper quantile is to substitute it into the survival probability. For example, the Pareto 95th percentile satisfies \((x_m/q_{0.95})^\alpha=0.05\), and the Weibull 95th percentile satisfies \(\exp\{-(q_{0.95}/\lambda)^k\}=0.05\). Such checks catch errors in signs, complements, or parameter placement.