Tutorials › VEE Mathematical Statistics › Comparing Exponential Means, Medians, and Percentiles

Home  /  Tutorials  /  VEE Mathematical Statistics / Uniform and Exponential Models

VEE Mathematical Statistics · Uniform and Exponential Models · Tutorial 191 of 880

Comparing Exponential Means, Medians, and Percentiles

Learn to calculate exponential means, medians, and percentiles, and explain how the distribution’s right tail makes them differ.

Beginner 9 min read

What You'll Learn

  • Calculate the mean and median from an exponential rate.
  • Derive and apply the formula for the pth percentile.
  • Compare percentiles with the mean to describe right-skewed outcomes.
  • Recover an exponential rate from a known median and use it to find other quantiles.
  • Convert rates to consistent time units before comparing distributions.
  • Recognize the sign error that gives the wrong percentile.

Why the Mean and Median Differ

For an exponential random variable, the mean and median are both measures of a typical value, but they do not describe the distribution in the same way. The mean is pulled upward by relatively large values in the long right tail. The median marks the halfway point: half the probability is below it and half is above it. Percentiles beyond the median show how far that right tail extends.

This tutorial compares these summaries for a single exponential distribution. The rate must be positive and expressed in the reciprocal of the time unit used for \(X\). For example, if \(X\) is measured in days, its rate is measured per day. All numerical examples below use hypothetical waiting-time models.

Definition: Let \(X\sim\operatorname{Exp}(\lambda)\), where \(\lambda>0\). Its density is \(f_X(x)=\lambda e^{-\lambda x}\) for \(x\geq 0\), and its cumulative distribution function is \(F_X(x)=1-e^{-\lambda x}\) for \(x\geq 0\). The mean is \(E[X]=1/\lambda\). For \(0<p<1\), the \(p\)th percentile, or \(p\)th quantile, is the value \(q_p\) satisfying \(F_X(q_p)=p\). For this continuous exponential distribution, that value is unique.

Deriving the Percentile Formula

To find a percentile, set the cumulative probability at the unknown value equal to the desired probability \(p\), then solve for that value. This works because the exponential CDF is continuous and strictly increasing from 0 toward 1 on \(x\geq 0\).

$$ \begin{aligned} F_X(q_p)&=p,\\ 1-e^{-\lambda q_p}&=p,\\ e^{-\lambda q_p}&=1-p,\\ -\lambda q_p&=\ln(1-p),\\ q_p&=\frac{-\ln(1-p)}{\lambda}. \end{aligned} $$

The median is the 50th percentile, so substituting \(p=0.5\) gives \(q_{0.5}=\ln(2)/\lambda\). In contrast, the mean is \(1/\lambda\). Since \(\ln(2)<1\), the median is always less than the mean for a nondegenerate exponential distribution. More generally, all percentiles are multiples of the same scale \(1/\lambda\); changing the rate stretches or compresses the distribution without changing the relative positions of its percentiles.

Theorem: For \(X\sim\operatorname{Exp}(\lambda)\), the \(p\)th percentile is \(q_p=-\ln(1-p)/\lambda\), the median is \(\ln(2)/\lambda\), and the mean is \(1/\lambda\). In particular, \(q_p/E[X]=-\ln(1-p)\). These formulas require an exponential model with a positive rate; they do not assert that a non-exponential waiting time has the same quantiles.

The median is about \(0.6931\) times the mean. The 90th percentile is about \(2.3026\) times the mean, while the 10th percentile is about \(0.1054\) times the mean. This asymmetry is the quantile pattern of a right-skewed distribution: many values can be relatively small even though a smaller number of large values extend the upper tail and raise the mean.

Worked Example: Comparing the Mean, Median, and 90th Percentile

A hypothetical sensor reports the time \(X\), in days, until a particular signal is detected. Assume \(X\sim\operatorname{Exp}(0.2)\), with rate \(\lambda=0.2\) per day. Find the mean, median, and 90th percentile, and determine what proportion of waiting times are below the mean. Round final time values to four decimal places and the probability to four decimal places.

Method. Use the rate-based formulas for the mean and quantiles. Keep the time unit in days throughout.

Mean. The reciprocal rate is \(1/\lambda=1/0.2=5\) days.

Median. Substitute \(p=0.5\): \(q_{0.5}=-\ln(1-0.5)/0.2=-\ln(0.5)/0.2=\ln(2)/0.2=0.693147\ldots/0.2=3.4657\) days.

90th percentile. Substitute \(p=0.9\): \(q_{0.9}=-\ln(1-0.9)/0.2=-\ln(0.1)/0.2=\ln(10)/0.2=2.302585\ldots/0.2=11.5129\) days.

Proportion below the mean. Evaluate the CDF at 5 days: \(F_X(5)=1-e^{-0.2(5)}=1-e^{-1}=1-0.367879\ldots=0.6321\). Thus about 63.21% of waiting times are below the mean, not 50%.

The results are ordered as \(3.4657<5<11.5129\) days, as expected for the median, mean, and 90th percentile. As a check, \(F_X(11.5129)\) is approximately \(1-e^{-2.302585}=0.9000\). The common mistake to avoid is treating the mean as the midpoint: here the mean is above the median because the right tail pulls it upward.

Reading Percentiles as Positions in the Distribution

A percentile is a probability statement about a cutoff, not a statement that an observation has that probability of occurring exactly at the cutoff. For example, \(q_{0.9}\) is a value with 90% of the distribution at or below it and 10% above it. Because \(X\) is continuous, the probability of an exact value such as \(P(X=q_{0.9})\) is zero.

For an exponential model, percentiles also have useful comparisons across rates. If two exponential variables have different rates, their quantiles may differ in time units, but each quantile is proportional to \(1/\lambda\). The ratio of two percentiles does not depend on the rate:

$$ \frac{q_{p_1}}{q_{p_2}} = \frac{-\ln(1-p_1)/\lambda}{-\ln(1-p_2)/\lambda} = \frac{\ln(1-p_1)}{\ln(1-p_2)}. $$

This rate cancellation is useful when comparing the relative spread of exponential models. It depends on both models being exponential; equal percentile ratios should not be assumed for arbitrary distributions.

Worked Example: Recovering a Rate from the Median

In a hypothetical queue, a customer’s waiting time \(X\), measured in hours, is modeled as exponential. The model’s median waiting time is 4 hours. Find the rate, mean, 75th percentile, and 90th percentile. Round the rate to four significant figures and the final time values to four decimal places.

Method. Use the median formula to solve for the rate, then use that rate for the mean and other percentiles.

The median equation is \(4=\ln(2)/\lambda\), so \(\lambda=\ln(2)/4=0.693147\ldots/4=0.1732868\ldots\) per hour. To four significant figures, \(\lambda=0.1733\) per hour. Retain the unrounded value in the following calculations.

Mean. \(E[X]=1/\lambda=4/\ln(2)=5.770780\ldots\), or 5.7708 hours.

75th percentile. \(q_{0.75}=-\ln(0.25)/\lambda=\ln(4)/( \ln(2)/4)=4\ln(4)/\ln(2)=8\) hours, since \(\ln(4)=2\ln(2)\).

90th percentile. \(q_{0.9}=-\ln(0.1)/\lambda=4\ln(10)/\ln(2)=13.2877\ldots\) hours, or 13.2877 hours.

The ordering is \(4<5.7708<8<13.2877\) hours. Check the median model: \(F_X(4)=1-e^{-(\ln(2)/4)4}=1-e^{-\ln(2)}=0.5\). The 75th-percentile check is \(F_X(8)=1-e^{-2\ln(2)}=1-1/4=0.75\). A common trap is to round the estimated rate too early; keeping \(\ln(2)/4\) during the calculations avoids needless rounding differences.

Unit Consistency When Comparing Models

The rate is measured in the reciprocal of the time unit. A rate stated per hour must be converted before using a formula with times in minutes. If a waiting-time model has rate 6 per hour, its rate per minute is \(6/60=0.1\) per minute. Its mean in minutes is then \(1/0.1=10\), not \(1/6\) minutes.

The next example compares two hypothetical waiting-time models. It illustrates both the effect of different rates and why rates and quantiles must use matching time units.

Worked Example: Comparing Two Waiting-Time Models

Two hypothetical service desks have exponential waiting times measured in minutes. Desk A has rate 6 per hour, and Desk B has rate 5 per hour. For each desk, find the mean and 90th percentile in minutes, then compare the 90th percentiles. Round final times to four decimal places.

Method. Convert both rates from per hour to per minute, then apply the same mean and percentile formulas.

For Desk A, \(\lambda_A=6/60=0.1\) per minute. Its mean is \(1/0.1=10\) minutes. Its 90th percentile is \(-\ln(0.1)/0.1=\ln(10)/0.1=23.025850\ldots\), or 23.0259 minutes.

For Desk B, \(\lambda_B=5/60=1/12\) per minute. Its mean is \(1/(1/12)=12\) minutes. Its 90th percentile is \(-\ln(0.1)/(1/12)=12\ln(10)=27.631021\ldots\), or 27.6310 minutes.

Desk B’s 90th percentile exceeds Desk A’s by \(27.6310-23.0259=4.6051\) minutes using the rounded displayed values. Using unrounded values, the difference is \(12\ln(10)-10\ln(10)=2\ln(10)=4.6052\) minutes to four decimal places. The 90th-percentile check for Desk A is \(F_X(23.02585\ldots)=1-e^{-2.302585\ldots}=0.9\); for Desk B it is \(F_X(27.63102\ldots)=1-e^{-2.302585\ldots}=0.9\).

Thus Desk B has both the larger mean and the larger 90th percentile. The common mistake is to insert 6 and 5 directly as rates per minute: those values are per hour, so doing so would produce times in hours rather than the requested minutes.

Common Percentile Errors

The most common algebra error is to use \(-\ln(p)/\lambda\) for the \(p\)th percentile. The CDF equation is \(1-e^{-\lambda q_p}=p\), so the quantity inside the logarithm after rearranging is \(1-p\), not \(p\). For example, with \(p=0.9\), the incorrect expression gives \(-\ln(0.9)/\lambda=0.1054/\lambda\). Its CDF probability is \(1-e^{-0.1054}=0.1\), so it is the 10th percentile, not the 90th percentile. The correct 90th-percentile expression is \(-\ln(0.1)/\lambda=2.3026/\lambda\).

Other useful checks are that \(q_p\) must be positive for \(0<p<1\), it must increase as \(p\) increases, and it must have the same time unit as \(1/\lambda\). Also, the mean is not the median: for an exponential distribution, the mean is greater, and its position is the 63.21st percentile because \(F_X(1/\lambda)=1-e^{-1}=0.6321\).

Key takeaway: For an exponential model, calculate the mean as \(1/\lambda\) and the \(p\)th percentile as \(-\ln(1-p)/\lambda\). The median is \(\ln(2)/\lambda\), below the mean, while upper percentiles extend farther into the right tail. Check the probability level, rate units, and ordering before interpreting a result.

Check Your Understanding

For each question, assume the stated exponential model is appropriate. Keep rate and time units consistent.

  1. A hypothetical inspection time \(X\), in days, has rate \(0.5\) per day. Find its mean, median, and 80th percentile. Round times to four decimal places. (Answer: mean 2.0000 days; median 1.3863 days; 80th percentile 3.2189 days.)
  2. A hypothetical machine’s exponential operating time has median 10 hours. Find its rate per hour and mean operating time. Round the rate to four significant figures and the mean to four decimal places. (Answer: rate \(0.06931\) per hour; mean 14.4270 hours.)
  3. A hypothetical dispatch delay \(X\), in hours, is exponential with mean 3 hours. Find its median and 95th percentile. Round times to four decimal places. (Answer: median 2.0794 hours; 95th percentile 8.9872 hours.)
  4. Two hypothetical queues have exponential waiting times in minutes. Queue A has rate 2 per hour, and Queue B has rate 0.025 per minute. Find the mean waiting time for each in minutes and identify which is larger. (Answer: Queue A mean 30 minutes; Queue B mean 40 minutes; Queue B’s mean is larger.)
  5. For any exponential variable \(X\), determine the probability that \(X\) is below its mean. Express the result as a probability to four decimal places and as a percentile to two decimal places. (Answer: \(1-e^{-1}=0.6321\), so the mean is the 63.21st percentile.)

Previous

Next

Back to Tutorials
Clinical Biostats