The chapter's second named density and the first that is not flat. The exponential distribution is concerned with the amount of time until some specific event occurs, and its shape is a declining curve: there are fewer large values and more small ones. It has a single parameter, the decay parameter m, which is the reciprocal of the mean, and unusually its standard deviation equals its mean. Probabilities come from a closed form rather than a table, with the area to the left given by one minus e to the minus m x and the area to the right by e to the minus m x. Its defining property is memorylessness: the probability of waiting a further t, given that you have already waited r, equals the unconditional probability of waiting t. The lesson closes on the equivalence between exponential gaps between events and Poisson counts of events.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 5 — Continuous Random Variables
The Exponential Distribution
Objectives
Five outcomes, and the fourth is the one that makes this distribution strange.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 302-312 — the section these objectives are drawn from
Warm-up
Section 5.2's conditioning raised a probability: knowing a smile had lasted past 8 seconds made passing 12 more likely.
Discussion prompt
A computer part lasts ten years on average. If a part has already run for ten years without failing, is it more likely, less likely, or equally likely to fail in the coming year than a brand new one?
Hint: Two arguments pull in opposite directions. Say what each one is before choosing.
Answer:
One argument says less likely: it has proved itself, so it must be a good one. Another says more likely: it is worn out and due to fail. Both are reasonable about real machinery, and neither is what the exponential model says.
The exponential says exactly equally likely. Its whole character is that the time already elapsed carries no information about the time remaining — the book puts it as the part staying as good as new until it suddenly breaks.
That is a strong assumption and often a wrong one, since real components do wear out. But it is exactly right for events that arrive at random with no ageing mechanism, like calls at a switchboard or the next earthquake, and this section is about the distribution that captures it.
Concept
The exponential distribution is often concerned with the amount of time until some specific event occurs. Values occur in the following way: there are fewer large values and more small values. Its density is m times e to the minus m x, where m is the decay parameter, and m is the reciprocal of the mean.
decay parameter — The single parameter m of an exponential distribution, equal to one divided by the mean. It is also the height of the density at zero, which is the largest value the curve takes.
\[ X \sim \text{Exp}(m), \quad f(x) = m e^{-mx}, \quad m = \frac{1}{\mu} \]
The book's examples are worth listing because they show the range: the time until an earthquake, the length of long-distance business calls, how long a car battery lasts, the amount customers spend on one supermarket trip, and — it notes — the value of the change in your pocket. What they share is that small values are common and large ones rare, which is what a declining curve encodes. Exponential distributions are commonly used in calculations of product reliability.
Figure (svg): A declining exponential curve starting at nought point two five and falling toward the horizontal axis
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 302-303
Section
Section 1
Concept
To do any calculations you must know m, the decay parameter, which is one divided by the mean. The standard deviation is the same as the mean, so a single number fixes the centre, the spread and the whole density.
the standard deviation equals the mean — For an exponential, sigma equals mu. A mean wait of four minutes carries a standard deviation of four minutes, which is a very wide spread relative to the centre and is a direct consequence of the long right tail.
\[ m = \frac{1}{\mu}, \qquad \sigma = \mu \]
The height of the density at zero is m itself, since e to the zero is one, and that is the maximum value on the vertical axis. So a large m gives a tall curve that falls away fast — short waits — while a small m gives a low curve with a long slow tail. Units matter here in a way they did not for the uniform: a mean of four minutes gives an m of 0.25 per minute, and mixing minutes with hours anywhere in a problem will produce a wrong answer that looks entirely reasonable.
Figure (svg): A card giving the decay parameter, the density and the fact that the mean and standard deviation are equal
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, p. 303 — Example 5.7, the notation and the density
Picture it
Example 5.7: a postal clerk averaging four minutes per customer.
Figure (svg): A declining exponential curve starting at nought point two five and falling toward the horizontal axis
The book checks the height at zero explicitly and finds 0.25, which is m — a useful anchor, because it means you can read the decay parameter straight off the vertical intercept of any exponential graph. Note that this height is not a probability, exactly as in section 5.1: only areas are.
Worked example
Example 5.7. The mean is given and everything else follows.
\[ \text{service times average } 4 \text{ minutes} \]
Note the mean
Why: Four minutes.
\[ \mu = 4 \]
Take the reciprocal
Why: One over four.
\[ m = 0.25 \]
State the standard deviation
Why: Equal to the mean.
\[ \sigma = 4 \]
Write the density
Why: m times e to the minus m x.
\[ f(x) = 0.25 e ^{-0.25 x} \]
Figure (svg): The solution to Worked example the postal clerk shown as a ladder of expressions, one row per legal move
\[ X \sim \text{Exp}(0.25), \quad f(x) = 0.25e^{-0.25x} \]
Verify: confirm the height at zero equals m
Why: Setting x to zero gives 0.25 times e to the zero, which is 0.25 times 1, or 0.25 — exactly m. The book performs this check on the graph, and it is worth repeating because it is the fastest way to confirm you have not accidentally used the mean where the decay parameter belongs. A curve starting at height 4 would signal that m and mu had been swapped.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, p. 303
Faded example
A car battery lasts 40 months on average.
Fill in the blanks
m = \frac400.025} = ___ \text___
Why: One over 40 is 0.025 per month, and the standard deviation is also 40 months. A small decay parameter like this means a long slow tail, which is right for something lasting years.
Worked example
Example 5.11(a) and (c), where the mean has to be derived first.
\[ 30 \text{ customers arrive per hour; time between arrivals is exponential} \]
Convert the rate to a gap
Why: Thirty per 60 minutes.
\[ \text{one every } 2\text{ minutes} \]
State the mean
Why: The average gap.
\[ \mu = 2\text{ minutes} \]
Take the reciprocal
Why: One over two.
\[ m = 0.5 \]
Write the distribution
Why: In the book's notation.
Figure (svg): The solution to Worked example reading m from a rate shown as a ladder of expressions, one row per legal move
\[ \mu = \frac{60}{30} = 2, \quad m = \frac{1}{2} = 0.5 \]
Verify: confirm that m and the arrival rate are the same number in compatible units
Why: Thirty customers an hour is 0.5 customers a minute, and m came out as 0.5 per minute — the decay parameter IS the event rate once the units agree. That is not a coincidence but the first sign of the Poisson connection this lesson closes on, and it also gives a quick unit check: if m and the stated rate disagree numerically, the units have been mixed.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, p. 308
Trap
\[ \mu = 4 \;\Rightarrow\; f(x) = 4e^{-4x} \]
Put the given number straight into the density
Why: Four is the number the problem supplies.
\[ \text{a mean of } \tfrac{1}{4} \text{ minute, not } 4 \]
That density describes a clerk averaging fifteen seconds per customer, which is sixteen times too fast.
\[ m = \frac{1}{\mu} = 0.25 \;\Rightarrow\; f(x) = 0.25e^{-0.25x} \]
Invert the mean first, and only then substitute
Why: The density is written in terms of m, never mu.
The height check settles it in one line: an exponential's curve starts at m, so a density starting at 4 would have to belong to a distribution with a mean of a quarter. Writing the words mean equals 4, so m equals 0.25 as an explicit line before any substitution is what prevents this, and it is the single commonest error in the section.
Sorting
Ask whether small values are commoner than large ones.
Sort into buckets
Sort each situation.
Items (a), (c) and (e) are the book's own examples. Item (d) is section 5.2's uniform, and the contrast is instructive: a bus on a strict timetable gives a flat wait distribution, while buses arriving at random would give an exponential one.
Two truths and a lie
All three concern the parameter.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false: m is the RECIPROCAL of the mean. For Example 5.7 the mean is 4 and m is 0.25, and swapping them gives a distribution sixteen times too fast. This is the section's commonest error.
Prediction
Commit before reasoning.
Predict first
Two exponentials have decay parameters 0.1 and 2. Which has the longer average wait?
Correct: The one with m = 0.1.
Why: Its mean is 1 over 0.1, which is 10, against 0.5 for the other — a factor of twenty. A large decay parameter means the curve falls away quickly, so long waits are rare; a small one means a long slow tail. The inverse relationship is the thing to hold on to, since it runs against the intuition that a bigger parameter means bigger values.
Section
Section 2
Concept
The cumulative distribution function gives the area to the left, and for the exponential it has a closed form: P(X < x) is one minus e to the minus m x. The area to the right is what remains, which simplifies to e to the minus m x, and the probability of an interval is the difference of two left areas.
the exponential CDF — P(X < x) equals one minus e to the minus m x. Because the total area is one, the right tail P(X > x) is exactly e to the minus m x, with no subtraction left to do.
\[ P(X < x) = 1 - e^{-mx}, \qquad P(X > x) = e^{-mx} \]
The right tail simplifying to a single exponential term is worth noticing, because it makes greater-than questions easier here than less-than ones — the reverse of the usual situation. The book derives it in one line: P(X > x) equals one minus one minus e to the minus m x, which is e to the minus m x. It is also the form that makes the memoryless property visible, as the fourth idea shows.
Figure (svg): The exponential curve with the narrow strip between four and five minutes shaded
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 303-307 — Examples 5.8 and 5.9
Picture it
Example 5.8(a): four to five minutes with the clerk.
Figure (svg): The exponential curve with the narrow strip between four and five minutes shaded
The answer of 0.0814 is small, and the picture says why: the strip is one minute wide on a curve that has already declined a long way by four minutes. Comparing a shaded region's size against the answer is as useful here as it was for the uniform, even though the areas are no longer proportional to width.
Worked example
Example 5.8(a), as a difference of two left areas.
\[ X \sim \text{Exp}(0.25); \text{ find } P(4 < x < 5) \]
Left area at 5
Why: One minus e to the minus 1.25.
\[ 0.7135 \]
Left area at 4
Why: One minus e to the minus 1.
\[ 0.6321 \]
Subtract
Why: The larger minus the smaller.
\[ 0.0814 \]
State it
Why: The book's answer.
\[ 0.0814 \]
Figure (svg): The solution to Worked example four to five minutes shown as a ladder of expressions, one row per legal move
\[ P(4 < x < 5) = 0.7135 - 0.6321 = 0.0814 \]
Verify: confirm the subtraction order against the shape of the curve
Why: The left area at 5 must exceed the left area at 4, since moving right only adds area — so the larger value belongs first, exactly as in section 5.1. A negative answer would signal the reverse order. The book also notes the shortcut of computing e to the minus one minus e to the minus 1.25 directly, which gives the same number with one operation fewer.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 303-304
Faded example
Phone calls have decay parameter m = 1/12. Find P(x > 5).
Fill in the blanks
P(x > 5) = e^-0.4167 = e^0.6592} \approx ___
Why: Five twelfths is about 0.4167, and e to the minus that is 0.6592 — the book's answer for Example 5.10. A right tail is a single exponential term with no subtraction.
Worked example
Example 5.9(a), where the right tail needs no subtraction.
\[ \mu = 10 \text{ years}; \text{ find } P(x > 7) \]
Find m
Why: One over ten.
\[ m = 0.1 \]
Recall the right tail
Why: e to the minus m x.
\[ e ^{-0.1 x} \]
Substitute
Why: At x = 7.
\[ e ^{-0.7} \]
Evaluate
Why: The book's value.
\[ 0.4966 \]
Figure (svg): The solution to Worked example a part lasting more than seven years shown as a ladder of expressions, one row per legal move
\[ P(x > 7) = e^{-(0.1)(7)} = e^{-0.7} \approx 0.4966 \]
Verify: confirm the answer against the median rather than the mean
Why: The answer is just under a half, which says seven years is close to the median lifetime — and indeed the median is ln 2 over m, about 6.93 years. That the median is well below the mean of ten is characteristic of the exponential, and it is worth checking a right tail against the median rather than the mean, since comparing 7 with the mean of 10 would wrongly suggest an answer well above a half.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 305-306
Error analysis
The decay parameter is 0.1 and the answer is 0.4966.
Annotate
On: \( \begin{aligned} &(1)\; e^{-(10)(7)} = e^{-70} \\ &(2)\; 1 - e^{-(0.1)(7)} = 0.5034 \\ &(3)\; 1 - e^{-0.7} \text{ then subtracted from } 1 \text{ again} \\ &(4)\; e^{-(0.1)(7)} = 0.4966 \end{aligned} \)
Error (2) is the dangerous one because the two answers straddle a half and differ by less than a hundredth. Sketching the curve and shading which side is wanted before substituting is the only reliable guard, and it costs a few seconds.
Discrimination
Decide whether each question needs the left area, the right tail, or a difference.
Sort into buckets
Sort each question for an exponential X.
Two truths and a lie
All three concern the closed form.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and swaps the two tails: the LEFT area is one minus e to the minus m x. Since the exponential term falls from 1 toward 0 as x grows, it can only be the right tail — a left area has to rise toward one.
Estimation
A component has a mean life of 10 years.
Predict first
Without computing, is P(x > 20) closer to 0.14 or to 0.50?
Correct: About 0.14.
Why: Twenty years is twice the mean, and e to the minus 2 is about 0.135. A useful anchor is that the right tail at the mean itself is e to the minus one, about 0.37 — so roughly 37 percent of items outlast the mean, and only about 14 percent last twice it. The mean of an exponential is NOT the halfway point, which is what makes 0.50 the tempting wrong answer.
Section
Section 3
Concept
A percentile is found by setting the left area equal to the given proportion and solving, which needs a natural logarithm. The book gives the formula directly: k is the natural log of one minus the area to the left, divided by minus m.
the percentile formula — k equals ln(1 minus p) over minus m, where p is the proportion below k. Because the exponential is right-skewed, the median falls below the mean rather than on it.
\[ k = \frac{\ln(1 - p)}{-m} \]
The median is worth computing once in general: setting p to a half gives k equal to ln 2 over m, or about 0.693 times the mean. So the median of an exponential is always about 69 percent of its mean, which is a large gap — for Example 5.7 that is 2.77 minutes against 4. Section 2.6 said a right-skewed distribution has its mean above its median, and this is that statement made exact.
Figure (svg): The exponential curve with the median at two point eight and the larger mean at four both marked
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 304-306 — Example 5.8(b) and (c), and Example 5.9(c)
Picture it
Example 5.8: the median at 2.8 minutes against a mean of 4.
Figure (svg): The exponential curve with the median at two point eight and the larger mean at four both marked
The shaded half of the area ends well to the left of the mean, and the reason is visible in the picture: the thin tail stretching far to the right contributes little area but a great deal of leverage. Half of all customers are finished within 2.8 minutes even though the average is four.
Worked example
Example 5.8(b), solved the book's long way.
\[ X \sim \text{Exp}(0.25); \text{ find } k \text{ with } P(x < k) = 0.50 \]
Write the equation
Why: The left area equals a half.
\[ 0.50 = 1 - e ^{-0.25 k} \]
Isolate the exponential
Why: Rearrange.
\[ e ^{-0.25 k} = 0.50 \]
Take natural logs
Why: Of both sides.
\[ -0.25 k = \ln(0.50) \]
Solve for k
Why: Divide through.
\[ k = 2.77 \]
Figure (svg): The solution to Worked example the median service time shown as a ladder of expressions, one row per legal move
\[ k = \frac{\ln(0.50)}{-0.25} \approx 2.77 \text{ minutes} \]
Verify: confirm the median falls below the mean, as it must
Why: The mean is four minutes and the median came out at 2.8, which is the right order for a right-skewed distribution — the book asks which is larger for exactly this reason. The general check is that the median should be about 69 percent of the mean, and 2.77 over 4 is 0.693. A median above the mean would signal that the logarithm had been applied to the wrong quantity.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 304-305
Faded example
Travellers buy tickets 15 days ahead on average, so m = 1/15. Find the median.
Fill in the blanks
k = \frac0.5010.4 = \frac___})}___ \approx ___ \text___
Why: The natural log of 0.5 is about minus 0.693, and dividing by minus one fifteenth gives 10.4 days. As always the median is about 69 percent of the mean of 15.
Worked example
Example 5.9(c), using the formula directly.
\[ m = 0.1; \text{ eighty percent of parts last at most how long?} \]
Identify the left area
Why: Eighty percent below k.
\[ p = 0.80 \]
Apply the formula
Why: ln of one minus p, over minus m.
\[ \ln(0.20) / (-0.1) \]
Evaluate the log
Why: The natural log of 0.2.
\[ \text{about } -1.609 \]
Divide
Why: By minus 0.1.
\[ 16.09 \]
Figure (svg): The solution to Worked example eighty percent of computer parts shown as a ladder of expressions, one row per legal move
\[ k = \frac{\ln(1 - 0.80)}{-0.1} = \frac{\ln(0.20)}{-0.1} \approx 16.1 \]
Verify: confirm the answer sits far above the mean, and why that is right
Why: The mean is ten years and the 80th percentile is 16.1, well beyond it — which is characteristic of a long right tail. For a symmetric distribution the 80th percentile would sit much closer to the centre. A useful cross-check is that the 80th percentile of an exponential is always about 1.61 times the mean, since ln 5 is about 1.609.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, p. 306
Trap
\[ k = \frac{\ln(0.80)}{-0.1} \approx 2.23 \]
Put the given eighty percent straight into the logarithm
Why: The formula has a log and the question supplies 0.80.
\[ 2.23 \text{ years, well BELOW the mean of ten} \]
The 80th percentile cannot lie below the mean of a right-skewed distribution, let alone at a fifth of it.
\[ k = \frac{\ln(1 - 0.80)}{-0.1} = \frac{\ln(0.20)}{-0.1} \approx 16.1 \]
Subtract from one first: the formula needs the area to the RIGHT inside the log
Why: One minus the left area is what remains above k.
The structure is worth understanding rather than memorising: the equation being solved is e to the minus m k equals one minus p, and the left-hand side is the right tail. So whatever goes inside the logarithm is always the proportion ABOVE k. Checking the answer against the mean catches the error whenever the percentile is above the 63rd, which is where the mean sits.
Two truths and a lie
All three concern percentiles and skew.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The mean sits at about the 63rd percentile, since the left area at the mean is one minus e to the minus one, about 0.632. Assuming the mean splits the distribution in half is the reason 0.50 feels like the right answer for a tail at the mean, and it is not.
Prediction
Commit before reasoning.
Predict first
For any exponential distribution, what proportion of values fall below the mean?
Correct: About 63 percent.
Why: The left area at the mean is one minus e to the minus one, which is about 0.632, and it is the same for every exponential because m cancels. So most values fall below the mean and a minority of long waits pull it upward — the signature of right skew. That the answer does not depend on m is worth noting: the exponential has the same shape at every scale.
Explain it
A classmate cannot see how half of all customers finish in 2.8 minutes when the average is 4.
Discussion prompt
In three sentences or fewer, explain it.
Hint: Ask what a few very long visits do to an average.
Answer:
Point out that the average is a balance point, and a handful of very long visits sit far out to the right where they exert a lot of leverage.
Most visits are short — the curve is highest near zero — so the middle value is pulled down toward the crowd while the mean is pulled up by the tail.
This is section 2.6's rule for right-skewed data, and for the exponential it is always the same gap: the median is about 69 percent of the mean, whatever the units.
Section
Section 4
Concept
The additional time spent waiting for the next event does not depend on how much time has already elapsed. The memoryless property says that the probability of waiting more than r plus t, given that you have already waited more than r, equals the unconditional probability of waiting more than t.
the memoryless property — P(X > r + t given X > r) equals P(X > t), for all r and t at least zero. The book's gloss for a component is that it stays as good as new until it suddenly breaks.
\[ P(X > r + t \mid X > r) = P(X > t) \]
The property is a direct consequence of the closed form. The conditional probability is e to the minus m times r plus t, over e to the minus m r, and the exponentials divide to leave e to the minus m t — which is P(X > t) with the r gone. Nothing else has that behaviour: the uniform of section 5.2 emphatically does not, since conditioning there raised the probability from 0.478 to 0.733.
Figure (svg): Two exponential curves of identical shape, one starting at zero and one starting at five minutes, showing that the wait ahead looks the same whenever you start measuring
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, p. 310 — the memorylessness section and Example 5.12
Picture it
Five minutes have passed with no arrival. What does the wait ahead look like?
Figure (svg): Two exponential curves of identical shape, one starting at zero and one starting at five minutes, showing that the wait ahead looks the same whenever you start measuring
The dashed curve is the solid one translated, not reshaped — that identity of shape is the whole property. It is why the book says an old part is not any more likely to break down at any particular time than a brand new one, and why the exponential is the wrong model for anything that genuinely wears out.
Worked example
Example 5.12. A customer has already been served four minutes.
\[ X \sim \text{Exp}(0.25); \text{ find } P(X > 7 \mid X > 4) \]
Identify r and t
Why: Four already, three more.
\[ r = 4, t = 3 \]
Apply the property
Why: The r drops out.
\[ P(X > 3) \]
Use the right tail
Why: e to the minus m t.
\[ e ^{-0.75} \]
Evaluate
Why: The book's answer.
\[ 0.4724 \]
Figure (svg): The solution to Worked example three more minutes with the clerk shown as a ladder of expressions, one row per legal move
\[ P(X > 7 \mid X > 4) = P(X > 3) = e^{-0.75} \approx 0.4724 \]
Verify: confirm the shortcut against the full conditional formula
Why: Doing it the long way gives P(X > 7) over P(X > 4), which is e to the minus 1.75 over e to the minus 1 — and subtracting exponents leaves e to the minus 0.75, the same 0.4724. The exponents subtract, which is exactly why the r disappears. Checking it once this way is worth the time, because the property is surprising enough that a shortcut alone rarely convinces.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, p. 310
Faded example
A light bulb has mean life 8 years and has already lasted 12. Find the chance it lasts past 19.
Fill in the blanks
P(X > 19 \mid X > 12) = P(X > 7) = e^0.4169 \approx ___
Why: Nineteen minus twelve is seven, so only the extra seven years matter and the twelve already elapsed drop out. The answer is e to the minus seven eighths, about 0.4169.
Worked example
The book's reliability reading of the same property.
\[ \text{a part with mean life } 10 \text{ years has already lasted } 10 \]
State the question
Why: Another seven years.
\[ P(X > 17 | X > 10) \]
Apply the property
Why: The ten drops out.
\[ P(X > 7) \]
Evaluate
Why: e to the minus 0.7.
\[ 0.4966 \]
Compare with a new part
Why: The same computation.
\[ 0.4966 \]
Figure (svg): The solution to Worked example an old part against a new one shown as a ladder of expressions, one row per legal move
\[ P(X > 17 \mid X > 10) = P(X > 7) = 0.4966 \]
Verify: confirm whether this is a reasonable assumption about real parts
Why: It usually is not. Real components wear out, so an old part is typically MORE likely to fail soon than a new one, and reliability engineers use distributions with an increasing hazard rate for exactly that reason. The exponential is right for failures caused by external shocks arriving at random rather than by accumulated wear — and knowing which situation you have is what decides whether the model may be used at all.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, p. 310
Trap
\[ \text{five minutes have passed, so an arrival is overdue} \]
Reason that a long gap makes the next event more likely soon
Why: The average gap is two minutes and five have gone by.
\[ P(X > 1 \mid X > 5) = P(X > 1) = 0.6065 \]
The chance of waiting another full minute is unchanged, exactly as if the last arrival had just happened.
\[ P(X > r + t \mid X > r) = P(X > t) \text{ for every } r \]
Discard the elapsed time entirely and ask only about the additional wait
Why: The past carries no information under this model.
The intuition being overridden here is the gambler's fallacy in its waiting-time form, and it is strong: the book acknowledges that it would SEEM more likely for a customer to arrive within the next minute. Whether the intuition or the model is right depends on the process — for genuinely random arrivals the model is right, and for anything on a schedule it is not.
Two truths and a lie
All three concern memorylessness.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false, and it overcorrects. Memorylessness means the old part faces exactly the SAME outlook as a new one — neither better nor worse. The book's phrase is that the part stays as good as new until it suddenly breaks, which is a statement of equality rather than advantage.
Prediction
Commit before reasoning.
Predict first
For a uniform on [0, 23], conditioning on x > 8 raised P(x > 12) from 0.478 to 0.733. What does the same conditioning do to an exponential?
Correct: Leaves it unchanged.
Why: For the exponential the chance of waiting a further four minutes is the same whether or not eight have already passed, because the elapsed time cancels out of the conditional. The uniform behaves differently because it is bounded above: surviving past 8 genuinely uses up part of a fixed budget of 23, whereas an exponential has no upper limit to use up.
Matching
Match each situation to whether memorylessness is a reasonable assumption.
Match the pairs
Why: The two reasonable cases share a feature: events arrive from outside with no internal clock. The two unreasonable ones each have a mechanism that makes the past informative — accumulated wear in one, a timetable in the other. Deciding which kind of process you have is what licenses the model.
Section
Section 5
Concept
Suppose the time between two successive events follows an exponential distribution with mean mu, and the times are independent. Then the number of events per unit time follows a Poisson distribution with mean lambda equal to one over mu. Conversely, if the counts are Poisson then the gaps are exponential.
the exponential-Poisson equivalence — One process, two descriptions. Exponential gaps with mean mu correspond to Poisson counts with mean one over mu per unit time. The book states the implication in both directions.
\[ \lambda = \frac{1}{\mu} \]
This closes a loop opened in section 4.6. That section's first characteristic required events to occur with a known average rate and independently of the time since the last event — and independence of the time since the last event is precisely memorylessness. So the Poisson's defining condition and the exponential's defining property are the same statement, seen from the count side and the gap side. That is why the two families always appear together.
Figure (svg): Two panels showing the same sequence of events described first by the gaps between them and second by the count in each unit of time
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 311-312 — the relationship and Example 5.13
Picture it
The same events, described by their gaps and by their counts.
Figure (svg): Two panels showing the same sequence of events described first by the gaps between them and second by the count in each unit of time
Being able to switch sides is genuinely useful, because some questions are far easier on one side than the other. How long until the next call is an exponential question; how many calls in the next eight minutes is a Poisson one — and Example 5.13 asks both of the same switchboard.
Worked example
Example 5.13(a) to (c). Four calls a minute on average.
\[ 4 \text{ calls per minute; find the mean gap, } P(T < \tfrac{1}{6}), \text{ and } P(X = 5) \]
Find the mean gap
Why: One minute over four calls.
Get the decay parameter
Why: One over the mean gap.
\[ m = 4 \]
Next call within ten seconds
Why: Ten seconds is a sixth of a minute.
\[ 1 - e ^{-\frac{4}{6}} = 0.4866 \]
Exactly five calls in a minute
Why: Poisson with mean 4.
\[ 0.1563 \]
Figure (svg): The solution to Worked example a police switchboard, both ways shown as a ladder of expressions, one row per legal move
\[ T \sim \text{Exp}(4), \quad X \sim \text{Poisson}(4) \]
Verify: confirm the two parameters are the same number, and why
Why: The exponential's decay parameter is 4 per minute and the Poisson's mean is 4 per minute — they coincide because lambda equals one over mu and m also equals one over mu. That coincidence only holds when the Poisson's interval is one unit of the same time in which the gap was measured; for an eight-minute window the Poisson mean becomes 32 while m stays 4, which is exactly what part (e) requires.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 311-312
Faded example
Accidents occur at a Poisson rate of three per week. Find the mean gap between them.
Fill in the blanks
\mu = \frac32.33} \text___ \approx ___ \text___
Why: The mean gap is the reciprocal of the rate, a third of a week, which is about 2.33 days. The exponential's decay parameter for gaps measured in weeks is 3 per week.
Worked example
Example 5.13(e), which must be done on the count side.
\[ \text{four calls a minute; find } P(Y > 40) \text{ over eight minutes} \]
Scale the mean to the window
Why: Eight times four.
\[ 32\text{ calls} \]
State the distribution
Why: Poisson on that window.
Use the complement
Why: More than forty.
\[ 1 - P(Y \le 40) \]
Evaluate
Why: The book's answer.
\[ 0.0707 \]
Figure (svg): The solution to Worked example more than forty calls in eight minutes shown as a ladder of expressions, one row per legal move
\[ P(Y > 40) = 1 - P(Y \le 40) \approx 0.0707 \]
Verify: confirm why this question resists the exponential side
Why: Asking for more than forty calls in eight minutes would, on the gap side, mean asking whether forty-one gaps sum to less than eight minutes — a question about a sum of forty-one random quantities, which this book has no tool for. Counting is easy where summing is hard, which is the practical reason for keeping both descriptions available. Note also the mean scales with the window, exactly as section 4.6 required.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, p. 312
Trap
\[ Y \sim \text{Poisson}(4) \text{ for an eight-minute period} \]
Keep the stated rate as the Poisson mean
Why: Four calls a minute is the number given.
\[ P(Y > 40) \approx 0 \text{ instead of } 0.0707 \]
Forty calls would be ten times the mean of four, which is essentially impossible — but over eight minutes the mean is 32 and forty is quite ordinary.
\[ \lambda = (8)(4) = 32, \quad Y \sim \text{Poisson}(32) \]
Scale the mean to the window the question asks about
Why: Section 4.6's rescaling rule, unchanged.
Notice the asymmetry between the two sides: the exponential's m never changes, since it describes the gaps whatever window you are looking at, while the Poisson's lambda must be rescaled for every new window. Keeping that straight is the main bookkeeping cost of working with both descriptions.
Sorting
Ask whether the question is about a duration or a count.
Sort into buckets
Sort each question about a switchboard taking four calls a minute.
The tell is the units of the answer: a time on the left, a whole number on the right. All five describe the same switchboard, which is the point — one process, and you pick whichever description makes the question easy.
Two truths and a lie
All three concern the equivalence.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The decay parameter describes the gaps between events and does not depend on what window you happen to be examining, so it stays at 4 per minute throughout. Only the count's mean scales, because only the count depends on how long you watch.
Prediction
Commit before reasoning.
Predict first
Which Poisson characteristic from section 4.6 corresponds to the exponential's memorylessness?
Correct: That events occur independently of the time since the last event.
Why: That phrase in the Poisson's first characteristic and the exponential's memoryless property are the same requirement, stated once about counts and once about gaps. Recognising them as one condition is what makes the equivalence between the two families feel inevitable rather than coincidental.
Comparison
Fill the blanks. Both are continuous; almost everything else differs.
Comparison matrix
| Feature | Uniform | Exponential |
|---|---|---|
| Shape | flat | declining |
| Parameters | a and b | m alone |
| Mean against median | equal | mean is larger |
| Conditioning on x > c | raises the probability | changes nothing |
The last row is the one worth carrying forward. Memorylessness singles the exponential out among every distribution in this book, and it is the reason it pairs with the Poisson rather than with anything in chapter 5.
Pattern
Six steps, and the first two are where the errors live.
If the question asks how many events rather than how long until one, switch to the Poisson with mean one over mu — and rescale that mean to the window being asked about.
OpenStax Introductory Business Statistics 2e, §5.3 The Exponential Distribution §5.3 The Exponential Distribution
Check
The parameter.
Check your understanding
A component lasts 25 months on average. What is m?
Answer: A
Why: The decay parameter is the reciprocal of the mean, and one over 25 is 0.04 per month.
Check
A right tail.
Check your understanding
For X ~ Exp(0.1), what is P(x > 7)?
Answer: A
Why: The right tail is e to the minus m x, with no subtraction needed: e to the minus 0.7 is about 0.4966.
Check
Memorylessness.
Check your understanding
A machine's time to failure is exponential with mean 10 years. It has run 6 years. What is the chance it runs at least 4 more?
Answer: A
Why: Memorylessness means the six years already elapsed drop out, leaving P(X > 4) — exactly the probability faced by a brand new machine.
Real world
A hospital's IT team reports that a server has run 400 days without an unplanned reboot, against a historical mean time between failures of 200 days. One manager argues the server is overdue and should be replaced this week; another argues it has proved itself and should be left alone.
Discussion prompt
Assess both arguments under an exponential model, then say what evidence would justify departing from that model.
Hint: Both managers are reasoning from the elapsed 400 days. What does the model say that is worth?
Answer:
Under an exponential model both managers are wrong, and for the same reason. With a mean of 200 days, m is 0.005 per day, and the chance of surviving another 30 days is e to the minus 0.15, about 0.861 — whether the server is brand new or has already run 400 days. The elapsed time carries no information at all.
\[ P(X > 430 \mid X > 400) = P(X > 30) = e^{-(0.005)(30)} \approx 0.861 \]
The 400 days is also not as surprising as it sounds. Surviving twice the mean has probability e to the minus 2, about 0.135 — roughly one machine in seven does it, so a fleet of any size will contain several such servers. Treating an ordinary outcome as a signal is the error both managers are making, in opposite directions.
What would justify departing from the model is evidence about the failure mechanism, not about this server's age. If failures come from wearing parts — disks, fans, thermal paste — then the hazard rises with age, the first manager is right in substance, and a distribution with an increasing hazard rate should replace the exponential. If they come from external shocks such as power events or software faults arriving at random, memorylessness is appropriate and neither age argument carries weight.
The practical recommendation follows from that: rather than arguing from one server's age, look at the fleet's failure times and check whether the empirical hazard rate rises, falls or stays flat with age. A flat hazard is exactly what an exponential means, and it is a testable claim rather than an assumption — which is the honest way to settle a disagreement of this kind.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does the elapsed time drop out of an exponential conditional probability?
Correct: Because the exponents subtract and leave only the additional time.
\[ \frac{e^{-m(r+t)}}{e^{-mr}} = e^{-mt} = P(X > t) \]
Why: The conditional is e to the minus m times r plus t, divided by e to the minus m r, and dividing powers subtracts exponents — leaving e to the minus m t, which is P(X > t) with r gone. The property is a consequence of the algebra of the closed form, not a separate assumption, which is why no other distribution in this book shares it.
Explain it
They computed the 80th percentile of a mean-ten-year part as ln(0.80) over minus 0.1, getting 2.23 years.
Discussion prompt
In three sentences or fewer, locate the error.
Hint: Ask whether the 80th percentile should sit above or below the mean.
Answer:
Ask them where the 80th percentile ought to lie relative to the mean of ten years: well above it, so an answer of 2.23 is wrong before the algebra is checked.
The equation being solved is e to the minus m k equals one minus p, and the left-hand side is the RIGHT tail — so what goes inside the logarithm is the proportion above k, which is 0.20 rather than 0.80.
That gives ln(0.20) over minus 0.1, about 16.1 years, which sits sensibly above the mean as a long right tail requires.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, m is one over the mean, and the curve's height at zero must equal m. For the second, sketch and shade before substituting. For the third, the proportion ABOVE k goes inside the log. For the fourth, discard the elapsed time and keep only the additional wait, and switch to the Poisson whenever the answer should be a whole number. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Twenty minutes.
Draw it
At the top, write mean equals 4 and m equals 0.25 for Example 5.7's postal clerk, then sketch the curve f(x) = 0.25e^(-0.25x) from 0 to 20, marking the height at zero as 0.25 and checking it equals m. Shade the strip from 4 to 5 and compute its area as the difference of two left areas, aiming for 0.0814. On the same sketch, mark the median at 2.8 and the mean at 4, and write one sentence on why they differ and in which direction. Below that, redo the whole thing for Example 5.9's computer part with a mean of ten years: compute m, the right tail past seven years, the 80th percentile, and the probability of lasting between nine and eleven years, checking against 0.4966, 16.1 and 0.0737. Beside those, write the memoryless property as a formula and verify it once by long division of the two tails, showing that the exponents subtract. At the bottom, draw a timeline with events marked on it, label the gaps as exponential with parameter m and the counts per unit time as Poisson with mean lambda, and write lambda equals one over mu between them. Finish with a two-column table of the uniform against the exponential, with rows for shape, parameters, mean against median, and what conditioning on x greater than c does.
Check the median against the mean by dividing: 2.77 over 4 is 0.693, and that ratio should come out the same for the computer part, 6.93 over 10. If it does not, the logarithm has been applied to the wrong quantity. And check that your memoryless long division really leaves e to the minus m t with no r anywhere in it.
Recap
Five things, and the fourth is unique to this distribution.
| If you see | Then |
|---|---|
| Time until an event, with small values commoner | Exponential: X follows Exp(m) |
| A mean given | m is its reciprocal, and sigma equals the mean |
| A greater-than question | e to the minus m x, with no subtraction |
| A proportion given, a time wanted | A percentile: the proportion ABOVE k goes in the log |
| Time already elapsed | Discard it: only the additional wait matters |
| A question wanting a whole number | Switch to the Poisson, and rescale its mean |
| A process that wears out with age | The exponential is the wrong model |
That closes chapter 5. Chapter 6 takes the same machinery to the distribution the rest of the book runs on: the normal, whose bell shape has no elementary area formula at all, so its probabilities must come from a table or a calculator. The z-score of section 2.7 returns there as the device that makes one table serve every normal distribution.
OpenStax Introductory Statistics 2e, §5.3 The Exponential Distribution §5.3, pp. 302-312 — everything on these slides traces back here
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