Probability Distributions: Poisson, Normal & More

How to choose between probability distributions

Start by defining the quantity you want to predict. A count of customers differs from the time until a customer arrives. Similarly, the number of purchases among 30 visitors differs from the number of visitors during an hour.

Next, identify the mechanism behind the observations. Fixed independent trials suggest a binomial model; independent arrivals over a fixed exposure suggest a Poisson model. Meanwhile, a continuous measurement may justify a normal model if its shape and context support that choice.

Source: The NIST distribution gallery distinguishes discrete and continuous families. The comparison above organizes those families around practical questions.

Discrete probability versus continuous density

A discrete model assigns probability to individual outcomes. For example, a Poisson variable can equal exactly three customers, and that event can have positive probability. By contrast, a continuous model assigns probabilities to intervals through areas under a density.

Consequently, a density height is not a probability. A continuous density may exceed one without violating probability rules, provided its total area equals one. Consider a uniform waiting time between zero and half an hour: its density equals two per hour, but its total probability remains one.

Moreover, P(X = a) equals zero for a variable with a continuous density. Thus, including or excluding a single endpoint does not change its interval probability. Integer counts behave differently, so P(X > 3) and P(X ≥ 3) usually differ.

The Poisson distribution: counting events

The Poisson distribution describes nonnegative integer counts with parameter λ. Here, λ represents the expected number of events in the exposure you selected. Therefore, six arrivals per hour implies λ = 6 for one hour, but λ = 3 for half an hour.

In this formula, k is the target count, e is approximately 2.71828, and k! is a factorial. Also, 0! equals one. The model gives E(X) = λ, Var(X) = λ, and SD(X) = √λ.

Reference: Penn State: The Poisson Distribution develops the model and its connection to rare binomial events.

Assumptions that deserve more attention than the formula

For a homogeneous Poisson arrival process, disjoint intervals have independent counts and the rate stays constant. Over a sufficiently short interval, the chance of one arrival grows approximately in proportion to the interval length. Meanwhile, the chance of multiple arrivals becomes negligible relative to that length.

However, “two possible outcomes” is not the defining assumption of a Poisson count. That wording belongs naturally to a Bernoulli trial. An hourly Poisson count can take zero, one, two, or many values.

In practice, lunch-hour surges, group arrivals, and advertising campaigns can break the simple constant-rate model. Therefore, define comparable time windows before you estimate an average. A daily average alone does not establish that hourly arrivals follow a homogeneous Poisson process.

Worked example: exactly three customers in one hour

Suppose a fuel station receives an average of six customers per hour. Assume a homogeneous Poisson process for this teaching example. Then the hourly count X has λ = 6.

Therefore, exactly three customers occur in about 8.92% of comparable hours under this model. This statement describes a long-run probability, not a schedule for the next hour.

Worked example: three or fewer customers

Now include every allowed count from zero through three. Because these outcomes cannot happen simultaneously, their probabilities add.

In addition, the mean equals six customers and the standard deviation equals √6 ≈ 2.4495 customers. Notice that the mean need not describe the most likely count uniquely. For λ = 6, counts of five and six have equal maximum probability.

Curiosity: the recurrence behind the two peaks

You can generate adjacent probabilities without recalculating factorials. Specifically, divide the formula for k + 1 by the formula for k.

For λ = 6 and k = 5, this ratio equals one. Consequently, P(X = 5) and P(X = 6) match. This small identity also provides a useful arithmetic check when you build a probability table.

Binomial versus Poisson: a defect example

Suppose each bulb independently has a 1% defect probability, and a sample contains 30 bulbs. Because the number of trials is fixed, the exact model is binomial: X ~ Binomial(30, 0.01). We want the probability of more than one defect.

Rather than sum 29 terms, use the complement. “More than one” means two or more, so only zero and one belong to the complementary event.

Alternatively, approximate the count with a Poisson variable whose mean is np = 0.3. This substitution gives a nearby answer, but it does not make the two distributions identical.

Thus, the approximation overstates the probability by about 2.18% relative to the exact answer. Although 0.0788 percentage points looks small, relative error answers a different question. Both measures help you judge whether an approximation suits a decision.

Reference: SciPy’s binomial documentation defines the exact model used for the numerical verification. These figures are calculations from the stated inputs, not observed factory defect data.

Why “large n” is not enough

A Poisson approximation to a binomial variable works best when individual success probabilities are small and the number of trials is large. However, no single classroom cutoff guarantees accuracy for every tail event. Therefore, compare with an exact calculation whenever software makes that easy.

If p is close to one, count failures instead. For example, a 99% success probability corresponds to a 1% failure probability. Then approximate the rare failure count and translate the event back carefully.

The normal distribution: measurements and z-scores

A normal distribution has a symmetric bell-shaped density. Its mean μ locates the center, while its standard deviation σ controls the spread. In the notation X ~ N(μ, σ²), the second parameter represents variance.

Here, exp(u) means e raised to the power u. It is a function, not another name for the constant e. Consequently, writing “exp = 2.7183” loses essential information about the exponent.

Standardize before looking up an area

Subtract the mean and divide by the standard deviation to obtain a z-score. A value of z = 2 lies two standard deviations above the mean. Similarly, z = −0.5 lies half a standard deviation below it.

For a normal X, the transformed variable Z follows the standard normal distribution. Define Φ(z) as P(Z ≤ z). Then an interval probability equals the cumulative probability at its upper boundary minus the cumulative probability at its lower boundary.

Reference: SciPy’s normal distribution documentation explains the location and scale parameters and the cumulative function used here.

The empirical rule, with more precise numbers

The familiar 68–95–99.7 rule summarizes central normal probabilities. However, those percentages depend on the normal model. A mean and standard deviation alone do not justify them.

For example, a three-standard-deviation event is unusual but possible. The two tails together contain approximately 0.26998% of the distribution. Therefore, “outside three sigma” does not mean an impossible observation.

Worked normal areas from the study pages

Use the same cumulative function for every region below. In particular, an interval crossing zero needs no special formula if your table or software returns Φ directly.

Also, check which area a printed z-table reports. Some tables give the area between zero and z, while others give the full area to the left. Confusing these conventions can shift an answer by 0.5.

Normal distribution examples: lifetimes, income, and grades

Bulb lifetime between 110 and 120 hours

Assume bulb lifetimes follow a normal model with μ = 100 hours and σ = 8 hours. First, convert both limits into standard units: 110 gives z = 1.25, and 120 gives z = 2.50. Then subtract the corresponding cumulative areas.

However, this assumption serves a mathematical exercise rather than a universal model of bulb failure. A normal variable permits negative values, whereas actual lifetimes do not. Therefore, assess whether that impossible tail is negligible before using a normal lifetime model in practice.

Income probabilities under a hypothetical normal model

Suppose a teaching exercise specifies μ = $16,000 and σ = $2,000. These dollar amounts are hypothetical inputs from the supplied material, not current income statistics. Consequently, the resulting probabilities describe the assumed model only.

In an actual income study, inspect the distribution before applying these formulas. A long upper tail or a mixture of different populations can undermine the normal assumption. Therefore, the exercise teaches standardization without establishing an empirical income model.

The cutoff for the highest 10% of exam scores

Assume scores follow N(78, 8²). The top 10% begins at the 90th percentile, so we need z = Φ⁻¹(0.90) ≈ 1.28155157. Next, convert that standardized cutoff back to the score scale.

Thus, the continuous-model cutoff is about 88.25. However, rounding it to 88 changes the rule rather than preserving an exact 10% share. If scores only take integer values and the policy requires a score at least as high as the continuous cutoff, the smallest eligible integer is 89.

Alternatively, an instructor can rank actual students and handle ties explicitly. That approach may achieve the desired class share more directly. A fitted normal percentile cannot guarantee exactly 10% of a finite class.

Normal approximations and continuity correction

To begin, a continuous curve spreads probability across intervals, while a discrete variable concentrates probability at integers. Continuity correction bridges that difference by placing boundaries halfway between neighboring integers. For example, “at least 10” begins at 9.5 on the approximating continuous scale.

Reference: Penn State: Approximations for Discrete Distributions covers normal approximations to binomial and Poisson counts.

At least 10 purchases among 30 visitors

Assume 30 independent visitors each have a 30% purchase probability. Then X ~ Binomial(30, 0.30), with mean 9 and variance 6.3. The exact upper-tail calculation gives P(X ≥ 10) ≈ 0.41119131.

For the normal approximation, keep σ = √6.3 rather than rounding it prematurely. Next, apply the 9.5 boundary.

Therefore, the approximation gives 42.1051%, compared with the exact 41.1191%. The difference is about 0.9860 percentage points, or 2.40% of the exact probability. Without continuity correction, the normal result drops to about 34.52%.

Notice that np = 9 and n(1 − p) = 21. Those values satisfy a common rule requiring both to exceed five, yet they do not guarantee a negligible error. Consequently, a rule of thumb should begin an accuracy check, not replace it.

Four or fewer defects when the expected count is 10

Now suppose X follows a Poisson distribution with λ = 10. We want P(X ≤ 4), a lower-tail event. Directly summing the first five probabilities gives the benchmark.

For comparison, use a normal variable with mean 10 and standard deviation √10. Continuity correction places the upper limit at 4.5.

Although the absolute difference is only about 1.17 percentage points, the approximation exceeds the exact probability by roughly 40.14%. This example exposes a weakness that a rounded percentage alone can hide. Therefore, verify small tail probabilities directly when the decision depends on them.

Two error measures worth keeping separate

Multiply the first result by 100 to express a difference in percentage points. Meanwhile, multiply the second by 100 to express relative error as a percent. In particular, a tiny exact probability can make a modest absolute error large in relative terms.

The exponential distribution: waiting for an event

The exponential distribution connects directly to a homogeneous Poisson process. Counts answer how many events occur during an interval; waiting times answer how long until the next event. To keep units clear, use r for the arrival rate and λ = rt for the expected count over duration t.

For example, no arrival by time t means the count in that interval equals zero. Therefore, P(T > t) = P(N(t) = 0). This connection derives the waiting-time formula directly from the Poisson model.

Source: NIST: Exponential life distribution connects exponential interarrival times with Poisson event counts. Also, NIST’s exponential reference gives the density and cumulative distribution.

First customer within half an hour

Return to the station with six expected arrivals per hour. Here r = 6 per hour and t = 0.5 hour, so rt = 3. Consequently, the probability of a first arrival within 30 minutes is:

Conversely, the probability of no customer during the same interval is e⁻³ ≈ 0.04978707, or 4.9787%. These events complement each other, so their probabilities sum to one.

Mean, variance, and units

When you convert hours into minutes, multiply the mean and standard deviation by 60. However, multiply the variance by 60². This distinction prevents a common dimensional mistake.

Waiting-time percentiles reveal more than the average

An average wait of 10 minutes does not mean that half of all waits exceed 10 minutes. In fact, the exponential distribution is right-skewed. Solve q = 1 − e⁻ʳᵗ to find its qth quantile.

Thus, the median is only about 6.93 minutes, but the 95th percentile approaches 30 minutes. A service promise based solely on the average can therefore misrepresent the customer experience. The relevant percentile depends on the promise you want to make.

Memorylessness does not mean an arrival becomes “due”

Under the exponential model, an elapsed wait does not change the distribution of the remaining wait. Specifically, P(T > s + t | T > s) = P(T > t). Therefore, waiting 20 minutes does not make a customer more likely to arrive during the next minute.

However, this conclusion relies on the constant-rate model. A known opening time, a bus schedule, or a lunchtime rush can invalidate that assumption. Penn State’s exponential distribution lesson explains the memoryless property.

Chebyshev’s inequality: a guarantee without a shape

Sometimes you know a mean and a finite variance but lack a defensible distributional shape. Chebyshev’s inequality then provides a probability bound. Unlike a normal calculation, it does not require a bell-shaped distribution.

Reference: UC Berkeley: normal approximation and probability inequalities distinguishes shape-based approximations from bounds that use limited information.

Schooling between six and ten years

Suppose schooling has mean eight years and standard deviation one year. The interval from six to ten extends two standard deviations from the mean. Consequently, Chebyshev guarantees at least 1 − 1/2² = 75% strictly between those limits.

The inclusive interval also contains at least 75%. Similarly, the probability of falling below six or above ten is at most 25%. Neither result establishes an exact probability.

Why the unknown shape matters: an original counterexample

Consider a hypothetical population where half the people have seven years of schooling and half have nine. Its mean is eight and its variance is one. Yet everyone lies between six and ten, so the interval probability equals 100%.

Now consider a different population: 1/18 have five years, 8/9 have eight years, and 1/18 have eleven years. Its mean also equals eight, and its variance also equals one. However, only 8/9, or 88.8889%, lie between six and ten.

Therefore, identical means and standard deviations can coexist with different interval probabilities. Both populations satisfy the 75% guarantee. This construction shows exactly why the available summary statistics cannot determine a unique answer.

What the calculation audit reveals

The supplied pages cover exercises 3.28 through 3.40. Their sequence offers a useful learning path, but several details deserve correction or qualification. The following audit separates small rounding differences from substantive interpretation errors.

Why early rounding changes the whole Poisson tail

Every term in the λ = 10 lower-tail sum contains e⁻¹⁰. Replacing that shared factor with 0.00005 multiplies all five terms by the same inflation factor. Therefore, the relative error carries through to their sum.

This issue differs from the normal approximation’s roughly 40.14% relative error. The first comes from rounding inside an otherwise exact calculation. By contrast, the second comes from substituting a different distribution.

How to use probability distributions responsibly

Define exposure before you compare counts

Ten defects per hour and ten defects per thousand items describe different exposures. If production speed changes, equal hourly counts can imply different defect rates per item. Therefore, choose the denominator that matches the operational question.

Similarly, compare customer arrivals across windows of equal duration and similar context. Pooling quiet mornings with busy evenings may obscure the pattern you need to predict. A single average can conceal more than it explains.

Inspect variation as well as the average

By definition, a Poisson model sets its theoretical variance equal to its mean. Consequently, much greater empirical variation can signal rate changes, dependence, or another mismatch. A finite sample will not match the equality perfectly, so treat the comparison as a diagnostic rather than an automatic verdict.

For continuous measurements, inspect outliers, shape, and time patterns before choosing a normal model. NIST’s guidance on testing assumptions emphasizes graphical checks of location, variation, randomness, and distributional form.

Separate assumptions, estimates, and predictions

The worked examples treat parameters such as six arrivals per hour as known inputs. In practice, you usually estimate those parameters from data. Therefore, uncertainty about the rate adds another layer beyond the conditional probabilities calculated here.

Also, a good fit in one period does not guarantee a stable future pattern. A promotion can change purchasing behavior, and a process adjustment can change defects. Revisit the model when the underlying conditions change.

Practice problems with explained answers

Problem 1: no customer within 15 minutes

Use the same six-per-hour arrival process. Since 15 minutes equals 0.25 hour, the interval mean is 1.5. Thus, the probability of no customer is e⁻¹·⁵ ≈ 0.22313016, or 22.3130%.

Problem 2: at least one customer within 10 minutes

Ten minutes equals 1/6 hour, so the expected count is one. Apply the complement to obtain 1 − e⁻¹ ≈ 0.63212056. Therefore, an average waiting time of 10 minutes corresponds to a 63.2121% chance of an arrival by that time.

Problem 3: a normal measurement above its mean

Any nondegenerate normal distribution puts half its probability above its mean. Consequently, the answer is 50%, regardless of μ and σ. Symmetry gives this result without a table.

Problem 4: a bound within three standard deviations

Assume only a finite mean and positive finite variance. Chebyshev gives at least 1 − 1/9 = 8/9, or 88.8889%, within three standard deviations. However, the normal model would give about 99.7300%, which requires an additional shape assumption.

Problem 5: exactly four instead of at most four

For X ~ Poisson(10), calculate only the k = 4 term. The result is e⁻¹⁰10⁴/4! ≈ 0.01891664, or 1.8917%. In contrast, the at-most-four probability includes the outcomes zero through four and equals 2.9253%.

Reproduce the numerical results in Python

The following calculations use SciPy’s probability functions. Here, pmf returns an individual discrete probability, cdf returns a cumulative probability, and sf returns the upper tail. For discrete X, sf(k) means P(X > k), so “at least 10” requires sf(9).

from math import exp, sqrt, log
from scipy.stats import poisson, binom, norm

# Customer counts
poisson.pmf(3, mu=6)                 # 0.0892350784
poisson.cdf(3, mu=6)                 # 0.1512038828

# More than one defective bulb
binom.sf(1, n=30, p=0.01)            # 0.0361479983
poisson.sf(1, mu=0.3)                # 0.0369363131

# Normal examples
norm.cdf(2.5) - norm.cdf(1.25)        # 0.0994401087
norm.ppf(0.90, loc=78, scale=8)       # 88.2524125244

# Exact and approximate purchase probabilities
binom.sf(9, n=30, p=0.30)            # 0.4111913148
norm.sf((9.5 - 9) / sqrt(6.3))        # 0.4210512848

# Poisson lower tail and normal approximation
poisson.cdf(4, mu=10)                # 0.0292526881
norm.cdf((4.5 - 10) / sqrt(10))       # 0.0409951605

# Exponential waiting times, with time in hours
1 - exp(-6 * 0.5)                    # 0.9502129316
-log(0.05) / 6 * 60                  # 29.9573227355 minutes

To check implementation details, consult the official documentation for Poisson, binomial, and normal functions. In particular, a survival function can avoid loss of precision when subtracting a cumulative probability very close to one.

Frequently asked questions about probability distributions

What is the difference between Poisson and binomial?

A binomial variable counts successes in a fixed number of independent trials with a common success probability. By comparison, a Poisson variable counts events over an exposure and has no fixed upper count limit. However, Poisson can approximate binomial counts when individual success probabilities are small.

Does a z-score make any variable normally distributed?

No. Standardization changes the center and scale but preserves the distribution’s basic shape. Therefore, a standardized skewed variable remains skewed.

Can a Poisson distribution describe events over distance?

Yes, a suitable Poisson model can describe counts over length, area, or another exposure. For example, the exposure might be a kilometer of material rather than an hour. However, the independence and rate assumptions still require justification.

Is Chebyshev’s inequality an estimate of the true percentage?

It provides a bound, not a point estimate. Consequently, an at-least-75% statement can remain valid when the actual probability equals 90% or 100%.

Why not always use a normal approximation?

A normal curve may fit the center reasonably while missing a tail that matters to your decision. Moreover, exact discrete functions are readily available for these examples. Use approximations to understand structure, and verify their accuracy when precision matters.

Conclusion: choose the model before the formula

Probability distributions help answer different kinds of questions: Poisson for counts, binomial for fixed trials, normal for suitable continuous measurements, and exponential for constant-rate waiting times. Meanwhile, Chebyshev provides guarantees when you know little about the shape.

Before calculating, define the event, check the units, and state the assumptions. Then preserve numerical precision, use continuity correction when appropriate, and distinguish exact results from approximations and bounds. Those habits turn a memorized formula into a defensible statistical answer.