Poisson Distribution Exercises in R: 16 Real-World Practice Problems

Explanation: dpois(x, lambda) evaluates the Poisson probability mass function $P(X = x) = \lambda^{x} e^{-\lambda} / x!$ at a single integer. Here $5^{2} e^{-5} / 2!$ collapses to roughly 0.0842, an 8.4 percent chance. A common mistake is calling ppois() instead, which would give the cumulative probability of 0, 1, or 2 calls (about 0.125). Whenever the question says "exactly k", reach for dpois(); "at most k" maps to ppois().

Explanation: The Poisson mass function peaks near floor(lambda). Router A with lambda = 3 sits one step below four, so dpois(4, 3) is still close to the mode and returns 0.168. Router B at lambda = 7 sits three steps above four, so the tail probability drops to 0.091. A frequent pitfall is comparing tail probabilities directly without normalising for the underlying rate; here the comparison is valid because both queries evaluate the same x at different lambdas.

Explanation: dpois() is vectorised over its first argument, so passing 0:10 returns the whole PMF in one call rather than looping. When lambda is an integer, the mass function has two equal modes at lambda - 1 and lambda; here that means k of 3 and 4 both peak at 0.1954. Naming the vector by its k values makes downstream filtering (for instance ex_1_3[as.character(5:7)]) read clearly. For a non-integer lambda, the mode collapses to the single integer floor(lambda).

Explanation: ppois(q, lambda) returns $P(X \le q)$ by default. Setting lower.tail = FALSE flips it to $P(X > q)$, which is more numerically stable than computing 1 - ppois(...) when the lower tail is close to one. Because the two events partition the sample space, the sum is exactly 1.0 up to floating-point error. A frequent mistake is reading "more than 4" as q = 5; the correct call is q = 4, lower.tail = FALSE, which excludes 4 itself.

Explanation: "20 or more tickets" is $P(X \ge 20)$, which equals $P(X > 19)$, so the upper-tail call uses q = 19 not q = 20. This off-by-one is the single biggest source of bugs when translating English statements into ppois() arguments. With overload sitting at roughly 1.2 percent the manager can comfortably staff for fifteen with confidence, but a separate Erlang-C calculation would also account for service time, not just arrival rate.

Explanation: Vectorising ppois() over cutoffs returns the CDF at each gridpoint in one call, which is the right way to inspect tail behaviour without a loop. Notice the CDF is non-decreasing and climbs from 0.19 at k=5 (well below the mean) to 0.99 at k=15 (almost two standard deviations above). A useful sanity check is that ppois(floor(lambda), lambda) for lambda = 8 should sit near 0.59, reflecting the slight right skew of the distribution.

Explanation: qpois(p, lambda) returns the smallest integer k with $P(X \le k) \ge p$. Because the Poisson is discrete the realised coverage almost always overshoots p; here 0.9621 is the first integer that clears 0.95. Reusing the same logic, qpois(0.5, lambda) returns the median, which for integer lambda equals lambda. A common pitfall is interpreting qpois() as a continuous inverse-CDF: the result is always integer-valued and stepwise, never interpolated.

Explanation: With weekly mean demand of 30 the standard deviation is $\sqrt{30} \approx 5.48$, so a 99-percent cushion sits roughly $2.33 \times 5.48 \approx 12.8$ units above the mean, giving 43. This matches the integer answer from qpois() exactly. The same call structure underpins safety-stock decisions across e-commerce and pharmacy inventory, where ordering "mean plus a margin" leaves you stocking out half the time and ordering by quantile keeps you above your target service level.

Explanation: rpois(n, lambda) returns n independent draws from a Poisson with the given rate. For large n the sample mean is consistent for lambda, which is why 7.046 sits very close to the true 7. The sample variance should be similar (here roughly 7.0) because the Poisson has the equidispersion property var = mean. A real-world departure from this equality is the classic test for overdispersion and motivates the negative binomial as a richer alternative.

Explanation: The Monte Carlo estimate's standard error scales like $\sqrt{p(1-p)/n}$, so 100,000 draws of a probability near 0.084 gives a standard error around 0.0009 and the two estimates agreeing to three decimals is expected. Note the asymmetry in arguments: mean(draws >= 10) is the empirical fraction at or above 10, which equals $P(X > 9) = $ ppois(9, lower.tail = FALSE). Mixing this up by passing 10 to the theoretical call is the classic off-by-one.

Explanation: rpois() recycles its lambda argument elementwise, so passing a length-30 vector draws each entry from a Poisson with its own rate in a single C-level call rather than three separate calls. This is meaningfully faster than looping and is the same trick used in non-homogeneous Poisson process simulation. Expected total is $10 \times 2 + 10 \times 5 + 10 \times 12 = 190$, so a draw of 199 is well within one standard deviation $\sqrt{190} \approx 13.8$ of the expectation.

Explanation: Solving $\partial / \partial \lambda \log L = 0$ for the Poisson likelihood gives $\hat{\lambda}_{\text{MLE}} = \bar{x}$, so mean() is the maximum-likelihood estimate and no optimiser is needed. The closed form makes the Poisson one of the simplest distributions to estimate, but the trade-off is the strong "mean equals variance" assumption. A quick diagnostic is var(counts_5_1) / mean(counts_5_1); values much above 1 signal overdispersion and a negative binomial may fit better.

Explanation: The Wald interval relies on the asymptotic normality of the MLE with variance $\lambda / n$, so the standard error is $\sqrt{\hat{\lambda}/n}$. For moderate to large samples this is accurate, but if your counts are small (mean below about 2) the interval can dip below zero and you should switch to the exact poisson.test() interval shown in Section 6. Using qnorm(0.975) (about 1.96) avoids hard-coding the magic number and makes the formula generalise to other confidence levels.

Explanation: MASS::fitdistr() is a general MLE fitter that recognises common distributions by name and skips the optimiser when a closed-form solution exists. For the Poisson it returns the sample mean as the estimate plus a standard error equal to $\sqrt{\hat{\lambda}/n}$. The wrapper is more useful when you need a quick AIC or log-likelihood comparison: fit_5_3$loglik and AIC(fit_5_3) work out of the box. For richer count models (negative binomial, zero-inflated) use MASS::glm.nb() or the pscl package instead.

Explanation: poisson.test(x, T, r) performs an exact binomial-based test for the rate parameter against the hypothesised value r. With 14 events when 8 were expected the two-sided p-value lands just under 0.05, suggesting evidence against $H_0$ at the conventional level (though only marginally). The T argument is the exposure (time, area, person-years); always set it explicitly so the test interprets the rate per unit correctly. The exact test is preferable to a normal approximation when the count is small.

Explanation: With two counts and two exposures, poisson.test() switches to a conditional binomial test on the first count given the total, testing the null rate ratio r. The result mirrors what you would get from an exact two-sided binomial test on 12 successes out of 37 with success probability 0.5. The 95-percent confidence interval for the rate ratio (test_6_2$conf.int) usefully complements the p-value because it tells you how big or small the true ratio could plausibly be, not just whether it differs from 1.

Explanation: The conf.int element of a htest object holds the exact confidence interval for the parameter, here the rate ratio $\lambda_A / \lambda_B$. Because the interval (0.23, 0.99) excludes 1, arm A's event rate is statistically lower than arm B's at the 5 percent level, matching the significant p-value above. For regulatory reporting it is good practice to quote both the point estimate (test_6_3$estimate) and the interval rather than the p-value alone, because the interval communicates effect size and uncertainty in one shot.