Biostatistics Quiz: Poisson Distribution

20 questions · exam conditions

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Poisson DistributionQuestion 1 of 20

A quality control inspector finds that defective microchips occur at a rate of 2.1 per batch of 1000. If the manufacturing process remains stable, what is the probability that a randomly selected batch will contain more than 3 defective chips?

1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] \approx 0.208

P(X=4) + P(X=5) + P(X=6) + ... \text{ (sum of first 10 terms)} \approx 0.185

1 - [P(X=0) + P(X=1) + P(X=2)] \approx 0.384

\frac{e^{-2.1} \cdot 2.1^4}{4!} \approx 0.095

1 - \Phi\left(\frac{3-2.1}{\sqrt{2.1}}\right) \approx 0.267

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Biostatistics Quiz

Biostatistics Quiz: Poisson Distribution

Practice Poisson Distribution in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Poisson Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] \approx 0.208$ (correct answer)
$P(X=4) + P(X=5) + P(X=6) + ... \text{ (sum of first 10 terms)} \approx 0.185$
$1 - [P(X=0) + P(X=1) + P(X=2)] \approx 0.384$
$\frac{e^{-2.1} \cdot 2.1^4}{4!} \approx 0.095$
$1 - \Phi\left(\frac{3-2.1}{\sqrt{2.1}}\right) \approx 0.267$

Explanation: When you encounter problems about rare events occurring at a known average rate, you're dealing with a Poisson distribution. This scenario fits perfectly: defective chips are rare events (2.1 per 1000 chips), and you need to find the probability of "more than 3" defects. For a Poisson distribution with λ = 2.1, the probability of "more than 3" means P(X > 3) = P(X ≥ 4). The most efficient approach uses the complement rule: P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]. This gives approximately 0.208. Choice A correctly applies this complement approach and provides the right probability calculation. Choice B attempts the direct method by summing P(X=4) + P(X=5) + P(X=6) + ..., which is conceptually correct but arrives at 0.185, suggesting either computational error or insufficient terms in the sum. Choice C uses the complement rule but stops at P(X ≤ 2) instead of P(X ≤ 3), answering "more than 2" rather than "more than 3." Choice D calculates only P(X=4), which gives the probability of exactly 4 defects, not more than 3. The key insight is recognizing that "more than 3" requires careful boundary thinking—it means 4 or greater, so you need P(X ≤ 3) for the complement. Always double-check whether the question asks for "more than," "at least," or "exactly," as these subtle word differences completely change your calculation approach in probability problems.

Question 2

In a study of rare genetic mutations, researchers observe an average of 1.8 mutations per 100,000 births in a population. If a hospital delivers 2,500 babies in a year, what is the expected number of babies with this mutation, and what distribution best models this scenario?

Expected value = 0.045; Poisson distribution with λ = 0.045 (correct answer)
Expected value = 0.045; Binomial distribution with n = 2500, p = 0.000018
Expected value = 1.8; Poisson distribution with λ = 1.8
Expected value = 45; Poisson distribution with λ = 45
Expected value = 4.5; Normal distribution with μ = 4.5

Explanation: When you encounter problems involving rare events occurring in large populations, you're typically dealing with scenarios that call for the Poisson distribution. The key indicators here are "rare genetic mutations" and a specific rate per large population. First, let's calculate the expected number of mutations. The rate is 1.8 per 100,000 births, so for 2,500 births:

\frac{1.8}{100,000} \times 2,500 = 0.045

. This becomes our expected value and our λ (lambda) parameter for the Poisson distribution. The Poisson distribution is ideal here because we have a rare event (low probability) occurring in a large population with a known average rate. When np < 5 in a binomial situation, Poisson provides an excellent approximation and is the preferred model. Looking at the wrong answers: Option B correctly calculates the expected value but suggests a binomial distribution, which is unnecessarily complex when Poisson is more appropriate for rare events. Option C uses λ = 1.8, which represents the rate per 100,000 births, not per 2,500 births—this ignores the scaling needed for the actual hospital population. Option D calculates 45 as the expected value, likely from an error like

1.8 \times 25

instead of properly scaling the rate. Study tip: For rare disease problems, always check if the rate needs to be scaled to match your population size, and remember that when you see "rare events" with known rates, think Poisson distribution first—it's almost always the right choice over binomial in these biostatistics contexts.

Question 3

A pharmaceutical company monitors adverse drug reactions, which occur at a rate of 0.8 per 1000 patients per month. During a clinical trial involving 5000 patients over one month, what is the probability of observing exactly 3 adverse reactions?

$e^{-4} \cdot \frac{4^3}{3!} \approx 0.195$ (correct answer)
$e^{-0.8} \cdot \frac{0.8^3}{3!} \approx 0.021$
$\binom{5000}{3} \cdot (0.0008)^3 \cdot (0.9992)^{4997} \approx 0.195$
$e^{-3} \cdot \frac{3^3}{3!} \approx 0.224$
$\frac{4^3}{3^4} \cdot e^{-4} \approx 0.146$

Explanation: When you encounter a problem about rare events occurring at a known rate over time, you're dealing with a Poisson distribution. This distribution models scenarios where events happen independently at a constant average rate, like adverse drug reactions, equipment failures, or arrivals at a service center. The correct approach requires calculating the expected number of events (λ) first. With a rate of 0.8 per 1000 patients per month and 5000 patients, you get: λ = (0.8/1000) × 5000 = 4 adverse reactions expected. Using the Poisson formula

P(X = k) = e^{-λ} \cdot \frac{λ^k}{k!}

, the probability of exactly 3 events is

e^{-4} \cdot \frac{4^3}{3!} ≈ 0.195

, making A correct. Option B makes a critical scaling error by using λ = 0.8 instead of properly scaling for 5000 patients. This treats the entire population as if it were just 1000 patients. Option C attempts to use the binomial distribution, which could work theoretically but is computationally impractical here. While the calculation setup is correct, the Poisson approximation (option A) is the standard approach when n is large and p is small. Option D arbitrarily uses λ = 3 without any justification from the given data, ignoring the proper calculation of the expected rate. Study tip: For rare events with known rates, always use Poisson. Calculate λ by multiplying the given rate by your population size and time period. If you see very large n with very small p, think Poisson approximation rather than exact binomial calculations.

Question 4

A research laboratory experiences equipment failures at an average rate of 1.5 per month. If equipment failures are independent and follow a Poisson process, what is the probability that the lab experiences no failures in a 2-week period?

$e^{-0.75} \approx 0.472$ (correct answer)
$e^{-1.5} \approx 0.223$
$e^{-3.0} \approx 0.050$
$(1-1.5)^{0.5} \approx 0.354$
$1 - 1.5 \times \frac{2}{4} = 0.25$

Explanation: When you encounter a Poisson process problem, you're dealing with events that occur randomly over time at a constant average rate. The key is recognizing that you need to adjust the rate parameter (λ) to match your time period of interest. The Poisson probability formula is

P(X = k) = \frac{e^{-λ}λ^k}{k!}

, where λ is the average rate for your specific time period. Here, you're given 1.5 failures per month, but you need the probability for a 2-week period. Since 2 weeks = 0.5 months, your adjusted rate is

λ = 1.5 × 0.5 = 0.75

failures per 2-week period. For zero failures (k = 0), the formula simplifies to

P(X = 0) = e^{-λ}

. Substituting λ = 0.75 gives you

P(X = 0) = e^{-0.75} ≈ 0.472

, which is answer A. Looking at the wrong answers: B uses

e^{-1.5}

, which incorrectly applies the monthly rate without adjusting for the 2-week timeframe. C uses

e^{-3.0}

, which appears to double the monthly rate instead of halving it for the shorter period. D uses

(1-1.5)^{0.5}

, which isn't even a valid probability formula and results in a negative base raised to a power. The critical study tip for Poisson problems: always convert your rate parameter to match the time period in question. If the rate is monthly but you need weekly probabilities, scale accordingly. This time-matching step is where most students make errors on biostatistics exams.

Question 5

A call center receives complaint calls following a Poisson distribution with an average of 4.5 calls per hour. During a randomly selected 20-minute period, what is the probability of receiving exactly 2 complaint calls?

$e^{-1.5} \cdot \frac{1.5^2}{2!} \approx 0.251$ (correct answer)
$e^{-4.5} \cdot \frac{4.5^2}{2!} \approx 0.113$
$e^{-3.0} \cdot \frac{3.0^2}{2!} \approx 0.224$
$\frac{4.5 \times 20}{60} \times e^{-2} \approx 0.203$
$\binom{20}{2} \cdot (4.5)^2 \cdot (0.5)^{18} \approx 0.001$

Explanation: When you encounter Poisson distribution problems, the critical first step is adjusting the rate parameter (λ) to match your time period. The Poisson formula is

P(X = k) = e^{-λ} \cdot \frac{λ^k}{k!}

, but λ must correspond to your specific time interval. Here, you're given 4.5 calls per hour, but you need the probability for a 20-minute period. Since 20 minutes = 1/3 hour, you must scale the rate: λ = 4.5 × (1/3) = 1.5 calls per 20 minutes. Now you can calculate:

P(X = 2) = e^{-1.5} \cdot \frac{1.5^2}{2!} ≈ 0.251

, which is answer A. Looking at the wrong answers: B uses the original hourly rate (4.5) without adjusting for the 20-minute period—a common mistake that ignores the time scaling requirement. C uses λ = 3.0, which would correspond to a 40-minute period (4.5 × 2/3), showing confusion about the time conversion. D attempts some time adjustment in the formula itself rather than in the rate parameter, which fundamentally misunderstands how Poisson distributions work. Study tip: Always convert your rate parameter to match the question's time period before plugging into the Poisson formula. Set up a proportion: if the given rate is R per time T₁, then for time T₂, your new rate is R × (T₂/T₁). This time-scaling step is where most students make errors on Poisson problems.

Question 6

A laboratory processes blood samples and occasionally encounters contaminated specimens at a rate of 1.2 per day. Assuming contamination events follow a Poisson process, what is the probability that the lab will encounter its first contaminated sample on the third day?

$P(\text{0 on day 1}) \times P(\text{0 on day 2}) \times P(\text{≥1 on day 3}) = e^{-1.2} \times e^{-1.2} \times (1-e^{-1.2}) \approx 0.065$ (correct answer)
$P(\text{exactly 1 on day 3}) = e^{-1.2} \times 1.2 \approx 0.361$
$P(\text{0 in first 2 days}) \times P(\text{1 on day 3}) = e^{-2.4} \times e^{-1.2} \times 1.2 \approx 0.033$
$(1.2)^3 \times e^{-3.6} \times \frac{1}{3!} \approx 0.072$
$1 - e^{-3.6} \approx 0.973$

Explanation: When you encounter questions about waiting times until the first occurrence of an event in a Poisson process, you're dealing with the geometric-like behavior of successive time intervals. The key insight is that for the first contaminated sample to appear on day 3, two conditions must be met: no contamination on days 1 and 2, AND at least one contamination on day 3. The correct approach breaks this into independent daily probabilities. With rate λ = 1.2 per day, the probability of zero contaminations on any single day is

e^{-1.2}

. Since days are independent, the probability of zero contaminations on both days 1 and 2 is

e^{-1.2} \times e^{-1.2}

. The probability of at least one contamination on day 3 is

1 - P(\text{0 on day 3}) = 1 - e^{-1.2}

. Answer A correctly multiplies these components. Answer B only calculates the probability of exactly one contamination on day 3, ignoring the requirement that days 1 and 2 have zero contaminations. Answer C makes a calculation error by including an extra

e^{-1.2}

term and using 2.4 instead of the correct exponent. Answer D treats this as a single Poisson calculation over 3 days with rate 3.6, which doesn't address the specific timing requirement. Study tip: For "first occurrence" problems in Poisson processes, always think about the waiting time structure: all previous periods must have zero events, and the target period must have at least one event. Set up the problem as a product of these independent probabilities.

Question 7

A pharmacy tracks medication errors, which occur at an average rate of 0.9 per 1000 prescriptions. On a day when 800 prescriptions are filled, what is the variance in the number of expected medication errors?

$\text{Variance} = 0.72$ (correct answer)
$\text{Variance} = 0.9$
$\text{Variance} = 0.81$
$\text{Variance} = \sqrt{0.72} \approx 0.849$
$\text{Variance} = \frac{0.9 \times 800}{1000} \times \left(1 - \frac{0.9}{1000}\right) \approx 0.720$

Explanation: When you encounter questions about rare events occurring at a constant rate over time or volume, you're dealing with a Poisson distribution. The key insight is that for Poisson distributions, the variance always equals the mean (λ). First, calculate the expected number of errors for 800 prescriptions. The rate is 0.9 errors per 1000 prescriptions, so for 800 prescriptions:

\lambda = 0.9 \times \frac{800}{1000} = 0.9 \times 0.8 = 0.72

Since this follows a Poisson distribution, the variance equals the mean:

\text{Variance} = \lambda = 0.72

Looking at the wrong answers: Answer B (0.9) incorrectly uses the original rate per 1000 prescriptions without adjusting for the actual volume of 800 prescriptions. Answer C (0.81) appears to come from squaring 0.9, which might stem from confusing variance with the square of the rate parameter. Answer D (

\sqrt{0.72} \approx 0.849

) calculates the standard deviation rather than the variance – this is a common mix-up since standard deviation is the square root of variance. The correct answer is A:

\text{Variance} = 0.72

. Study tip: For Poisson distribution problems, remember that variance = mean = λ. Always adjust the rate parameter for your specific sample size, and don't confuse variance with standard deviation. When you see "rare events at a constant rate," think Poisson and recall this fundamental property.

Question 8

A veterinary clinic sees cases of a specific parasitic infection at a rate of 2.5 cases per month. The clinic wants to determine the probability that they will see exactly 2 cases in each of the next 3 months. Assuming independence between months, what is this probability?

$\left[e^{-2.5} \cdot \frac{2.5^2}{2!}\right]^3 \approx 0.069$ (correct answer)
$e^{-7.5} \cdot \frac{7.5^6}{6!} \approx 0.103$
$e^{-2.5} \cdot \frac{2.5^6}{6!} \approx 0.006$
$3 \times e^{-2.5} \cdot \frac{2.5^2}{2!} \approx 0.768$
$\binom{3}{2} \times e^{-2.5} \cdot \frac{2.5^2}{2!} \approx 0.768$

Explanation: When you encounter questions about events occurring at a fixed rate over time with independence between periods, you're dealing with the Poisson distribution. This distribution models rare events happening at a constant average rate. The key insight here is recognizing that you need exactly 2 cases in each of three separate months. Since the months are independent, you calculate the probability for one month, then raise it to the third power for all three months. For a Poisson distribution, the probability of exactly k events when the average rate is λ is:

P(X = k) = e^{-λ} \cdot \frac{λ^k}{k!}

With λ = 2.5 cases per month and k = 2 cases, the probability for one month is:

P(X = 2) = e^{-2.5} \cdot \frac{2.5^2}{2!}

For all three months:

\left[e^{-2.5} \cdot \frac{2.5^2}{2!}\right]^3

This makes A correct. B treats this as a single 3-month period with 6 total cases (λ = 7.5, k = 6), ignoring the "exactly 2 per month" requirement. C incorrectly uses the 3-month rate (λ = 7.5) but keeps k = 2, which doesn't match the scenario. D multiplies by 3, suggesting you're choosing which month has 2 cases rather than requiring all three months to have exactly 2 cases. Study tip: For Poisson problems involving multiple time periods, always check whether you need a specific pattern in each period (multiply probabilities) or just a total count across periods (single calculation with combined rate).

Question 9

A medical device company tracks device malfunctions that occur at a rate of 0.4 per 1000 units manufactured. During a production run of 12,500 units, what is the probability of observing between 3 and 7 malfunctions (inclusive)?

$\sum_{k=3}^{7} e^{-5} \cdot \frac{5^k}{k!} \approx 0.867$ (correct answer)
$\sum_{k=3}^{7} e^{-0.4} \cdot \frac{0.4^k}{k!} \approx 0.005$
$e^{-5} \cdot \left(\frac{5^3}{3!} + \frac{5^4}{4!} + \frac{5^5}{5!} + \frac{5^6}{6!} + \frac{5^7}{7!}\right) \times 5 \approx 4.335$
$\Phi\left(\frac{7.5-5}{\sqrt{5}}\right) - \Phi\left(\frac{2.5-5}{\sqrt{5}}\right) \approx 0.867$
$\frac{7!}{3! \times 4!} \times e^{-5} \times 5^5 \approx 0.525$

Explanation: When you encounter questions about rare events occurring at a known rate, think Poisson distribution. This distribution models the number of events in a fixed interval when events occur independently at a constant average rate. Here, you need to identify the correct parameter λ (lambda). The malfunction rate is 0.4 per 1000 units, so for 12,500 units: λ = (0.4/1000) × 12,500 = 5. This represents the expected number of malfunctions in this production run. The probability of observing between 3 and 7 malfunctions is

P(3 ≤ X ≤ 7) = \sum_{k=3}^{7} e^{-5} \cdot \frac{5^k}{k!}

, which equals approximately 0.867. Answer A correctly uses λ = 5 and provides the right calculation. Answer B incorrectly uses λ = 0.4, failing to scale the rate to the actual production volume. This gives an unrealistically low probability of 0.005. Answer C appears to use the correct Poisson formula with λ = 5 but then mysteriously multiplies by 5, yielding an impossible probability greater than 1. Answer D applies a normal approximation to the Poisson distribution with continuity correction, which can be appropriate for large λ, but this approach isn't what the question is testing. Study tip: Always scale your rate parameter to match your observation period. If the rate is "per 1000 units" but you're analyzing 12,500 units, multiply the base rate by 12.5. The Poisson parameter λ must reflect the expected count for your specific scenario, not the general rate.

Question 10

A research team studies the occurrence of a specific genetic variant that appears in 1.4 out of every 10,000 newborns. In a cohort study following 25,000 newborns, what is the probability that fewer than 2 children will have this variant?

$P(X=0) + P(X=1) = e^{-3.5}(1 + 3.5) \approx 0.136$ (correct answer)
$P(X=0) + P(X=1) = e^{-1.4}(1 + 1.4) \approx 0.592$
$\binom{25000}{0}(0.00014)^0(0.99986)^{25000} + \binom{25000}{1}(0.00014)^1(0.99986)^{24999} \approx 0.136$
$P(X=1) = e^{-3.5} \times 3.5 \approx 0.106$
$\Phi\left(\frac{1.5-3.5}{\sqrt{3.5}}\right) \approx 0.143$

Explanation: When you encounter problems about rare events in large populations, you're typically dealing with Poisson distribution scenarios. The key indicators are: a low probability event (1.4 per 10,000), a large sample size (25,000), and questions about specific counts of occurrences. For Poisson problems, you first calculate λ (lambda), the expected number of events: λ = probability × sample size = (1.4/10,000) × 25,000 = 3.5. Then use the Poisson formula:

P(X=k) = \frac{e^{-λ}λ^k}{k!}

Since we want "fewer than 2," we need P(X < 2) = P(X = 0) + P(X = 1). Using λ = 3.5:

P(X = 0) = $e^{-3.5}$
P(X = 1) = $e^{-3.5} × 3.5$

Therefore: P(X < 2) =

e^{-3.5}(1 + 3.5) ≈ 0.136

Answer A correctly identifies this calculation and result. Answer B uses the wrong λ value (1.4 instead of 3.5), likely from confusing the rate per 10,000 with the expected count. Answer C attempts to use the binomial distribution, which is technically correct but unnecessarily complex—when n is large and p is small, Poisson approximates binomial much more efficiently. Answer D only calculates P(X = 1) instead of P(X < 2), missing the P(X = 0) component. Strategy tip: For rare events in large populations, immediately think Poisson. Calculate λ = np, then apply Poisson formulas rather than wrestling with unwieldy binomial calculations.

Question 11

A quality assurance team monitors production defects that occur at a rate of 1.9 defects per 100 items. If they inspect batches of 500 items each, what is the probability that a randomly selected batch will have exactly the expected number of defects?

$e^{-9.5} \cdot \frac{9.5^{10}}{10!} \approx 0.125$ (using 10 as the rounded expected value) (correct answer)
$e^{-9.5} \cdot \frac{9.5^{9.5}}{9.5!} \text{ (undefined for non-integer factorial)}$
$e^{-1.9} \cdot \frac{1.9^{1.9}}{1.9!} \text{ (undefined for non-integer factorial)}$
$e^{-9.5} \cdot \frac{9.5^9}{9!} \approx 0.132$ (using 9 as the rounded expected value)
$\frac{1}{\sqrt{2\pi \times 9.5}} \approx 0.129$ (using normal approximation)

Explanation: When you encounter a problem about defects occurring at a fixed rate over a specified number of items, you're dealing with a Poisson distribution scenario. The key insight is recognizing that defects happen randomly and independently at a constant average rate. First, you need to scale the rate to your sample size. With 1.9 defects per 100 items, a batch of 500 items has an expected number of defects of

\lambda = 1.9 \times \frac{500}{100} = 9.5

defects. The Poisson probability formula is

P(X = k) = e^{-\lambda} \cdot \frac{\lambda^k}{k!}

, where

k

must be a non-negative integer since you can't have fractional defects. Since the expected value is 9.5, you need to round to the nearest integer to find the probability of exactly that many defects occurring. Rounding 9.5 gives us 10 defects. Choice A correctly applies the formula with

k = 10

e^{-9.5} \cdot \frac{9.5^{10}}{10!}

. Choice B attempts to use 9.5 directly in the factorial, but factorials are undefined for non-integers. Choice C makes the same factorial error while also incorrectly using the original rate of 1.9 instead of scaling to the batch size of 500. Choice D uses the correct approach but rounds 9.5 down to 9 instead of using standard rounding rules. Remember: when asked for the probability of "exactly the expected number" in a Poisson context, round the expected value to the nearest integer since the actual count must be whole.

Question 12

A biosafety laboratory experiences containment breaches at an average rate of 0.3 per month. Laboratory protocols require immediate shutdown if more than 1 breach occurs in any 3-month period. What is the probability that the lab will need to shut down during a randomly selected 3-month period?

$1 - P(X=0) - P(X=1) = 1 - e^{-0.9}(1 + 0.9) \approx 0.227$ (correct answer)
$1 - P(X=0) - P(X=1) = 1 - e^{-0.3}(1 + 0.3) \approx 0.037$
$P(X \geq 2) = \sum_{k=2}^{\infty} e^{-0.9} \cdot \frac{0.9^k}{k!} \approx 0.227$
$P(X > 1) = 1 - e^{-0.9} \approx 0.593$
$\left[1 - e^{-0.3}(1 + 0.3)\right]^3 \approx 0.000$

Explanation: When you encounter a question about rare events occurring at a constant average rate over time, you're dealing with a Poisson distribution. The key is identifying the rate parameter for your specific time period and then calculating the complement of what you don't want. Here, breaches occur at 0.3 per month, so over 3 months the expected number is

\lambda = 0.3 \times 3 = 0.9

. The lab shuts down if "more than 1 breach" occurs, meaning 2 or more breaches. It's easier to calculate the complement: find the probability of 0 or 1 breach, then subtract from 1. Using the Poisson formula

P(X = k) = e^{-\lambda} \frac{\lambda^k}{k!}

$P(X = 0) = e^{-0.9} \frac{0.9^0}{0!} = e^{-0.9}$
$P(X = 1) = e^{-0.9} \frac{0.9^1}{1!} = 0.9e^{-0.9}$

Therefore:

P(X > 1) = 1 - P(X = 0) - P(X = 1) = 1 - e^{-0.9}(1 + 0.9) \approx 0.227

Option B incorrectly uses

\lambda = 0.3

instead of scaling to the 3-month period. Option C shows the correct setup for direct calculation of

P(X \geq 2)

but is unnecessarily complex. Option D forgets to subtract

P(X = 1)

, calculating only

1 - P(X = 0)

. Study tip: Always verify your rate parameter matches the time period in question, and remember that "more than k" problems are often easier solved using complements rather than direct calculation.

Question 13

A public health department monitors cases of a waterborne illness that typically affects 2.8 people per 10,000 during summer months. In a city of 200,000 people, what is the probability of observing between 50 and 60 cases (inclusive) during the summer?

$\sum_{k=50}^{60} e^{-56} \cdot \frac{56^k}{k!} \approx 0.245$ (correct answer)
$\sum_{k=50}^{60} e^{-2.8} \cdot \frac{2.8^k}{k!} \approx 0.000$
$\Phi\left(\frac{60.5-56}{\sqrt{56}}\right) - \Phi\left(\frac{49.5-56}{\sqrt{56}}\right) \approx 0.245$
$\sum_{k=50}^{60} \binom{200000}{k} \cdot (0.00028)^k \cdot (0.99972)^{200000-k} \approx 0.245$
$e^{-28} \cdot \frac{28^{55}}{55!} \times 11 \approx 0.198$

Explanation: When you encounter a question about rare events occurring in large populations, you're typically dealing with a Poisson distribution. This distribution models the probability of a specific number of independent events occurring within a fixed interval when you know the average rate. First, calculate the expected number of cases:

\lambda = \frac{2.8}{10,000} \times 200,000 = 56

cases. Since we want the probability of observing between 50 and 60 cases inclusive, we need

P(50 \leq X \leq 60) = \sum_{k=50}^{60} P(X = k)

, where each individual probability follows the Poisson formula:

P(X = k) = e^{-\lambda} \cdot \frac{\lambda^k}{k!}

. Answer A correctly uses the Poisson distribution with

\lambda = 56

, giving us

\sum_{k=50}^{60} e^{-56} \cdot \frac{56^k}{k!}

. Answer B incorrectly uses

\lambda = 2.8

instead of scaling up to the population of 200,000. This would be the rate per 10,000 people, not the expected total cases. Answer C applies a normal approximation with continuity correction, which could work given the large

\lambda

, but the question asks for the exact Poisson calculation, not an approximation. Answer D uses the binomial distribution

\binom{n}{k}p^k(1-p)^{n-k}

, which is theoretically correct but computationally impractical with

n = 200,000

. The Poisson distribution is specifically designed as an approximation for binomial distributions with large

n

and small

p

. Study tip: For rare diseases or events (small rates) in large populations, always think Poisson first. Remember to scale the rate to match your population size.

Question 14

A hospital's infection control team tracks surgical site infections, which occur at a rate of 0.6 per 100 procedures. During a month with 450 procedures, what is the probability that exactly 4 infections will occur?

$e^{-2.7} \cdot \frac{2.7^4}{4!} \approx 0.141$ (correct answer)
$e^{-0.6} \cdot \frac{0.6^4}{4!} \approx 0.003$
$\binom{450}{4} \cdot (0.006)^4 \cdot (0.994)^{446} \approx 0.141$
$e^{-4} \cdot \frac{4^4}{4!} \approx 0.195$
$\frac{450 \times 0.6}{100} \times \frac{1}{4!} \approx 0.113$

Explanation: When you encounter questions about rare events occurring at a known rate over a specific time period, you're dealing with a Poisson distribution. This is the go-to model for infection rates, equipment failures, or any low-probability events with a constant average rate. The key insight is calculating the expected number of events (λ) for your specific time period. Here, infections occur at 0.6 per 100 procedures, so with 450 procedures: λ = (0.6/100) × 450 = 2.7 expected infections. Using the Poisson formula

P(X=k) = e^{-λ} \cdot \frac{λ^k}{k!}

, you get

P(X=4) = e^{-2.7} \cdot \frac{2.7^4}{4!} ≈ 0.141

, which is answer A. Answer B incorrectly uses the original rate (0.6) without scaling it to 450 procedures—a common mistake when students forget to adjust λ for the actual sample size. Answer C applies a binomial distribution, treating each procedure as having a 0.006 probability of infection. While this could theoretically work, it's unnecessarily complex and the Poisson approximation is more appropriate for rare events. Answer D arbitrarily uses λ = 4, perhaps confusing the target outcome (4 infections) with the expected rate. Remember: for rare events with known rates, always calculate the expected number for your specific scenario first (rate × sample size), then apply the Poisson formula. Don't get trapped by using the original rate without scaling, and recognize that Poisson is simpler than binomial for low-probability events.

Question 15

A manufacturing plant experiences power outages that follow a Poisson distribution with a rate of 1.6 per month. If the plant operates continuously, what is the probability that the time until the next power outage exceeds 3 weeks?

$e^{-1.6 \times \frac{3}{4}} = e^{-1.2} \approx 0.301$ (correct answer)
$1 - e^{-1.6 \times 3} = 1 - e^{-4.8} \approx 0.992$
$e^{-1.6 \times 3} = e^{-4.8} \approx 0.008$
$\sum_{k=0}^{0} e^{-1.2} \cdot \frac{1.2^k}{k!} = e^{-1.2} \approx 0.301$
$1 - \sum_{k=0}^{3} e^{-1.2} \cdot \frac{1.2^k}{k!} \approx 0.034$

Explanation: When you encounter a Poisson process problem asking about time until the next event, you're dealing with the exponential distribution. If events occur at rate λ per unit time (Poisson), then the time between events follows an exponential distribution with the same rate parameter. The key insight is recognizing that "time until next power outage exceeds 3 weeks" means we want P(T > 3 weeks). For an exponential distribution, this probability is

e^{-\lambda t}

, where λ = 1.6 per month and t must be converted to months. Since 3 weeks =

\frac{3}{4}

months (there are 4 weeks in a month), we calculate: P(T > 3 weeks) =

e^{-1.6 \times \frac{3}{4}} = e^{-1.2} \approx 0.301

. This makes A correct. B uses the wrong formula

1 - e^{-\lambda t}

, which gives P(T ≤ t) rather than P(T > t), and incorrectly treats 3 weeks as 3 months. C makes the same unit error (3 weeks ≠ 3 months) but uses the right exponential formula structure. D attempts to use the Poisson probability mass function rather than the exponential survival function, which is conceptually wrong for this "time until" question. Study tip: Remember the connection between Poisson and exponential distributions. When asked about the number of events in a fixed time period, use Poisson. When asked about the time until the next event, use exponential with P(T > t) =

e^{-\lambda t}

. Always check your time units carefully!

Question 16

A hospital's emergency department typically sees 3.5 cardiac arrests per week. However, during a particular week, they observe 7 cardiac arrests. Assuming cardiac arrests follow a Poisson distribution, what is the probability of observing exactly 7 cardiac arrests in this week?

$e^{-3.5} \cdot \frac{3.5^7}{7!} \approx 0.0284$ (correct answer)
$e^{-7} \cdot \frac{7^7}{7!} \approx 0.1490$
$\frac{3.5^7}{7^7} \cdot e^{-3.5} \approx 0.0041$
$e^{-3.5} \cdot \frac{7^{3.5}}{7!} \approx 0.0156$
$\frac{7!}{3.5^7} \cdot e^{-7} \approx 0.0003$

Explanation: When you encounter a question about rare events occurring at a known average rate, you're dealing with the Poisson distribution. This distribution models scenarios like disease outbreaks, equipment failures, or in this case, cardiac arrests in a hospital setting. The Poisson probability formula is:

P(X = k) = e^{-\lambda} \cdot \frac{\lambda^k}{k!}

, where λ (lambda) is the expected average rate and k is the observed number of events. Here, λ = 3.5 cardiac arrests per week (the historical average), and k = 7 (the observed events this particular week). Substituting into the formula:

P(X = 7) = e^{-3.5} \cdot \frac{3.5^7}{7!} \approx 0.0284

. This matches answer choice A perfectly. Looking at the incorrect options: Answer B uses λ = 7 instead of 3.5, treating the observed value as the expected rate—a common misconception. Answer C incorrectly places 3.5 and 7 in a ratio format that doesn't follow the Poisson formula structure. Answer D substitutes the values backwards, using

7^{3.5}

instead of

3.5^7

, which reverses the relationship between the rate parameter and observed events. Remember this key principle: in Poisson problems, always use the historical average or expected rate as your λ parameter, not the current observation. The observation is what you're calculating the probability for, while the historical average defines the distribution's shape.

Question 17

A hospital's emergency department tracks cardiac arrests, which follow a Poisson distribution with λ = 2.8 per week. The department wants to calculate the probability that the number of cardiac arrests in a week will be within one standard deviation of the mean. What is this probability?

$\sum_{k=1}^{4} e^{-2.8} \cdot \frac{2.8^k}{k!} \approx 0.570$ (correct answer)
$\sum_{k=2}^{4} e^{-2.8} \cdot \frac{2.8^k}{k!} \approx 0.469$
$\Phi(1) - \Phi(-1) \approx 0.683$
$\sum_{k=0}^{5} e^{-2.8} \cdot \frac{2.8^k}{k!} \approx 0.935$
$\sum_{k=2}^{3} e^{-2.8} \cdot \frac{2.8^k}{k!} \approx 0.348$

Explanation: When dealing with Poisson distributions, remember that the mean and variance are both equal to λ, so the standard deviation is

\sqrt{λ}

. Here, with λ = 2.8, the standard deviation is

\sqrt{2.8} ≈ 1.67

. To find the probability within one standard deviation of the mean, you need P(μ - σ ≤ X ≤ μ + σ), which translates to P(2.8 - 1.67 ≤ X ≤ 2.8 + 1.67) = P(1.13 ≤ X ≤ 4.47). Since X must be a whole number (you can't have fractional cardiac arrests), this becomes P(1 ≤ X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4), which equals

\sum_{k=1}^{4} e^{-2.8} \cdot \frac{2.8^k}{k!}

. Choice A correctly captures this calculation. Choice B uses the same approach but incorrectly excludes k = 1, starting the sum at k = 2 instead. This represents a boundary error in determining which values fall within one standard deviation. Choice C applies the normal approximation rule (68% within one standard deviation), but this only works well for large λ values or when using continuity correction—not appropriate here with λ = 2.8. Choice D sums from k = 0 to k = 5, incorrectly including the probability of zero arrests and extending beyond the upper boundary. Study tip: For Poisson problems involving "within X standard deviations," always calculate the exact boundaries using

μ ± X\sqrt{λ}

, then round to appropriate integer values before summing the probabilities.

Question 18

In epidemiological surveillance, cases of a rare disease occur at a rate of 3.2 per 100,000 people per year in a specific region. A county with 75,000 residents wants to estimate the probability of observing at least 3 cases in the next year. Which approach is most appropriate?

Use Poisson with λ = 2.4; calculate 1 - P(X=0) - P(X=1) - P(X=2) (correct answer)
Use binomial with n = 75,000 and p = 0.000032; calculate P(X ≥ 3)
Use Poisson with λ = 3.2; calculate 1 - P(X=0) - P(X=1) - P(X=2)
Use normal approximation with μ = 2.4; calculate P(X ≥ 2.5)
Use hypergeometric distribution since the population is finite at 75,000

Explanation: When analyzing rare disease occurrence in populations, you're dealing with a classic Poisson process scenario. The key insight is correctly calculating the expected number of cases (λ) for your specific population, then using the complement rule for "at least" probabilities. Why A is correct: First, calculate λ for this county: (3.2 cases per 100,000) × (75,000 residents) ÷ 100,000 = 2.4 expected cases. The Poisson distribution is ideal for rare events in large populations, and P(X ≥ 3) = 1 - P(X ≤ 2) = 1 - P(X=0) - P(X=1) - P(X=2). This approach correctly scales the rate to the population size. Why the others are wrong: Option B uses the binomial distribution, which is theoretically correct but computationally impractical with n = 75,000 and such a small p-value. More importantly, it would give virtually the same result as the Poisson approximation but with unnecessary complexity. Option C incorrectly uses λ = 3.2, failing to adjust the rate from 100,000 people to the actual population of 75,000. Option D has the right λ = 2.4 but inappropriately applies normal approximation for small expected values (the rule of thumb requires λ ≥ 10), and uses the wrong continuity correction boundary. Study tip: For rare disease surveillance problems, always: (1) scale the given rate to your actual population size to find λ, (2) use Poisson when λ < 10, and (3) remember that P(X ≥ k) = 1 - P(X ≤ k-1) requires summing all probabilities from 0 to k-1.

Question 19

Emergency room visits for food poisoning follow a Poisson distribution with an average of 2.4 cases per day. What is the probability that on a randomly selected day, the hospital will see fewer than 2 cases of food poisoning?

$P(X=0) + P(X=1) = e^{-2.4}(1 + 2.4) \approx 0.308$ (correct answer)
$P(X=0) + P(X=1) + P(X=2) = e^{-2.4}\left(1 + 2.4 + \frac{2.4^2}{2}\right) \approx 0.570$
$P(X=1) = e^{-2.4} \cdot 2.4 \approx 0.218$
$1 - P(X=2) = 1 - e^{-2.4} \cdot \frac{2.4^2}{2} \approx 0.844$
$\Phi\left(\frac{1.5-2.4}{\sqrt{2.4}}\right) \approx 0.279$

Explanation: When you encounter questions about counting rare events over time intervals, you're dealing with the Poisson distribution. The key phrase here is "fewer than 2 cases," which means you need

P(X < 2) = P(X = 0) + P(X = 1)

. For a Poisson distribution with parameter

\lambda = 2.4

, the probability formula is

P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}

. So:

$P(X = 0) = \frac{e^{-2.4} \cdot 2.4^0}{0!} = e^{-2.4}$
$P(X = 1) = \frac{e^{-2.4} \cdot 2.4^1}{1!} = e^{-2.4} \cdot 2.4$

Therefore:

P(X < 2) = e^{-2.4}(1 + 2.4) \approx 0.308

Answer A correctly calculates this probability. Answer B incorrectly includes

P(X = 2)

, giving you

P(X \leq 2)

instead of

P(X < 2)

– this is the classic "less than vs. less than or equal to" trap. Answer C only calculates

P(X = 1)

, missing the

P(X = 0)

component entirely. Answer D uses the complement rule incorrectly;

1 - P(X = 2)

gives you

P(X \neq 2)

, which includes all values except exactly 2. Remember that "fewer than" means strictly less than, so don't include the boundary value. Always double-check whether the question asks for "less than," "at most," "more than," or "at least" – each requires a different calculation approach in discrete probability distributions.

Question 20

An environmental monitoring station detects radiation spikes at an average rate of 3.1 per week. If these events follow a Poisson distribution, what is the probability of detecting more than 5 spikes in a 10-day period?

$1 - \sum_{k=0}^{5} e^{-\frac{31}{7}} \cdot \frac{(\frac{31}{7})^k}{k!} \approx 0.126$ (correct answer)
$1 - \sum_{k=0}^{5} e^{-3.1} \cdot \frac{3.1^k}{k!} \approx 0.084$
$\sum_{k=6}^{\infty} e^{-4.43} \cdot \frac{4.43^k}{k!} \approx 0.126$
$1 - \Phi\left(\frac{5.5-4.43}{\sqrt{4.43}}\right) \approx 0.309$
$e^{-5} \cdot \sum_{k=6}^{10} \frac{5^k}{k!} \approx 0.238$

Explanation: When you encounter a Poisson distribution problem with different time periods, the key is properly adjusting the rate parameter (λ) to match the time frame of your question. The Poisson distribution models rare events occurring at a constant average rate over time. Here, radiation spikes occur at 3.1 per week, but you need the probability for a 10-day period. First, convert the rate: 10 days = 10/7 weeks, so λ = 3.1 × (10/7) = 31/7 ≈ 4.43 spikes per 10-day period. To find P(X > 5), you calculate 1 - P(X ≤ 5), which equals

1 - \sum_{k=0}^{5} e^{-\frac{31}{7}} \cdot \frac{(\frac{31}{7})^k}{k!}

. This makes choice A correct. Choice B uses λ = 3.1 without adjusting for the 10-day period—a common error of forgetting to scale the rate parameter to match the question's timeframe. Choice C correctly calculates the rate as 4.43 but uses the complement approach

\sum_{k=6}^{\infty}

instead of the more practical

1 - \sum_{k=0}^{5}

form. While mathematically equivalent, it doesn't match the given format and approximation. Choice D applies a normal approximation with continuity correction, which isn't appropriate here since we're working directly with the Poisson distribution. Remember: always verify that your λ parameter matches the time period in the question. When given a rate for one time unit, multiply by the ratio of your target period to the given period to get the correct λ value.