College Statistics Quiz: Poisson Distribution
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Poisson DistributionQuestion 1 of 20

The number of goals scored by a soccer team in a game follows a Poisson distribution with a mean of 1.5. A fan buys a ticket for two upcoming games. What is the probability that the team scores a total of exactly 2 goals in these two games?

0.2240
0.2510
0.2565
0.4481
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College Statistics Quiz

College Statistics Quiz: Poisson Distribution

Practice Poisson Distribution in College Statistics 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 College Statistics.

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

The number of goals scored by a soccer team in a game follows a Poisson distribution with a mean of 1.5. A fan buys a ticket for two upcoming games. What is the probability that the team scores a total of exactly 2 goals in these two games?

  1. 0.2240 (correct answer)
  2. 0.2510
  3. 0.2565
  4. 0.4481
Explanation: The total number of goals in two independent games will also follow a Poisson distribution. The rate for the combined period (two games) is the sum of the rates for each game: λ_total = 1.5 + 1.5 = 3. We now need to calculate the probability of k=2 goals with this new rate λ=3. P(X=2)=e−3322!=e−3⋅92=4.5e−3≈0.2240P(X=2) = \frac{e^{-3}3^2}{2!} = \frac{e^{-3} \cdot 9}{2} = 4.5e^{-3} \approx 0.2240. Distractor B is P(X=2) with λ=1.5 (for a single game). Distractor C is P(X=1) with λ=3. Distractor D is P(X=1) for one game squared, which would calculate P(1 goal in game 1 AND 1 goal in game 2), not the total number of goals.

Question 2

The number of customers arriving at a small coffee shop follows a Poisson process. However, the manager notes that during the morning rush, customers often arrive in groups of two or three. Which assumption of the Poisson distribution is most clearly violated in this scenario?

  1. The average rate of arrivals is constant over time.
  2. The probability of an arrival in a very small time interval is proportional to the length of the interval.
  3. The number of arrivals in two disjoint time intervals are independent.
  4. The probability of more than one arrival in a very small time interval is negligible. (correct answer)
Explanation: The Poisson process assumes that events occur one at a time. The statement that the probability of more than one event in a small interval is negligible (effectively zero) captures this idea. Customers arriving in groups (simultaneous events) directly violates this assumption. Distractor A is also a potential issue (morning rush vs. rest of day), but the information about groups provides direct evidence against assumption D. Distractor C refers to independence between disjoint intervals, while group arrivals are about a lack of independence within an interval. Distractor B is a formal definition related to the rate, which is less directly violated than D.

Question 3

An email server receives messages at a Poisson rate of λ=10 per hour. Of these emails, 20% are high-priority. The arrivals of high-priority emails can be modeled as a separate, 'thinned' Poisson process. What is the probability of receiving exactly two high-priority emails in one hour?

  1. 0.1465
  2. 0.1839
  3. 0.2707 (correct answer)
  4. 0.5413
Explanation: This problem describes a 'thinning' of a Poisson process. If events from a Poisson(λ) process are independently kept with probability p, the resulting process of kept events is also a Poisson process with a new rate λ' = λ * p. Here, λ = 10 and p = 0.20. The rate for high-priority emails is λ' = 10 * 0.20 = 2 per hour. We want to find the probability of k=2 high-priority emails in one hour using λ'=2. P(X=2)=e−2222!=2e−2≈0.2707P(X=2) = \frac{e^{-2}2^2}{2!} = 2e^{-2} \approx 0.2707. Distractor A is P(X=2) using λ=4. Distractor B is P(X=2) using λ=10 (the original rate). Distractor D is P(X<=1) using λ=2.

Question 4

The number of claims filed with an insurance company per day follows a Poisson distribution with a mean of 5. The company has a staff that can handle at most 7 claims per day. What is the probability that on a given day, the number of claims will exceed the staff's capacity?

  1. 0.1334 (correct answer)
  2. 0.2378
  3. 0.7622
  4. 0.8666
Explanation: The staff's capacity is 7 claims, so we need to find the probability that the number of claims X is greater than 7, i.e., P(X>7)P(X > 7). This is calculated as 1−P(X≤7)1 - P(X \le 7). With λ=5, we need to sum P(X=0)+P(X=1)+...+P(X=7)P(X=0) + P(X=1) + ... + P(X=7). Using a calculator or software: P(X≤7) ≈ 0.8666. Therefore, P(X>7)=1−0.8666=0.1334P(X > 7) = 1 - 0.8666 = 0.1334. Distractor D is P(X≤7). Distractor B is P(X>6), an off-by-one error in interpreting 'exceed'. Distractor C is 1 - P(X>6).

Question 5

During business hours, a company's main server and backup server receive requests independently. The main server receives requests at a Poisson rate of 3 per minute, and the backup server receives requests at a Poisson rate of 1 per minute. What is the probability that a total of exactly 3 requests are received by the two servers in a given minute?

  1. 0.1804
  2. 0.1954 (correct answer)
  3. 0.2240
  4. 0.2707
Explanation: When two independent Poisson processes with rates λ₁ and λ₂ are combined, the total number of events also follows a Poisson distribution with a rate of λ_total = λ₁ + λ₂. In this case, λ_total = 3 + 1 = 4 requests per minute. We need to find the probability of k=3 total requests using λ=4. P(X=3)=e−4433!=e−4⋅646≈0.1954P(X=3) = \frac{e^{-4}4^3}{3!} = \frac{e^{-4} \cdot 64}{6} \approx 0.1954. Distractor A is P(X=3) with λ=2 (the average of the rates). Distractor C is P(X=3) with λ=3 (using only the main server's rate). Distractor D is P(X=2) with λ=4.

Question 6

A binomial random variable X has parameters n=100 and p=0.03. A Poisson distribution is used to approximate the probability P(X=4). Let P_B be the exact probability from the binomial distribution and P_P be the approximate probability from the Poisson distribution. Which statement is true?

  1. P_P > P_B, and the approximation is appropriate.
  2. P_P < P_B, and the approximation is appropriate. (correct answer)
  3. P_P ≈ P_B, but the approximation is inappropriate because p is too large.
  4. P_P ≈ P_B, but the approximation is inappropriate because n is too small.
Explanation: The Poisson distribution can approximate the binomial distribution when n is large and p is small. Here, n=100 is large and p=0.03 is small, so the approximation is appropriate. The mean for both distributions is λ = np = 100 * 0.03 = 3. Let's calculate the probabilities. Binomial: PB=(1004)(0.03)4(0.97)96≈0.1683P_B = \binom{100}{4}(0.03)^4(0.97)^{96} \approx 0.1683. Poisson: PP=e−3344!≈0.1680P_P = \frac{e^{-3}3^4}{4!} \approx 0.1680. Let me re-calculate with more precision. P_B = 0.16833. P_P = 0.16803. So P_P < P_B. Let me check my logic. The Poisson approximation tends to be slightly off. Let me re-examine the question and choices. Perhaps I made a mistake in my assumption of the relationship. Let's re-run a simple case. n=10, p=0.1. lambda=1. P_B(X=2) = C(10,2)(0.1)^2(0.9)^8 = 0.1937. P_P(X=2) = e^-1 * 1^2 / 2! = 0.1839. Here P_P < P_B. Let's try the original problem again. P_B = C(100,4)(0.03)^4(0.97)^96 = 3921225 * 8.1e-7 * 0.0535 = 0.1696. P_P = e^-3 * 81 / 24 = 0.1680. So P_P < P_B. This means option B is correct. Let me review the options and reasoning again. A well-known result is that the Poisson approximation overestimates the probability in the left tail and underestimates it in the right tail relative to the mean. Since the mean is 3, k=4 is in the right tail, so the Poisson probability should be an underestimate of the Binomial probability. Thus, P_P < P_B. The conditions n=100, p=0.03 (np=3) are well within the standard rules of thumb for a good approximation (e.g., n>20, p<0.05). Therefore, the approximation is appropriate. This supports choice B.

Question 7

The number of defects in a cable follows a Poisson distribution. The standard deviation of the number of defects per 100 feet of cable is 2. What is the probability that a 100-foot section of cable has exactly one defect?

  1. 0.0733 (correct answer)
  2. 0.1465
  3. 0.2707
  4. 0.5413
Explanation: This is a multi-step problem. For a Poisson distribution, the variance (σ²) is equal to the mean (λ). The problem gives the standard deviation (σ) as 2. Therefore, the variance is σ² = 2² = 4. This means the mean rate is λ = 4 defects per 100 feet. The question asks for the probability of k=1 defect in a 100-foot section, so we use λ=4. P(X=1)=e−4411!=4e−4≈0.0733P(X=1) = \frac{e^{-4}4^1}{1!} = 4e^{-4} \approx 0.0733. Distractor B is P(X=2) with λ=4. Distractor C calculates the probability using λ=2, mistakenly using the standard deviation as the mean. Distractor D incorrectly calculates P(X=1) with λ=ln(2).

Question 8

The number of typos on a newspaper page follows a Poisson distribution with a mean of 2.5 typos per page. A page is selected at random. Given that the page has at least one typo, what is the approximate probability that it has exactly two typos?

  1. 0.2565
  2. 0.2795 (correct answer)
  3. 0.3347
  4. 0.5438
Explanation: This is a conditional probability problem: P(X=2∣X≥1)P(X=2 | X \ge 1). The formula is P(X=2 and X≥1)P(X≥1)\frac{P(X=2 \text{ and } X \ge 1)}{P(X \ge 1)}. Since X=2X=2 is a case of X≥1X \ge 1, this simplifies to P(X=2)P(X≥1)\frac{P(X=2)}{P(X \ge 1)}. We are given λ=2.5. First, calculate P(X=2)=e−2.52.522!≈0.2565P(X=2) = \frac{e^{-2.5}2.5^2}{2!} \approx 0.2565. Next, calculate P(X≥1)=1−P(X=0)P(X \ge 1) = 1 - P(X=0). P(X=0)=e−2.52.500!=e−2.5≈0.0821P(X=0) = \frac{e^{-2.5}2.5^0}{0!} = e^{-2.5} \approx 0.0821. So, P(X≥1)=1−0.0821=0.9179P(X \ge 1) = 1 - 0.0821 = 0.9179. Finally, the conditional probability is 0.25650.9179≈0.2795\frac{0.2565}{0.9179} \approx 0.2795. Distractor A is just P(X=2), ignoring the condition. Distractor C is P(X=2)/P(X>1), a misreading of the condition. Distractor D is P(X≤2), another misinterpretation.

Question 9

A manufacturing process produces glass sheets. The number of bubbles per square meter follows a Poisson distribution. It is known that the probability of a square meter having no bubbles is 0.05. What is the probability that a square meter will have more than two bubbles?

  1. 0.4232
  2. 0.5768 (correct answer)
  3. 0.8753
  4. 0.9500
Explanation: First, we must find the rate parameter λ. We are given P(X=0)=0.05P(X=0) = 0.05. For a Poisson distribution, P(X=0)=e−λP(X=0) = e^{-\lambda}. So, e−λ=0.05e^{-\lambda} = 0.05, which means λ=−ln⁡(0.05)≈2.9957\lambda = -\ln(0.05) \approx 2.9957, which we can round to λ=3 for calculation. The question asks for P(X>2)P(X > 2), which is 1−P(X≤2)=1−[P(X=0)+P(X=1)+P(X=2)]1 - P(X \le 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]. Using λ=3: P(X=0)=e−3≈0.0498P(X=0) = e^{-3} \approx 0.0498, P(X=1)=e−3311!≈0.1494P(X=1) = \frac{e^{-3}3^1}{1!} \approx 0.1494, P(X=2)=e−3322!≈0.2240P(X=2) = \frac{e^{-3}3^2}{2!} \approx 0.2240. Summing these gives P(X≤2)≈0.0498+0.1494+0.2240=0.4232P(X \le 2) \approx 0.0498 + 0.1494 + 0.2240 = 0.4232. Finally, P(X>2)=1−0.4232=0.5768P(X > 2) = 1 - 0.4232 = 0.5768. Distractor A is P(X≤2). Distractor C is 1 - P(X=0) - P(X=1). Distractor D is 1 - P(X=0).

Question 10

A biologist is modeling the number of a certain bacteria in a 1 ml water sample. Which of the following findings from an initial data set would most strongly suggest that a Poisson distribution is NOT an appropriate model?

  1. The sample mean number of bacteria is 3.7, which is not an integer.
  2. The sample variance is 15.2, while the sample mean is 4.1. (correct answer)
  3. In 10% of the samples, there were more than 10 bacteria.
  4. The most frequently observed number of bacteria in a sample was 3.
Explanation: A key property of the Poisson distribution is that its mean and variance are equal (λ = σ²). The finding that the sample variance (15.2) is nearly four times the sample mean (4.1) is a strong indicator of overdispersion, which violates this fundamental property. Therefore, a Poisson distribution would not be a good fit. Distractor A is incorrect because the mean (λ) of a Poisson distribution does not need to be an integer. Distractor C is a statement about the tail of the distribution and does not by itself invalidate the model. Distractor D describes the mode of the data, which is consistent with a Poisson distribution where the mean is around 3 or 4.

Question 11

During business hours, a company's main server and backup server receive requests independently. The main server receives requests at a Poisson rate of 3 per minute, and the backup server receives requests at a Poisson rate of 1 per minute. What is the probability that a total of exactly 3 requests are received by the two servers in a given minute?

  1. 0.1804
  2. 0.1954 (correct answer)
  3. 0.2240
  4. 0.2707
Explanation: When two independent Poisson processes with rates λ₁ and λ₂ are combined, the total number of events also follows a Poisson distribution with a rate of λ_total = λ₁ + λ₂. In this case, λ_total = 3 + 1 = 4 requests per minute. We need to find the probability of k=3 total requests using λ=4. P(X=3)=e−4433!=e−4⋅646≈0.1954P(X=3) = \frac{e^{-4}4^3}{3!} = \frac{e^{-4} \cdot 64}{6} \approx 0.1954. Distractor A is P(X=3) with λ=2 (the average of the rates). Distractor C is P(X=3) with λ=3 (using only the main server's rate). Distractor D is P(X=2) with λ=4.

Question 12

The number of cars passing a checkpoint follows a Poisson distribution with a mean of 72 cars per hour. What is the probability that exactly one car passes in a one-minute interval?

  1. 0.2215
  2. 0.3007
  3. 0.3614 (correct answer)
  4. 0.7200
Explanation: First, the rate parameter λ must be adjusted to the one-minute interval. The rate is 72 cars per hour, and there are 60 minutes in an hour. The adjusted rate is λ = 72 / 60 = 1.2 cars per minute. The question asks for the probability of k=1 car in one minute. We use the Poisson formula with λ=1.2: P(X=1)=e−1.2(1.2)11!=1.2⋅e−1.2≈1.2⋅0.30119≈0.3614P(X=1) = \frac{e^{-1.2}(1.2)^1}{1!} = 1.2 \cdot e^{-1.2} \approx 1.2 \cdot 0.30119 \approx 0.3614. Distractor B is P(X=1) using λ=1 (a rounding error). Distractor A is P(X=2) using λ=1.2. Distractor D is an incorrect calculation possibly related to λ/k.

Question 13

A machine produces widgets with a Poisson-distributed number of minor flaws, with a mean of λ flaws per widget. If the probability of a widget having at least one flaw is 0.75, what is the approximate value of λ?

  1. 0.29
  2. 0.75
  3. 1.39 (correct answer)
  4. 2.08
Explanation: The probability of at least one flaw is P(X≥1)=1−P(X=0)P(X \ge 1) = 1 - P(X=0). We are given that this probability is 0.75. So, 1−P(X=0)=0.751 - P(X=0) = 0.75, which implies P(X=0)=0.25P(X=0) = 0.25. For a Poisson distribution, P(X=0)=e−λP(X=0) = e^{-\lambda}. Therefore, e−λ=0.25e^{-\lambda} = 0.25. To solve for λ, we take the natural logarithm of both sides: −λ=ln⁡(0.25)-\lambda = \ln(0.25). This gives λ=−ln⁡(0.25)=ln⁡(1/0.25)=ln⁡(4)≈1.386\lambda = -\ln(0.25) = \ln(1/0.25) = \ln(4) \approx 1.386. Distractor A is -ln(1-0.25). Distractor B confuses the probability with the rate. Distractor D results from an incorrect algebraic manipulation.

Question 14

A city experiences earthquakes at a Poisson rate of 2 per decade. What is the probability that there will be exactly one earthquake in each of two consecutive decades?

  1. 0.0733 (correct answer)
  2. 0.1465
  3. 0.2707
  4. 0.5413
Explanation: This question involves two independent events. The number of earthquakes in the first decade is independent of the number in the second. Let X₁ be the number of earthquakes in the first decade and X₂ be the number in the second. Both follow a Poisson(λ=2) distribution. We want to find P(X1=1 and X2=1)P(X_1=1 \text{ and } X_2=1). Because the events are independent, this is P(X1=1)×P(X2=1)P(X_1=1) \times P(X_2=1). First, calculate P(X=1)P(X=1) with λ=2: P(X=1)=e−2211!=2e−2≈0.2707P(X=1) = \frac{e^{-2}2^1}{1!} = 2e^{-2} \approx 0.2707. Now, square this probability: (0.2707)2≈0.073278(0.2707)^2 \approx 0.073278. Distractor C is the probability for just one of the decades. Distractor B is P(X=2) for a single decade with λ=2, which is incorrect reasoning. Distractor D might arise from adding probabilities instead of multiplying.

Question 15

A researcher models the number of occurrences of a rare word in a book's chapter, X, with a Poisson distribution. They calculate the probabilities P(X=0), P(X=1), and P(X=2). Which of the following relationships between these probabilities must be true if the mean λ is greater than 2?

  1. P(X=0) > P(X=1) > P(X=2)
  2. P(X=1) > P(X=2) > P(X=0)
  3. P(X=2) > P(X=1) > P(X=0) (correct answer)
  4. P(X=2) > P(X=0) > P(X=1)
Explanation: The ratio of consecutive Poisson probabilities is given by P(X=k)P(X=k−1)=λk\frac{P(X=k)}{P(X=k-1)} = \frac{\lambda}{k}. For P(X=1) vs P(X=0), the ratio is λ1=λ\frac{\lambda}{1} = \lambda. Since λ > 2, P(X=1) > P(X=0). For P(X=2) vs P(X=1), the ratio is λ2\frac{\lambda}{2}. Since λ > 2, this ratio is greater than 1, meaning P(X=2) > P(X=1). Combining these, we get P(X=2)>P(X=1)>P(X=0)P(X=2) > P(X=1) > P(X=0). This makes sense, as for λ>2, the distribution is still increasing at k=2.

Question 16

The number of cars passing a checkpoint follows a Poisson distribution with a mean of 72 cars per hour. What is the probability that exactly one car passes in a one-minute interval?

  1. 0.2215
  2. 0.3007
  3. 0.3614 (correct answer)
  4. 0.7200
Explanation: First, the rate parameter λ must be adjusted to the one-minute interval. The rate is 72 cars per hour, and there are 60 minutes in an hour. The adjusted rate is λ = 72 / 60 = 1.2 cars per minute. The question asks for the probability of k=1 car in one minute. We use the Poisson formula with λ=1.2: P(X=1)=e−1.2(1.2)11!=1.2⋅e−1.2≈1.2⋅0.30119≈0.3614P(X=1) = \frac{e^{-1.2}(1.2)^1}{1!} = 1.2 \cdot e^{-1.2} \approx 1.2 \cdot 0.30119 \approx 0.3614. Distractor B is P(X=1) using λ=1 (a rounding error). Distractor A is P(X=2) using λ=1.2. Distractor D is an incorrect calculation possibly related to λ/k.

Question 17

The number of typos on a newspaper page follows a Poisson distribution with a mean of 2.5 typos per page. A page is selected at random. Given that the page has at least one typo, what is the approximate probability that it has exactly two typos?

  1. 0.2565
  2. 0.2795 (correct answer)
  3. 0.3347
  4. 0.5438
Explanation: This is a conditional probability problem: P(X=2∣X≥1)P(X=2 | X \ge 1). The formula is P(X=2 and X≥1)P(X≥1)\frac{P(X=2 \text{ and } X \ge 1)}{P(X \ge 1)}. Since X=2X=2 is a case of X≥1X \ge 1, this simplifies to P(X=2)P(X≥1)\frac{P(X=2)}{P(X \ge 1)}. We are given λ=2.5. First, calculate P(X=2)=e−2.52.522!≈0.2565P(X=2) = \frac{e^{-2.5}2.5^2}{2!} \approx 0.2565. Next, calculate P(X≥1)=1−P(X=0)P(X \ge 1) = 1 - P(X=0). P(X=0)=e−2.52.500!=e−2.5≈0.0821P(X=0) = \frac{e^{-2.5}2.5^0}{0!} = e^{-2.5} \approx 0.0821. So, P(X≥1)=1−0.0821=0.9179P(X \ge 1) = 1 - 0.0821 = 0.9179. Finally, the conditional probability is 0.25650.9179≈0.2795\frac{0.2565}{0.9179} \approx 0.2795. Distractor A is just P(X=2), ignoring the condition. Distractor C is P(X=2)/P(X>1), a misreading of the condition. Distractor D is P(X≤2), another misinterpretation.

Question 18

A manufacturing process produces glass sheets. The number of bubbles per square meter follows a Poisson distribution. It is known that the probability of a square meter having no bubbles is 0.05. What is the probability that a square meter will have more than two bubbles?

  1. 0.4232
  2. 0.5768 (correct answer)
  3. 0.8753
  4. 0.9500
Explanation: First, we must find the rate parameter λ. We are given P(X=0)=0.05P(X=0) = 0.05. For a Poisson distribution, P(X=0)=e−λP(X=0) = e^{-\lambda}. So, e−λ=0.05e^{-\lambda} = 0.05, which means λ=−ln⁡(0.05)≈2.9957\lambda = -\ln(0.05) \approx 2.9957, which we can round to λ=3 for calculation. The question asks for P(X>2)P(X > 2), which is 1−P(X≤2)=1−[P(X=0)+P(X=1)+P(X=2)]1 - P(X \le 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]. Using λ=3: P(X=0)=e−3≈0.0498P(X=0) = e^{-3} \approx 0.0498, P(X=1)=e−3311!≈0.1494P(X=1) = \frac{e^{-3}3^1}{1!} \approx 0.1494, P(X=2)=e−3322!≈0.2240P(X=2) = \frac{e^{-3}3^2}{2!} \approx 0.2240. Summing these gives P(X≤2)≈0.0498+0.1494+0.2240=0.4232P(X \le 2) \approx 0.0498 + 0.1494 + 0.2240 = 0.4232. Finally, P(X>2)=1−0.4232=0.5768P(X > 2) = 1 - 0.4232 = 0.5768. Distractor A is P(X≤2). Distractor C is 1 - P(X=0) - P(X=1). Distractor D is 1 - P(X=0).

Question 19

The number of customers arriving at a small coffee shop follows a Poisson process. However, the manager notes that during the morning rush, customers often arrive in groups of two or three. Which assumption of the Poisson distribution is most clearly violated in this scenario?

  1. The average rate of arrivals is constant over time.
  2. The probability of an arrival in a very small time interval is proportional to the length of the interval.
  3. The number of arrivals in two disjoint time intervals are independent.
  4. The probability of more than one arrival in a very small time interval is negligible. (correct answer)
Explanation: The Poisson process assumes that events occur one at a time. The statement that the probability of more than one event in a small interval is negligible (effectively zero) captures this idea. Customers arriving in groups (simultaneous events) directly violates this assumption. Distractor A is also a potential issue (morning rush vs. rest of day), but the information about groups provides direct evidence against assumption D. Distractor C refers to independence between disjoint intervals, while group arrivals are about a lack of independence within an interval. Distractor B is a formal definition related to the rate, which is less directly violated than D.

Question 20

The number of claims filed with an insurance company per day follows a Poisson distribution with a mean of 5. The company has a staff that can handle at most 7 claims per day. What is the probability that on a given day, the number of claims will exceed the staff's capacity?

  1. 0.1334 (correct answer)
  2. 0.2378
  3. 0.7622
  4. 0.8666
Explanation: The staff's capacity is 7 claims, so we need to find the probability that the number of claims X is greater than 7, i.e., P(X>7)P(X > 7). This is calculated as 1−P(X≤7)1 - P(X \le 7). With λ=5, we need to sum P(X=0)+P(X=1)+...+P(X=7)P(X=0) + P(X=1) + ... + P(X=7). Using a calculator or software: P(X≤7) ≈ 0.8666. Therefore, P(X>7)=1−0.8666=0.1334P(X > 7) = 1 - 0.8666 = 0.1334. Distractor D is P(X≤7). Distractor B is P(X>6), an off-by-one error in interpreting 'exceed'. Distractor C is 1 - P(X>6).