# Poisson Distribution Worksheet ## Section A: Multiple Choice Questions
Q1: The Poisson distribution is used to model which type of variable? (a) Continuous random variables (b) Discrete random variables (c) Qualitative variables (d) Dependent variables
Solution:
Ans: (b) Explanation: The Poisson distribution is a probability distribution that models discrete random variables, specifically the number of events occurring in a fixed interval of time or space. Continuous variables would use distributions like the normal distribution, not Poisson.
Q2: What is the formula for the Poisson probability mass function, where \(X\) is the random variable, \(k\) is the number of events, and \(\lambda\) is the average rate? (a) \(P(X = k) = \frac{\lambda^k e^{-k}}{k!}\) (b) \(P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}\) (c) \(P(X = k) = \frac{k^{\lambda} e^{-\lambda}}{\lambda!}\) (d) \(P(X = k) = \frac{\lambda e^{-k}}{k!}\)
Solution:
Ans: (b) Explanation: The correct Poisson probability mass function is \(P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}\), where \(\lambda\) is raised to the power \(k\), \(e\) is raised to the power \(-\lambda\), and the denominator is \(k!\). Option (a) incorrectly has \(e^{-k}\), option (c) has the variables reversed, and option (d) is missing the exponent on \(\lambda\).
Q3: In a Poisson distribution with parameter \(\lambda = 5\), what is the mean of the distribution? (a) 2.5 (b) 5 (c) 10 (d) 25
Solution:
Ans: (b) Explanation: In a Poisson distribution, both the mean and the variance are equal to the parameter \(\lambda\). Therefore, if \(\lambda = 5\), the mean is also 5. This is a fundamental property of the Poisson distribution.
Q4: What is the variance of a Poisson distribution with \(\lambda = 8\)? (a) 4 (b) 8 (c) 16 (d) 64
Solution:
Ans: (b) Explanation: For a Poisson distribution, the variance equals the parameter \(\lambda\). Since \(\lambda = 8\), the variance is also 8. This equality of mean and variance is a unique characteristic of the Poisson distribution that distinguishes it from other discrete distributions.
Q5: If \(X\) follows a Poisson distribution with \(\lambda = 3\), what is \(P(X = 0)\)? (a) \(e^{-3}\) (b) \(3e^{-3}\) (c) \(\frac{1}{3}\) (d) 0
Solution:
Ans: (a) Explanation: Using the formula \(P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}\), when \(k = 0\): \(P(X = 0) = \frac{3^0 e^{-3}}{0!} = \frac{1 \cdot e^{-3}}{1} = e^{-3}\) Since \(3^0 = 1\) and \(0! = 1\), the probability simplifies to \(e^{-3}\).
Q6: Which of the following situations is best modeled by a Poisson distribution? (a) The number of heads in 10 coin flips (b) The number of customers arriving at a store per hour (c) The height of students in a class (d) The probability of selecting a red card from a deck
Solution:
Ans: (b) Explanation: The Poisson distribution is ideal for modeling the number of events occurring in a fixed interval of time or space when events occur independently at a constant average rate. Customer arrivals per hour fit this criterion. Option (a) is a binomial distribution, option (c) involves continuous data, and option (d) is a simple probability without a counting process.
Q7: For a Poisson distribution with \(\lambda = 4\), what is the standard deviation? (a) 2 (b) 4 (c) 8 (d) 16
Solution:
Ans: (a) Explanation: The standard deviation is the square root of the variance. Since variance = \(\lambda = 4\) in a Poisson distribution, the standard deviation is \(\sqrt{4} = 2\). This relationship comes from the property that variance equals \(\lambda\) in Poisson distributions.
Q8: If \(X\) follows a Poisson distribution with \(\lambda = 2\), what is \(P(X = 2)\)? (a) \(\frac{2e^{-2}}{2}\) (b) \(2e^{-2}\) (c) \(\frac{4e^{-2}}{2}\) (d) \(4e^{-2}\)
Solution:
Ans: (b) Explanation: Using \(P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}\) with \(k = 2\) and \(\lambda = 2\): \(P(X = 2) = \frac{2^2 e^{-2}}{2!} = \frac{4e^{-2}}{2} = 2e^{-2}\) The calculation simplifies to \(2e^{-2}\) after dividing 4 by 2.
## Section B: Fill in the Blanks Q9: In a Poisson distribution, the mean and the __________ are always equal to the parameter \(\lambda\).
Solution:
Ans: variance Explanation: A defining characteristic of the Poisson distribution is that both the mean and variance equal \(\lambda\). This property helps distinguish it from other probability distributions.
Q10: The Poisson distribution is a __________ probability distribution used to count the number of events in a fixed interval.
Solution:
Ans: discrete Explanation: The Poisson distribution is a discrete probability distribution because it models countable outcomes (0, 1, 2, 3, ...) rather than continuous values.
Q11: The parameter \(\lambda\) in a Poisson distribution represents the __________ number of events occurring in the interval.
Solution:
Ans: average (or mean) Explanation: The parameter \(\lambda\) represents the average or mean rate at which events occur in the specified interval. It is the expected value of the distribution.
Q12: The value of \(e\) used in the Poisson formula is approximately equal to __________.
Solution:
Ans: 2.718 (or 2.71828) Explanation: The constant \(e\) is Euler's number, approximately equal to 2.718. It is the base of the natural logarithm and appears in the Poisson probability mass function.
Q13: In a Poisson distribution, the probability of observing exactly \(k\) events is given by the formula \(P(X = k) = \frac{\lambda^k e^{-\lambda}}{__________}\).
Solution:
Ans: \(k!\) Explanation: The denominator in the Poisson probability mass function is \(k!\) (k factorial), which accounts for the number of ways \(k\) events can occur.
Q14: For a Poisson distribution to be an appropriate model, events must occur __________ of each other.
Solution:
Ans: independently Explanation: One of the key assumptions of the Poisson distribution is that events occur independently, meaning the occurrence of one event does not affect the probability of another event occurring.
## Section C: Word Problems Q15: A call center receives an average of 6 calls per hour. Assuming the number of calls follows a Poisson distribution, what is the probability that exactly 4 calls are received in a given hour? (Use \(e^{-6} \approx 0.00248\))
Solution:
Ans: Step 1: Identify the parameters: \(\lambda = 6\) and \(k = 4\) Step 2: Apply the Poisson formula: \(P(X = 4) = \frac{6^4 e^{-6}}{4!}\) Step 3: Calculate \(6^4 = 1296\) Step 4: Calculate \(4! = 24\) Step 5: Substitute: \(P(X = 4) = \frac{1296 \times 0.00248}{24} = \frac{3.21408}{24} \approx 0.1339\) Final Answer: The probability is approximately 0.1339 or 13.39%
Q16: A factory produces light bulbs, and defects occur at an average rate of 2 per day. Assuming defects follow a Poisson distribution, what is the probability that no defects occur on a given day? (Use \(e^{-2} \approx 0.1353\))
Solution:
Ans: Step 1: Identify the parameters: \(\lambda = 2\) and \(k = 0\) Step 2: Apply the Poisson formula: \(P(X = 0) = \frac{2^0 e^{-2}}{0!}\) Step 3: Simplify: \(2^0 = 1\) and \(0! = 1\) Step 4: Calculate: \(P(X = 0) = \frac{1 \times 0.1353}{1} = 0.1353\) Final Answer: The probability is approximately 0.1353 or 13.53%
Q17: On average, 3 students visit the school library during lunch period. If the number of students follows a Poisson distribution, find the mean and standard deviation of this distribution.
Solution:
Ans: Step 1: Identify \(\lambda = 3\) Step 2: For a Poisson distribution, mean = \(\lambda = 3\) Step 3: For a Poisson distribution, variance = \(\lambda = 3\) Step 4: Standard deviation = \(\sqrt{\text{variance}} = \sqrt{3} \approx 1.732\) Final Answer: Mean = 3 students, Standard deviation ≈ 1.732 students
Q18: A website experiences an average of 5 crashes per month. Assuming crashes follow a Poisson distribution, what is the probability of exactly 3 crashes in a given month? (Use \(e^{-5} \approx 0.00674\))
Solution:
Ans: Step 1: Identify the parameters: \(\lambda = 5\) and \(k = 3\) Step 2: Apply the Poisson formula: \(P(X = 3) = \frac{5^3 e^{-5}}{3!}\) Step 3: Calculate \(5^3 = 125\) Step 4: Calculate \(3! = 6\) Step 5: Substitute: \(P(X = 3) = \frac{125 \times 0.00674}{6} = \frac{0.8425}{6} \approx 0.1404\) Final Answer: The probability is approximately 0.1404 or 14.04%
Q19: A bookstore sells an average of 4 science fiction books per day. If sales follow a Poisson distribution, what is the probability that exactly 5 science fiction books are sold on a particular day? (Use \(e^{-4} \approx 0.0183\))
Solution:
Ans: Step 1: Identify the parameters: \(\lambda = 4\) and \(k = 5\) Step 2: Apply the Poisson formula: \(P(X = 5) = \frac{4^5 e^{-4}}{5!}\) Step 3: Calculate \(4^5 = 1024\) Step 4: Calculate \(5! = 120\) Step 5: Substitute: \(P(X = 5) = \frac{1024 \times 0.0183}{120} = \frac{18.7392}{120} \approx 0.1562\) Final Answer: The probability is approximately 0.1562 or 15.62%
Q20: Emergency calls to a fire station occur at an average rate of 2.5 calls per night. Assuming the number of calls follows a Poisson distribution, what is the variance of the number of calls per night?
Solution:
Ans: Step 1: Identify \(\lambda = 2.5\) Step 2: In a Poisson distribution, variance = \(\lambda\) Step 3: Therefore, variance = 2.5 Final Answer: The variance is 2.5 calls²
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