Binomial & Geometric Models - Business Statistics

1. A manufacturing process produces defective items with probability 0.06. Quality inspectors sample 50 items. Using normal approximation to the binomial, what is the probability that the number of defective items differs from the expected value by more than 2 standard deviations?
   Answer: Approximately 0.05, calculated as $2 \times P(Z > 2) = 2 \times 0.0228 = 0.0456$ With n=50, p=0.06, we have μ=3 and σ=√(50×0.06×0.94)≈1.68. We want P(|X-μ|>2σ) = P(Z<-2 or Z>2) = 2×P(Z>2) ≈ 2×0.0228 ≈ 0.0456. Choice A gives the right answer but wrong reasoning (not about Type I error). Choice B has wrong final calculation. Choice D gives the complement probability.
2. A software development team estimates a 60% chance of successfully completing a feature in a one-week sprint. Sprints are independent. What is the probability that the team's third successful feature completion occurs in the fifth sprint?
   Answer: 0.207 This is a negative binomial problem that can be solved with binomial reasoning. For the 3rd success to occur in the 5th sprint, two conditions must be met: 1) The 5th sprint must be a success (probability = 0.6). 2) The first four sprints must contain exactly two successes. We can calculate the probability of the second condition using the binomial formula with $n=4$, $k=2$, and $p=0.6$: $P(\text{2 successes in 4 sprints}) = \binom{4}{2}(0.6)^2(0.4)^2 = 6 \times 0.36 \times 0.16 = 0.3456$. Since the sprints are independent, we multiply the probabilities of the two conditions: $0.3456 \times 0.6 = 0.20736$.
3. A factory produces electronic chips in batches of 500. The long-run probability of a chip being defective is 0.02. A batch is flagged for having unusually high variability if the number of defects is more than two standard deviations above the mean. What is the minimum integer number of defects required for a batch to be flagged?
   Answer: 17 This is a binomial distribution with $n=500$ and $p=0.02$. The mean is $\mu = np = 500 \times 0.02 = 10$. The variance is $\sigma^2 = np(1-p) = 500 \times 0.02 \times 0.98 = 9.8$. The standard deviation is $\sigma = \sqrt{9.8} \approx 3.13$. The threshold for being flagged is $\mu + 2\sigma = 10 + 2(3.13) = 10 + 6.26 = 16.26$. Since the number of defects must be an integer and the condition is "more than" 16.26, the minimum integer number of defects is 17.
4. A customer service center finds that 25% of incoming calls are for technical support. Calls are independent events. An operator has just handled four consecutive non-technical calls. What is the probability that the very next call will be for technical support?
   Answer: 0.250 This question tests the memoryless property of Bernoulli trials, which are the foundation of the geometric and binomial distributions. Because the trials (calls) are independent, the outcome of previous calls has no impact on the outcome of the next call. The probability that any given call is for technical support remains constant at 0.25.
5. A company sources a component from two suppliers. Supplier A has a 5% defect rate and sends components in crates of 20. Supplier B has a 3% defect rate and sends components in crates of 30. A crate is rejected if it contains two or more defective components. Which supplier is more likely to have a crate rejected, and by approximately what margin?
   Answer: Supplier A, by a margin of 0.037 We need to calculate the probability of rejection, $P(X \ge 2)$, for each supplier using the binomial model. This is equal to $1 - [P(X=0) + P(X=1)]$. For Supplier A ($n=20, p=0.05$): $P(X_A \ge 2) = 1 - [\binom{20}{0}(0.05)^0(0.95)^{20} + \binom{20}{1}(0.05)^1(0.95)^{19}] = 1 - [0.3585 + 0.3774] = 1 - 0.7359 = 0.2641$. For Supplier B ($n=30, p=0.03$): $P(X_B \ge 2) = 1 - [\binom{30}{0}(0.03)^0(0.97)^{30} + \binom{30}{1}(0.03)^1(0.97)^{29}] = 1 - [0.4010 + 0.3721] = 1 - 0.7731 = 0.2269$. Supplier A has a higher rejection probability (0.2641 vs 0.2269). The margin is $0.2641 - 0.2269 = 0.0372$.
6. An inspector samples 5 items from a large batch with a 30% defect rate. It is known that the sample contains at least one defective item. Given this information, what is the probability that the sample contains exactly two defective items?
   Answer: 0.371 This is a conditional probability problem using the binomial distribution ($n=5, p=0.3$). Let A be the event $X=2$ (exactly two defects) and B be the event $X \ge 1$ (at least one defect). We want $P(A|B) = P(A \cap B) / P(B)$. Since A is a subset of B, $A \cap B = A$. So we need to calculate $P(X=2) / P(X \ge 1)$. Numerator: $P(X=2) = \binom{5}{2}(0.3)^2(0.7)^3 = 10(0.09)(0.343) = 0.3087$. Denominator: $P(X \ge 1) = 1 - P(X=0) = 1 - \binom{5}{0}(0.3)^0(0.7)^5 = 1 - 0.16807 = 0.83193$. Conditional probability: $0.3087 / 0.83193 \approx 0.3711$.
7. A streaming service models customer churn using a geometric distribution. There is a 15% probability that a customer will cancel their subscription in any given month, independent of previous months. A promotion is offered to customers who stay subscribed for at least one full year (12 months). What is the probability that a new customer will qualify for this promotion?
   Answer: 0.142 To qualify for the promotion, a customer must not cancel their subscription for each of the first 12 months. The probability of cancellation in a given month is $p=0.15$. The probability of not cancelling (i.e., renewing) in a given month is $1-p = 0.85$. Since the months are independent, the probability of not cancelling for 12 consecutive months is $(0.85)^{12}$. Calculating this value gives $(0.85)^{12} \approx 0.1422$. This is the probability that the first cancellation occurs after month 12, $P(X > 12)$.
8. An e-commerce site has a 8% conversion rate for visitors who view a product page. A data scientist is modeling the number of page views until the 3rd conversion occurs. If X represents this random variable, what is the coefficient of variation for X?
   Answer: $\text{CV} = \frac{\sqrt{3(1-0.08)/0.08^2}}{3/0.08} = \frac{\sqrt{431.25}}{37.5} \approx 0.555$ For negative binomial distribution with r=3 successes and p=0.08, E[X]=r/p=3/0.08=37.5 and Var(X)=r(1-p)/p²=3(0.92)/(0.08)²=431.25. Therefore CV=σ/μ=√431.25/37.5≈0.555. Choice A uses wrong variance formula. Choice C uses p instead of (1-p). Choice D oversimplifies the calculation incorrectly.
9. A marketing firm observes that 15% of visitors to a client's landing page make a purchase. The firm launches a campaign that drives 20 new, independent visitors to the page. The campaign is considered a failure if two or fewer visitors make a purchase. What is the approximate probability that the campaign is considered a failure?
   Answer: 0.405 This scenario follows a binomial distribution with $n = 20$ trials (visitors) and a success probability of $p = 0.15$ (purchase). The campaign is a failure if the number of successes, $X$, is $X \le 2$. We need to calculate $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$. $P(X=0) = \binom{20}{0}(0.15)^0(0.85)^{20} \approx 0.0388$ $P(X=1) = \binom{20}{1}(0.15)^1(0.85)^{19} \approx 0.1368$ $P(X=2) = \binom{20}{2}(0.15)^2(0.85)^{18} \approx 0.2293$ Summing these probabilities gives $0.0388 + 0.1368 + 0.2293 = 0.4049 \approx 0.405$.
10. A factory produces 10,000 plastic bottles per day, and the probability of a bottle having a minor defect is 0.08. What is the approximate probability that between 770 and 820 bottles (inclusive) are defective on a given day, using a normal approximation to the binomial distribution?
    Answer: 0.65 First, verify the normal approximation is appropriate: $np = 10000 \times 0.08 = 800 > 10$ and $n(1-p) = 9200 > 10$. The mean is $\mu = np = 800$. The standard deviation is $\sigma = \sqrt{np(1-p)} = \sqrt{10000 \times 0.08 \times 0.92} = \sqrt{736} \approx 27.13$. We want $P(770 \le X \le 820)$. Applying the continuity correction, we find the probability for the range [769.5, 820.5]. Convert to z-scores: $z_1 = (769.5 - 800) / 27.13 \approx -1.12$ and $z_2 = (820.5 - 800) / 27.13 \approx 0.76$. We need $P(-1.12 \le Z \le 0.76)$, which is $\Phi(0.76) - \Phi(-1.12) = 0.7764 - 0.1314 = 0.645$.
11. A digital marketing agency tracks email campaign performance where 14% of recipients click through to the website. For a campaign sent to 25 recipients, what is the probability that the number of click-throughs is at least 2 but no more than the expected value plus one standard deviation?
    Answer: $P(2 \leq X \leq 5) = P(X \leq 5) - P(X \leq 1) \text{ using binomial tables}$ When you encounter problems about independent trials with fixed success probability, you're dealing with the binomial distribution. Here, each email recipient either clicks (success, p = 0.14) or doesn't click (failure), and we have n = 25 independent trials. The key insight is interpreting "at least 2 but no more than the expected value plus one standard deviation." First, calculate the distribution parameters: expected value μ = np = 25(0.14) = 3.5, and standard deviation σ = √(np(1-p)) = √(25 × 0.14 × 0.86) ≈ 1.73. So μ + σ = 3.5 + 1.73 = 5.23. Since we need integer values, this becomes P(2 ≤ X ≤ 5). Answer D correctly identifies this range and uses the complement rule: P(2 ≤ X ≤ 5) = P(X ≤ 5) - P(X ≤ 1), which is computationally efficient when using binomial tables. Answer A incorrectly stops at X = 3, apparently misunderstanding that 3.5 represents the expected value, not the upper bound. It also shows the calculation stopping at a non-integer. Answer B uses the wrong upper bound of 6, likely from incorrectly rounding μ + σ = 5.23 up instead of recognizing that 5 is the correct integer boundary. Answer C uses 4 as the upper bound, probably from rounding μ + σ down incorrectly or miscalculating the standard deviation. Study tip: Always calculate both μ and σ for binomial problems, and remember that when dealing with ranges involving non-integers, you need to determine the appropriate integer boundaries based on the context.
12. An e-commerce site runs two conversion experiments. Experiment 1 consists of showing an ad to 100 users, with a conversion probability $p=0.1$. Experiment 2 shows a different ad to 200 users, also with a conversion probability $p=0.1$. Which of the following statements correctly compares the resulting binomial distributions of the number of conversions?
    Answer: The distribution for Experiment 2 will have a larger mean and be more spread out. When comparing binomial distributions, you need to examine how sample size (n) and probability (p) affect the three key characteristics: mean, variance, and shape. For any binomial distribution, the mean equals $np$ and the variance equals $np(1-p)$. In Experiment 1, with n=100 and p=0.1, the mean is 10 and variance is 9. In Experiment 2, with n=200 and p=0.1, the mean is 20 and variance is 18. Since Experiment 2 has both a larger mean and larger variance, it will be more spread out (higher standard deviation of $\sqrt{18} \approx 4.24$ versus $\sqrt{9} = 3$). Answer D correctly identifies that Experiment 2 has both a larger mean and greater spread. Answer A incorrectly claims Experiment 2 will be "less spread out" – this confuses the concept that while the distribution becomes more normal-looking with larger n, the actual spread (measured by standard deviation) still increases when variance increases. Answer B is wrong because skewness depends on both n and p; while both experiments have the same p=0.1, the different sample sizes create different degrees of skewness. Answer C makes a fundamental error by suggesting equal variances – variance depends on both n and p, not just p alone. Study tip: Remember the binomial formulas: mean = np, variance = np(1-p). When sample size increases while p stays constant, both mean and variance increase proportionally, making the distribution more spread out despite becoming more symmetric.
13. A market researcher is surveying 100 customers selected from a small, local store with only 250 loyalty members. Customers are not replaced in the sampling pool after being surveyed. The researcher is modeling the number of "highly satisfied" customers using a binomial distribution. Which of the following assumptions of the binomial model is most likely violated in this scenario?
    Answer: The trials are not independent of each other. When you encounter questions about binomial distributions, you need to verify that all four key assumptions are met: fixed number of trials, exactly two outcomes per trial, constant probability of success, and independence between trials. In this scenario, the researcher is sampling 100 customers from a small pool of only 250 loyalty members without replacement. This creates a dependency problem. When the first customer is surveyed and removed from the pool, it changes the composition of the remaining population. If that customer was "highly satisfied," there's now one fewer satisfied customer available for subsequent selections, which slightly alters the probability for all remaining trials. Each selection affects the probability of success for the next selection. Let's examine why the other options don't represent violations: Option A is incorrect because the number of trials is clearly fixed at 100 customers. Option B is wrong since each customer can only be classified as either "highly satisfied" or "not highly satisfied" – exactly two outcomes. Option C misses the point entirely; while the true probability might be unknown to the researcher, this doesn't violate the binomial assumption that the probability remains constant throughout the experiment. Option D correctly identifies the independence violation. Without replacement sampling from a small population creates dependencies between trials, making the binomial model inappropriate. Study tip: Remember the "10% rule" – binomial approximation is generally acceptable when sampling without replacement only if your sample size is less than 10% of the population. Here, 100/250 = 40%, far exceeding this threshold.
14. A company sends out promotional mailers. 2% of recipients make a purchase, and 5% request more information. These are mutually exclusive events. What is the expected number of mailers the company must send to get its first response of any kind (either a purchase or a request for information)?
    Answer: 14.3 This is a geometric distribution problem. A "success" is defined as getting any kind of response. Since the events (purchase, information request) are mutually exclusive, the total probability of success on any one trial is the sum of their individual probabilities: $p = 0.02 + 0.05 = 0.07$. The expected number of trials until the first success in a geometric distribution is given by $E[X] = 1/p$. Therefore, the expected number of mailers is $1 / 0.07 \approx 14.286$.
15. A sales team makes 50 cold calls a day. Historically, 21% of calls lead to a follow-up meeting. What is the single most likely number of follow-up meetings the team will schedule in a day?
    Answer: 10 The question asks for the mode of a binomial distribution with $n=50$ and $p=0.21$. The mode of a binomial distribution is the integer value $k$ that has the highest probability. It can be found by calculating $\lfloor(n+1)p\rfloor$. In this case, $(50+1) \times 0.21 = 51 \times 0.21 = 10.71$. The floor of 10.71 is 10. Therefore, 10 is the most likely number of follow-up meetings. The mean is $np=50 \times 0.21 = 10.5$, which is not a possible outcome and is different from the mode.
16. A company rates employee project outcomes as either "successful" or "unsuccessful". Four departments, each completing 50 projects, report their success rates ($p$). Department A: $p=0.10$, Department B: $p=0.45$, Department C: $p=0.85$, Department D: $p=0.50$. For which department would the theoretical binomial distribution of the number of successful projects be most skewed?
    Answer: Department A The skewness of a binomial distribution depends on the probability of success, $p$. The distribution is perfectly symmetric when $p=0.5$. The skewness increases as $p$ moves further away from 0.5. We compare the absolute difference between each department's $p$ and 0.5: Dept A: $|0.10 - 0.5| = 0.40$ Dept B: $|0.45 - 0.5| = 0.05$ Dept C: $|0.85 - 0.5| = 0.35$ Dept D: $|0.50 - 0.5| = 0.00$ The largest difference is 0.40, corresponding to Department A. Therefore, its distribution will be the most skewed (specifically, skewed to the right).
17. In a clinical trial for a new drug, the probability of a patient responding positively is 0.3. Patients are treated sequentially and independently. What is the probability that the first positive response is observed after the fourth patient?
    Answer: 0.240 This scenario is modeled by the geometric distribution, with a success probability of $p=0.3$. The question asks for the probability that the first success occurs after the 4th trial, i.e., $P(X > 4)$. This is equivalent to the probability of having 4 consecutive failures. The probability of a single failure (a non-response) is $1-p = 1-0.3 = 0.7$. The probability of four independent failures in a row is $(0.7)^4 = 0.2401$.
18. A call center has a 25% conversion rate (probability a call results in a sale). During a shift, an agent makes calls until achieving exactly 4 conversions, then stops. If the agent made exactly 20 calls total, what is the probability of this specific outcome?
    Answer: $\binom{19}{3}(0.25)^4(0.75)^{16}$ This is a negative binomial scenario where the agent stops after the 4th success. The 20th call must be a conversion (the 4th), and among the first 19 calls, exactly 3 must be conversions. This gives $\binom{19}{3}(0.25)^4(0.75)^{16}$. Choice A treats this as a regular binomial (fixed trials). Choice C ignores the combinatorial aspect. Choice D has incorrect binomial coefficient and powers.
19. A financial advisor finds that the probability of converting a prospect into a client is 0.10 on any given meeting. What is the probability that the first conversion occurs after the 3rd meeting but on or before the 6th meeting?
    Answer: 0.198 This scenario follows a geometric distribution with success probability $p=0.10$. We want the probability that the first success (conversion) occurs on trial $X=4, 5, \text{or } 6$. We can calculate these probabilities and sum them. The formula is $P(X=k) = (1-p)^{k-1}p$. $P(X=4) = (0.9)^3(0.1) = 0.0729$ $P(X=5) = (0.9)^4(0.1) = 0.06561$ $P(X=6) = (0.9)^5(0.1) = 0.059049$ The total probability is $0.0729 + 0.06561 + 0.059049 \approx 0.1976$. Alternatively, this is $P(X > 3) - P(X > 6) = (0.9)^3 - (0.9)^6 = 0.729 - 0.531441 \approx 0.198$.
20. A web designer is A/B testing two page layouts. The established conversion rate for the original layout (Layout A) is 20%. The designer tests a new layout (Layout B) on a sample of 10 visitors. The probability of observing zero conversions among these 10 visitors with Layout B was found to be 0.0282. Assuming the number of conversions follows a binomial distribution, what is the approximate improvement in the conversion rate (as a percentage point difference) offered by Layout B over Layout A?
    Answer: 10.0% Let $p_B$ be the conversion rate for Layout B. The probability of zero conversions in $n=10$ trials is given by $P(X=0) = \binom{10}{0} p_B^0 (1-p_B)^{10} = (1-p_B)^{10}$. We are given that this probability is 0.0282. So, $(1-p_B)^{10} = 0.0282$. To find $1-p_B$, we take the 10th root of 0.0282: $1-p_B = (0.0282)^{1/10} \approx 0.70$. This implies $p_B = 1 - 0.70 = 0.30$, or 30%. The conversion rate for Layout A is 20%. The improvement is the difference: $30\% - 20\% = 10\%$.

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