A Local Drugstore Owner Knows That On Average 20 People Ente

1 A Local Drugstore Owner Knows That On Average 20 People Enter His

A local drugstore owner knows that, on average, 20 people enter his store each hour, assume that the number of people that come into the store follows a Poisson process. (a) Find the probability that at least 10 people will come to the store in 30 minutes. (b) Find the probability that at least 5 people will come to the store in 15 minutes.

2. A store has 20 guitars in stock but 3 are defective. Claire buys 5 guitars from this lot. (a) Find the probability that Claire bought 2 defective guitars. (b) Suppose that 3% of a total of 5000 guitars are defective. Claire buys 10 guitars from this lot. Find the probability that Claire bought 2 defective guitars from this lot.

3. It is estimated that 5 out of every 8 houses give away candy for Halloween. (a) If we want to receive candy from 5 different houses, what is the probability that we will need to visit at least 10 of them? (b) If you plan to visit 26 houses, what is the probability that you will get candy in at least 15 of them?

Sample Size (n) = 186 Sample Mean = 3103 Sample Standard Deviation = 696 A. Construct a 95% Confidence Interval. B. Modify (change) your sample size to 18600. Construct a new 95% Confidence Interval.

Paper For Above instruction

The problems presented encompass a range of statistical concepts, including Poisson processes, hypergeometric and binomial distributions, and confidence interval construction. This paper will analyze each problem in detail, elucidating the relevant statistical methods, calculations, and interpretations involved in solving them.

Problem 1: Poisson Process in a Drugstore

The first problem involves modeling the number of customers entering a store as a Poisson process, with an average rate of 20 customers per hour. To determine the probabilities over different time intervals, we recognize that the Poisson distribution’s parameter λ scales proportionally with time. Specifically, for 30 minutes (half an hour), λ becomes 10 (since 20 per hour × 0.5 hour = 10). Similarly, for 15 minutes, λ becomes 5.

The general Poisson probability mass function (PMF) is given by:

P(X = k) = (λ^k * e^(-λ)) / k!

where λ is the expected number of occurrences in the interval, and k is the number of occurrences we are calculating the probability for.

Part (a): Probability of at least 10 people in 30 minutes

We seek P(X ≥ 10). Since the Poisson distribution is discrete, this can be computed as:

P(X ≥ 10) = 1 - P(X ≤ 9) = 1 - Σk=0^9 P(X = k)

Calculating P(X ≤ 9) involves summing Poisson probabilities for k = 0 to 9 with λ = 10. Using statistical software or tables simplifies this process. For example, using R:

ppois(9, lambda=10)

which yields approximately 0.457.

Therefore, P(X ≥ 10) ≈ 1 - 0.457 = 0.543.

Part (b): Probability of at least 5 people in 15 minutes

Similarly, λ = 20 × 0.25 = 5. Thus, we calculate:

P(X ≥ 5) = 1 - P(X ≤ 4) = 1 - ppois(4, lambda=5)

Using software:

ppois(4, lambda=5)

which yields approximately 0.440. So, P(X ≥ 5) ≈ 1 - 0.440 = 0.560.

Problem 2: Hypergeometric and Binomial Distributions in Guitar Stock

Part (a): Probability of buying exactly 2 defective guitars out of 5 from 20 guitars with 3 defective using hypergeometric distribution:

Hypergeometric PMF:

P(X = k) = [C(D, k) * C(N - D, n - k)] / C(N, n)

where:

  • N = 20 (total guitars)
  • D = 3 (defective guitars)
  • n = 5 (guitars bought)
  • k = 2 (defective guitars wanted)

Calculations:

P = (C(3,2) * C(17,3)) / C(20,5)

Expressed as:

C(3,2) = 3, C(17,3) = 680, C(20,5) = 15504

P = (3 * 680) / 15504 ≈ 204 / 15504 ≈ 0.0132

Part (b): Probability of buying exactly 2 defective guitars from a much larger lot with 3% defective among 5000 guitars.

This situation is better modeled with the binomial distribution, as the lot is large and sampling is without replacement, but approximation with binomial is acceptable (sampling fraction is small). The binomial probability with n=10, p=0.03, k=2:

P = C(10,2)  (0.03)^2  (1-0.03)^{8}

Calculating:

C(10,2) = 45, (0.03)^2 = 0.0009, (0.97)^8 ≈ 0.817

P ≈ 45 0.0009 0.817 ≈ 45 * 0.000735 ≈ 0.0331

Problem 3: Probability in House Visits for Halloween

In this context, each house has a 5/8 chance of giving candy. The number of successful houses in a series follows a binomial distribution: X ∼ Binomial(n, p), with p = 5/8 ≈ 0.625.

Part (a): Probability that at least 5 houses yield candy, needing to visit at least 10 houses.

For receiving candy from 5 houses, the event involves the number of houses visited until getting 5 candies. This situation resembles a negative binomial distribution where the count of trials until a fixed number of successes occurs, but the problem asks for the probability that at least 10 houses are needed to get 5 candies, meaning fewer than 5 candies in less than 10 visits. Alternatively, it can be approached by considering the probability that, after 10 visits, fewer than 5 candies are obtained:

However, based on interpretation, the question seems to be: "What is the probability that we need to visit at least 10 houses to collect 5 candies?" This equates to the probability that, in fewer than 10 visits, we have received 5 candies, which can be modeled with the negative binomial distribution. But since the question states "at least 10," the correct approach is to compute P(X ≥ 10) where X is the trial count needed for 5 successes in a Negative Binomial context, which simplifies to 1 minus the probability of getting the 5th success in fewer than 10 trials.

Alternatively, if assuming binomial for the total number of houses visited:

Probability of getting at least 5 candies in 10 houses is P(X ≥ 5), where X ∼ Binomial(10, 0.625). Using cumulative distribution functions:

P=1 - pbinom(4, size=10, prob=0.625)

This yields the probability that we get at least 5 candies in 10 visits.

Part (b): For visiting 26 houses, the probability that at least 15 yield candy:

P=1 - pbinom(14, size=26, prob=0.625)

Applying the binomial cumulative distribution function, these calculations provide approximate probabilities.

Confidence Interval Calculation

The sample data indicate a sample size of n=186, with a mean of 3103 and standard deviation of 696.

Constructing a 95% confidence interval for the population mean involves the formula:

CI = mean ± Z* (standard deviation / sqrt(n))

Where Z* for 95% confidence is approximately 1.96.

Calculating:

Standard Error (SE) = 696 / sqrt(186) ≈ 696 / 13.653 ≈ 50.99

Lower bound:

3103 - 1.96 * 50.99 ≈ 3103 - 100.0 ≈ 3003

Upper bound:

3103 + 1.96 * 50.99 ≈ 3103 + 100.0 ≈ 3203

Thus, the 95% confidence interval is approximately (3003, 3203).

For the larger sample size of 18,600, the standard error reduces because of the increased sample size:

SE = 696 / sqrt(18600) ≈ 696 / 136.11 ≈ 5.11

Recalculating the confidence interval:

3103 - 1.96 * 5.11 ≈ 3103 - 10.0 ≈ 3093

and

3103 + 1.96 * 5.11 ≈ 3103 + 10.0 ≈ 3113

Resulting in a tighter interval: (3093, 3113), reflecting increased precision with larger samples.

Conclusion

The statistical analyses showcased here demonstrate the application of different probability models, including the Poisson, hypergeometric, binomial, and negative binomial distributions, within practical scenarios. Calculations highlight how selecting appropriate models and formulas can effectively estimate probabilities and confidence intervals for various real-world problems, from customer arrival rates and product defect probabilities to Halloween house visit strategies and confidence interval estimations based on sample data.

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