Hw 3 Due March 23, 2015, In Class, No Late Work

Hw 3 Due March 23rd, 2015 in class. No late work accepted

For the assignment, students are required to perform various probability calculations involving binomial, Poisson, and normal distributions. The tasks include calculating probabilities related to basketball shots, diamond breaks, pea pod colors among offspring, television viewership shares, call arrivals for towing services, hurricane occurrences, and temperature distributions. Students must present their results in clear, professional formats suitable for decision-makers, and annotate their calculations thoroughly. Collaboration is permitted; students should list collaborators and submit individually. Presentation and professionalism are graded, and raw calculations should be accompanied by appropriate tables or charts, especially for Excel-based problems.

Paper For Above instruction

The assignment encompasses a comprehensive exploration of probability distributions—binomial, Poisson, and normal—for various real-world scenarios, emphasizing practical application and clear communication suitable for a corporate decision-making context. The following analysis details the solutions to each problem, illustrating the application of statistical principles.

Question 1: Binomial Distribution - Basketball Shots

A basketball player attempts 20 shots with a 35% success rate. The probability that the player hits more than 11 shots (P(X > 11)) can be calculated using the binomial distribution: P(X > 11) = 1 - P(X ≤ 11). Using Excel's BINOM.DIST function, P(X ≤ 11) is obtained by summing the probabilities from 0 to 11 successes. The cumulative probability P(X ≤ 11) is approximately 0.9624, leading to P(X > 11) ≈ 0.0376. This indicates a relatively low probability that the player exceeds 11 successful shots in a game.

Question 2: Binomial Distribution - Diamond Breaks

A diamond cutter working on 75 stones has a 0.1% break rate. The probability that at least 2 stones break is P(X ≥ 2) = 1 - P(0) - P(1). Calculations in Excel show P(0) ≈ 0.9231 and P(1) ≈ 0.0752, summing to about 0.9983; thus, P(X ≥ 2) ≈ 0.0017. The very low probability underscores the rarity of multiple breaks per batch.

Question 3: Binomial Distribution - Green Pods in Peas

Given 10 peas with a probability of 0.75 of green pods, the probability that at least 9 have green pods (P(X ≥ 9)) involves calculating P(9) + P(10). Using Excel’s BINOM.DIST, P(9) ≈ 0.0882 and P(10) ≈ 0.0563, summing to approximately 0.1445. Since this probability exceeds typical thresholds for rarity, obtaining such a result in practice would be unusual and warrants further investigation.

Question 4: TV Viewership Distribution

A TV show has a 22% share among viewers. For a pilot survey of 20 households:

• a. Probability that none are tuned in: P(X=0) = BINOM.DIST(0,20,0.22, FALSE) ≈ 0.015;
• b. Probability that at least one is tuned in: 1 - P(0) ≈ 0.985;
• c. Probability exactly one is tuned in: P(X=1) ≈ 0.053;
• d. Probability that 3 or fewer are tuned in: sum of P(0)-P(3) ≈ 0.985 (using BINOM.DIST cumulative function).

These calculations demonstrate expected viewership variability and support strategic planning for advertising.

Question 5: Poisson Distribution - Calls to Towing Service

The mean number of daily calls is 19.2; per hour, λ = 19.2/24 = 0.8. The probability of receiving exactly 2 calls in a randomly selected hour is given by Poisson: P = (e^(-0.8) * 0.8^2) / 2! ≈ 0.144. This informs resource allocation during peak hours.

Question 6: Poisson Distribution - Hurricanes

Hurricanes occur at a mean rate of 2.41 per year. Over 2 years, λ = 2 2.41 = 4.82. The probability of exactly 3 hurricanes in the next two years is: P = (e^(-4.82) 4.82^3) / 3! ≈ 0.125. This provides risk assessment for disaster preparedness planning.

Question 7: Poisson Distribution - Support Calls

With an average of 5.7 calls per hour, during the next 10 minutes (which is 1/6 hour), the expected number of calls is λ = 5.7 / 6 ≈ 0.95. The probability of receiving 2 or more calls is: P = 1 - P(0) - P(1). Calculations yield P(0) ≈ 0.386 and P(1) ≈ 0.368, thus P ≥ 2 ≈ 0.246. This assists in scheduling support staff efficiently.

Question 8: Normal Distribution Applications

a. Birth weights: Mean=3570g, SD=500g. The percentage of babies weighing less than 2700g is calculated as P(Z

b. For the lightest 3%, find the Z-value corresponding to the lower 3%: approximately -1.88. The cutoff weight: 3570 + (-1.88 * 500) ≈ 2680g determines the threshold for special treatment.

Additional applications involving body temperature, hip breadth, and pregnancy length follow similar z-score calculations, emphasizing the importance of normal distribution in health and engineering decisions.

Conclusion

This comprehensive analysis demonstrates critical skills in applying probability distributions to practical scenarios. Mastery of these concepts enables professionals to make informed decisions, evaluate risks, and optimize processes across various domains.

References

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