Stat 200 Week 3 Homework Problems 414a Project Conducted By

Stat 200 Week 3 Homework Problems 414a Project Conducted By The Austr

Analyze a series of statistical problems involving probability, data analysis, distributions, and basic inferential statistics based on provided data and scenarios. The problems include calculating probabilities from data tables, understanding game odds and payouts, counting arrangements, and computing probabilities for binomial and other distributions. Interpret data, draw histograms, and calculate measures of central tendency and variability. Additionally, explore concepts related to experiments, such as coin flips, manufactured products, and characteristics like handedness, applying appropriate probability models.

Paper For Above instruction

Introduction

Statistics plays a crucial role in understanding real-world phenomena by analyzing data, calculating probabilities, and interpreting distributions. This paper discusses multiple statistical problems rooted in data analysis, probability calculations, and inferential statistics. Problems include analyzing car preference data, defective lens inspections, roulette odds, combinatorial choices, binomial experiments, and probabilities concerning M&Ms and handedness. Through these examples, the importance of probability distributions, expectation, variance, and interpretation of statistical results are elucidated.

Car Preference Analysis

An Australian survey collected data on reasons for choosing a particular car, including safety, reliability, cost, performance, comfort, and looks. To analyze this data, the probability that a randomly chosen individual selects each reason can be computed by dividing the number of respondents citing each reason by the total number of respondents. For example, if 300 respondents out of 1500 cited safety as their reason, the probability for safety would be 300/1500 = 0.2. Calculating these probabilities offers insights into consumers' priorities, influencing marketing strategies and vehicle design.

Defective Lens Probabilities

The given data involves counting different types of defects in manufactured lenses. To find the probability that a randomly selected lens is scratched or flaked, sum the counts for 'scratch' and 'flaked' defects and divide by the total number of lenses inspected. Similarly, the probability of the lens being wrong PD or lost in the lab involves adding their counts and dividing by the total. Probabilities for lenses that are not scratched or not the wrong shape can be found by subtracting the respective defective counts from the total or by directly calculating the complement probabilities. These analyses help quality control processes identify defect rates and improve manufacturing quality.

Roulette Odds and Probabilities

In the roulette game, with numbers 0-36 and 00, the total possible outcomes are 38. The probability of winning by picking number 7, which must come up for a win, is 1/38. Odds against winning are expressed as the ratio of unfavorable outcomes to favorable ones, which is 37:1. If the casino pays \$20 for every dollar wagered when the number hits, the expected profit for the casino per dollar wager can be calculated as follows:

Expected Value = (Probability of win × payout) + (Probability of lose × loss). Assuming a \$1 bet, the casino's profit per game is the difference between winnings and payouts, which results in an expected profit of approximately \$19.47 per dollar wager, illustrating the house advantage.

Combinatorics: Choosing People

Calculating the number of ways to choose 7 individuals from a group of 20 involves the binomial coefficient, expressed as "20 choose 7" (denoted as C(20,7)). Using factorial notation, this is:

C(20,7) = 20! / (7! × 13!). This calculation provides the total number of possible combinations, which is essential in probability and statistical sampling plans.

Binomial Experiments and Distributions

In flipping a coin 3 times, the number of heads is a binomial random variable X with parameters n=3 and p=0.5. The probability distribution for X can be calculated using the binomial formula:

P(X=k) = C(3,k) × (0.5)^k × (0.5)^(3−k), for k=0,1,2,3.

The histogram of this distribution is symmetric, centered around the mean. The mean of this distribution is n×p=1.5, the variance is n×p×(1−p)=0.75, and the standard deviation is √0.75 ≈ 0.866.

The probability of getting two or more heads (k=2 or 3) is P(X≥2)=P(X=2)+P(X=3). Calculations show this probability is approximately 0.25, indicating that flipping two or more heads is not unusual in three trials.

Expected Value of Extended Warranty

For an LG dishwasher costing \$800 with a 20% chance of needing replacement within two years, the expected value of purchasing an extended warranty costing \$112.10 is:

Expected value = (Probability of replacement × replacement cost) + (Probability of no replacement × 0) - cost of warranty,

= 0.2 × \$800 + 0.8 × \$0 - \$112.10 = \$160 - \$112.10 = \$47.90.

This positive expected value suggests the warranty might be a worthwhile investment given the risk of replacement.

Binomial Probabilities with Technology

When dealing with binomial distributions with parameters n=6 and p=0.3, probabilities for specific outcomes can be calculated using statistical software or calculators running the binomial distribution. For example, the probability of exactly 2 successes out of 6 trials is computed using the binomial probability formula or appropriate functions.

Probability of Brown M&Ms

Given that approximately 14% of M&Ms are brown, in a packet of 52, the random variable X can represent the number of brown M&Ms. This situation follows a binomial distribution with n=52 and p=0.14.

- The probability that exactly 6 M&Ms are brown is calculated as P(X=6).

- For 25 brown M&Ms, P(X=25) is extremely low, indicating an unusual occurrence.

- The probability that all 52 are brown, P(X=52), is even more negligible.

- Analyzing whether scenarios like all M&Ms being brown are unusual helps assess variability and potential manufacturing issues.

Statistics of Left-Handedness

In a group of 15 people, with a 10% prevalence of left-handedness, the number of left-handed individuals (X) follows a binomial distribution with p=0.10.

- The mean number of left-handed people = n×p=1.5.

- Variance = n×p×(1−p)=1.35.

- Standard deviation = √1.35 ≈ 1.16.

A histogram of X would be right-skewed, with most groups having 0 or 1 left-handed individual, and the probability that exactly 5 are left-handed is very low, indicating this is an unusual occurrence.

Conclusion

The analyzed problems demonstrate fundamental statistical concepts such as probability calculations, distributions, expectation, variance, and the interpretation of data. Understanding these concepts allows researchers and analysts to make informed decisions based on data, evaluate risks, and understand variability within real-world contexts. Applying these statistical tools to practical scenarios enhances decision-making processes across diverse fields, from manufacturing quality control to gaming strategies and demographic studies.

References

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