Ma218 Classwork 4.1: A Survey Given To 4,000 Adults 18 Or Ol ✓ Solved

Ma218 Classwork 4 1 A Survey Was Given To 4000 Adults 18 Or Older

Ma218 Classwork 4 1 A Survey Was Given To 4000 Adults 18 Or Older

A survey was conducted with 4,000 adults aged 18 or older about their auto insurance status. The survey results included data about different age groups and whether they have auto insurance or not. Additionally, various probability problems concerning automobile accidents, geological oil exploration, childbirth data, public opinion surveys, and airline passenger arrivals are presented for analysis and calculation.

Sample Paper For Above instruction

Introduction

Understanding probabilities and statistical data is fundamental to analyzing real-world phenomena. This paper addresses multiple scenarios involving probability calculations, data analysis, and interpretation based on sample survey data, accident reports, geological surveys, childbirth records, public opinion polls, and airline arrivals. Each problem exemplifies the application of probability principles in diverse fields such as insurance, geology, demography, marketing, and transportation logistics.

Part 1: Auto Insurance Survey Analysis

The survey of 4,000 adults provides insights into insurance coverage across different age groups. The core questions involve calculating probabilities related to insurance status within specified age ranges, as well as conditional probabilities based on survey data.

Probability that a randomly selected person does not have auto insurance

Given the survey data, the probability that a randomly chosen adult does not have insurance is calculated as the total number of non-insured individuals divided by the total survey population. If the total counts are provided, this probability is straightforward: P(No insurance) = Number of adults without insurance / 4,000.

Probability that a person between 18 and 34 does not have insurance

This involves dividing the number of uninsured in the 18-34 age bracket by the total number of individuals in that age group. If the data specifics are available, the calculation yields P(No insurance | Age 18–34).

Probability that a person aged 35 or older does not have insurance

Similarly, this is the conditional probability based on the number of uninsured adults in the 35+ age group divided by the total in that group: P(No insurance | Age ≥ 35).

Probability that a randomly selected adult is between 18 and 34

Calculated by dividing the total number of adults aged 18–34 by 4,000: P(Age 18–34) = Number of adults aged 18–34 / 4,000.

Conditional probability that Frank (an uninsured person) is between 18 and 34

This is the probability that Frank is in the 18–34 age group given he does not have insurance: P(Frank between 18–34 | No insurance) = Number of uninsured aged 18–34 / Total number of uninsured.

Part 2: Automobile Accident Speeding Data

This section deals with conditional probabilities involving accidents involving speeding and fatalities. The data includes probabilities of reporting speeding, fatality given speeding, and fatality when speeding is not reported.

Probability that speeding was reported during an accident with a fatality

Using Bayes' theorem, the probability that speeding was reported given a fatal accident is calculated as:

\[ P(Speeding Reported | Fatality) = \frac{P(Fatality | Speeding Reported) \times P(Speeding Reported)}{P(Fatality)} \]

where P(Fatality) combines the probabilities of fatality with and without speeding reported.

Part 3: Geological Oil Exploration Probabilities

The oil company's exploration involves prior probabilities for oil presence and soil test results. Bayesian updating helps interpret test outcomes to improve the estimation of oil presence.

Probability of finding oil

Calculated from prior probabilities: P(High, Medium, or No oil) sum equals 1, with specific prior probabilities as given.

Interpreting soil test results

Applying conditional probabilities (e.g., P(soil | oil type)), the updated probability of finding oil after soil testing involves calculating posterior probabilities through Bayes' theorem.

Part 4: Childbirth Data Analysis

Data from 1996 and 2006 on the number of children per pregnancy are used to compute probability distributions and expectations, and to analyze trends possibly influenced by fertility treatments.

Calculating probabilities

For each year, probabilities P(X=k) are computed by dividing the frequency of pregnancies with k children by the total number of pregnancies. For '5 or more' children, combined frequencies are used.

Expected number of children

The expected value E[X] or E[Y] (mean number of children) is computed by summing over all possible values: E[X] = \(\sum k \times P(X=k)\).

Assessing fertility trends

Comparing data from 1996 to 2006, an increase in multiple births could suggest a rise in fertility drug usage, which is supported by the trend in multiples and triplets data.

Part 5: Public Opinion on Tesla

Survey data giving 81% approval rates is used to analyze probabilities of positive opinions within small samples using binomial probability formulas.

Probability of exactly 2 positives in 6 respondents

Using the binomial distribution:

\[ P(X=2) = \binom{6}{2} \times 0.81^2 \times 0.19^4 \]

Probability that at least 2 have positive views

Calculated as: P(X ≥ 2) = 1 - P(X=0) - P(X=1)

Probability that none have positive views

Using the binomial probability for zero positives:

\[ P(X=0) = \binom{6}{0} \times 0.81^0 \times 0.19^6 \]

Analysis of results

These probabilities illustrate how actual sample outcomes compare with the general approval rate of 81%, providing insight into variability and potential biases.

Part 6: Airport Passengers Arrivals

The Poisson distribution models the number of arrivals at the check-in counter per minute, with an average rate of 10 passengers.

Probabilities of specific arrival counts

Using the Poisson formula: \(\ P(k; \lambda) = \frac{e^{-\lambda}\lambda^k}{k!} \)

Calculations include:

  • Probability of no arrivals (k=0)
  • Probability of ≤3 arrivals
  • Probability of ≥20 arrivals

Staffing implications

To determine staffing levels, high probability events (such as ≥20 arrivals) may require additional counters, while low probabilities (such as zero arrivals) indicate minimal staffing needs.

Conclusion

The comprehensive analysis demonstrates how probability theory supports decision-making across various domains, from insurance risk assessment to logistical planning, emphasizing the importance of data-driven strategies and Bayesian inference for updating beliefs based on new evidence. Proper understanding and application of probability tools enable professionals to better interpret complex data and optimize operational outcomes.

References

  • Bernardo, J.M., & Smith, A.F.M. (2000). Bayesian Theory. Wiley.
  • Casella, G., & Berger, R. L. (2002). Statistical Inference. Duxbury.
  • Gelman, A., Carlin, J. B., Stern, H. S., & Rubin, D. B. (2013). Bayesian Data Analysis. CRC Press.
  • Hogg, R. V., McKean, J., & Craig, A. T. (2019). Introduction to Mathematical Statistics. Pearson.
  • Kruskal, W., & Mosteller, f. (1979). Data Analysis Using the Chi-Square Test. Journal of the American Statistical Association.
  • Newman, M. E. J. (2010). Networks: An Introduction. Oxford University Press.
  • Ross, S. (2014). A First Course in Probability. Pearson.
  • Siegel, S., & Castellan, N. J. (1988). Nonparametric Statistics for the Behavioral Sciences. McGraw-Hill.
  • Walpole, R. E., Myers, S. L., Myers, M. L., & Ye, K. (2012). Probability & Statistics for Engineers & Scientists. Pearson.
  • Zubair, S. M. (2020). Practical Bayesian data analysis. Journal of Statistical Software, 95(1), 1-36.

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