Stat 350 Spring 2017 Homework 4: Practice Problems

Stat 350 Spring 2017 Homework 4 20 Points 1practice Problems 57

In Illinois, a typical DUI (Driving Under the Influence) offender is a 34-year-old male, arrested between 11 p.m. and 4 a.m. on a weekend, with a BAC of 0.16. Eighty-five percent of all drivers arrested for DUI are first-time offenders. Suppose 40 people arrested for DUI in Illinois are selected at random. Determine the probability that exactly six are repeat offenders, the probability that at least two are repeat offenders, the mean number of repeat offenders, and the standard deviation of that number.

Paper For Above instruction

Driving Under the Influence (DUI) statistics and probability models are essential tools in understanding criminal behavior patterns and informing policy decisions. In Illinois, where the typical DUI offender profile involves a 34-year-old male arrested during late weekend hours with elevated BAC levels, statistical analysis of such incidents provides insights into offender demographics and arrest patterns. Given that 85% of DUI arrests involve first-time offenders, we can evaluate the likelihood of various arrest scenarios through the binomial distribution model.

In the problem, we consider 40 arrests, with the probability of a first-time offender being 0.85, and consequently, the probability of a repeat offender being 0.15 (since 1 - 0.85 = 0.15). Let X represent the number of repeat offenders in this sample, which follows a binomial distribution with parameters n=40 and p=0.15.

Part a: Probability that exactly six are repeat offenders

The probability of observing exactly k=6 repeat offenders can be calculated using the binomial probability formula:

P(X = k) = C(n, k) p^k (1 - p)^{n - k}

where C(n, k) is the combination of n items taken k at a time. Plugging in the values:

P(X = 6) = C(40, 6) (0.15)^6 (0.85)^{34}

Calculating C(40, 6):

C(40, 6) = 3,838,380

Thus,

P(X = 6) ≈ 3,838,380 (0.15)^6 (0.85)^{34}

Using calculator or statistical software, this probability is approximately 0.188.

Part b: Probability that at least two are repeat offenders

This requires finding P(X ≥ 2), which is equivalent to 1 - P(X

P(X=0) = C(40, 0) (0.15)^0 (0.85)^{40} = 1 1 (0.85)^{40} ≈ 0.013

P(X=1) = C(40, 1) (0.15)^1 (0.85)^{39} = 40 0.15 (0.85)^{39} ≈ 0.077

Therefore,

P(X ≥ 2) ≈ 1 - (0.013 + 0.077) = 0.91

This indicates a high likelihood that at least two individuals are repeat offenders among the 40 arrested.

Part c: Mean of the number of repeat offenders

The mean for a binomial distribution is given by:

μ = n p = 40 0.15 = 6

This suggests that, on average, 6 of the arrested individuals are repeat offenders in such a sample.

Part d: Standard deviation of the number of repeat offenders

The standard deviation is calculated as:

σ = √(n p (1 - p)) = √(40 0.15 0.85) ≈ √(5.1) ≈ 2.26

This measure quantifies the variability expected in the number of repeat offenders among samples.

Conclusion

The binomial distribution provides a robust framework for modeling the number of repeat DUI offenders in file samples. The probability calculations highlight that while the most probable number of repeat offenders is around 6, there is a very high chance that at least two are repeat offenders, and the variability in counts can be characterized by a standard deviation of approximately 2.26. Such statistical insights can aid law enforcement and policymakers in designing targeted intervention strategies, resource allocation, and understanding the underlying patterns in DUI recidivism.

References

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  • Bluman, A. G. (2017). Elementary Statistics: A Step By Step Approach. McGraw-Hill Education.
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