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Does crime pay? The FBI Standard Survey of Crimes showed that for about \(80 \%\) of all property crimes (burglary, larceny, car theft, etc.), the criminals are never found and the case is never solved (Source: True Odds, by James Walsh, Merrit Publishing). Suppose a neighborhood district in a large city suffers repeated property crimes, not always perpetuated by the same criminals. The police are investigating six property crime cases in this district. (a) What is the probability that none of the crimes will ever be solved? (b) What is the probability that at least one crime will be solved? (c) What is the expected number of crimes that will be solved? What is the standard deviation? (d) Quota Problem How many property crimes \(n\) must the police investigate before they can be at least \(90 \%\) sure of solving one or more cases?

Short Answer

Expert verified
(a) 0.2621, (b) 0.7379, (c) 1.2 expected, 0.98 std deviation, (d) at least 11 crimes.

Step by step solution

01

Understanding the Problem Context

We have a situation where the probability of a property crime not being solved is 80%. Therefore, the probability of a property crime being solved is 20%, or 0.2. We need to calculate different probabilities and expectations based on these probabilities.
02

Probability of None Solved

For six crimes, the probability that none are solved is given by multiplying the probability of not solving each crime: \[ P(\text{none solved}) = 0.8^6 \] Calculate \(0.8^6\) to find the probability of none of the crimes being solved.
03

Probability of At Least One Solved

The probability of at least one crime being solved is the complement of the probability that none are solved:\[ P(\text{at least one solved}) = 1 - P(\text{none solved}) \]Use the result from Step 2 in this formula to get the probability that at least one crime will be solved.
04

Expected Number of Solved Crimes

The expected number of crimes solved can be calculated using the formula for the expected value of a binomial distribution:\[ E(X) = n \cdot p = 6 \cdot 0.2 \]where \(n\) is the number of trials, and \(p\) is the probability of success (solving a crime). Calculate this to get the expected number.
05

Standard Deviation of Solved Crimes

For a binomial distribution, the standard deviation is given by:\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{6 \cdot 0.2 \cdot 0.8} \]Calculate this value to find the standard deviation of the number of crimes solved.
06

Quota Problem with 90% Certainty

We need to find the minimum number \(n\) such that the probability of solving at least one crime is 90%:\[ 1 - 0.8^n \geq 0.9 \]\[ 0.8^n \leq 0.1 \]Take the natural logarithm on both sides and solve for \(n\):\[ n \cdot \ln(0.8) \leq \ln(0.1) \]\[ n \geq \frac{\ln(0.1)}{\ln(0.8)} \]Calculate the result to find the minimum number of cases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculations
When dealing with probability calculations, particularly with binomial distributions, it's essential to understand the fundamental concept of probability— which is a measure of the likelihood of an event. In the given scenario, solving a crime is considered a successful event. To calculate different probabilities, such as none of the crimes being solved, you consider the complementary event, which is the crime not being solved. Given that the probability of not solving a crime is 0.8 (or 80%), the probability of none of the crimes (six crimes in this context) being solved is found by multiplying the probability of not solving a crime for each individual crime: - Formula: \[ P(\text{none solved}) = (0.8)^6 \] - Here, \(0.8^6\) gives us the probability that all six crimes are unsolved. This complements step-by-step approaches with simple multiplication rules suited for binomial distribution contexts.Moreover, another common probability calculation is finding the chance that at least one crime is solved. This is determined using the complement rule, which simplifies complex probability problems.
Expected Value
The expected value in probability and statistics gives us a measure of the central tendency or average outcome of a random process. For a binomial distribution, which is perfect for situations like ours involving crime solving, calculating the expected value helps in predicting the average number of successfully solved crimes. The formula used is: - \[ E(X) = n \cdot p \] - Where \(n\) is the number of trials (in this case, crimes to solve), and \(p\) is the probability of solving a crime (0.2).In our scenario: - \(E(X) = 6 \cdot 0.2 = 1.2\) This calculation suggests that, on average, out of six crimes, around 1.2 are expected to be solved. Expected values help in making informed assumptions and forming strategies.
Standard Deviation
Standard deviation is critical in understanding the variability or dispersion of a dataset from its mean or expected value. For a binomial distribution, standard deviation offers insights into how much the number of crimes solved may differ from the expected value.The formula for calculating standard deviation in a binomial distribution is: - \[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \] Breaking down the formula:- \(n\) is the total number of trials,- \(p\) is the probability of success,- \((1-p)\) accounts for the probability of failure.In our crime-solving context:- \(\sigma = \sqrt{6 \cdot 0.2 \cdot 0.8}\) Solving this gives a standard deviation that brings us closer to understanding the specific spread around the average number of crimes solved, signaling the likely range of variability.
Complement Rule
The complement rule is a powerful tool in probability calculations that simplifies complex problems by considering what is not happening instead of what is happening. Essentially, the rule states that the probability of an event occurring is 1 minus the probability of it not occurring.In binomial probabilities, such as solving crimes, using the complement rule helps find the probability of at least one crime being solved: - If the probability of none being solved is, say, \(0.8^6\), the probability of at least one being solved is: - \[ P(\text{at least one solved}) = 1 - P(\text{none solved}) \] Applying this: - \[ P(\text{at least one solved}) = 1 - 0.8^6 \] This highlights how the complement rule aids in finding probabilities efficiently and reduces complex processes to manageable ones, which is essential in statistical reasoning.

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Most popular questions from this chapter

Lost Bags USA Today reported that for all airlines, the number of lost bags was May: \(6.02\) per 1000 passengers \(\quad\) December: \(12.78\) per 1000 passengers Note: A passenger could lose more than one bag. (a) Let \(r=\) number of bags lost per 1000 passengers in May. Explain why the Poisson distribution would be a good choice for the random variable \(r\). What is \(\lambda\) to the nearest tenth? (b) In the month of May, what is the probability that out of 1000 passengers, no bags are lost? that 3 or more bags are lost? that 6 or more bags are lost? (c) In the month of December, what is the probability that out of 1000 passengers, no bags are lost? that 6 or more bags are lost? that 12 or more bags are lost? (Round \(\lambda\) to the nearest whole number.)

The Honolulu Advertiser stated that in Honolulu there was an average of 661 burglaries per 100,000 households in a given year. In the Kohola Drive neighborhood there are 316 homes. Let \(r=\) number of these homes that will be burglarized in a year. (a) Explain why the Poisson approximation to the binomial would be a good choice for the random variable \(r .\) What is \(n\) ? What is \(p ?\) What is \(\lambda\) to the nearest tenth? (b) What is the probability that there will be no burglaries this year in the Kohola Drive neighborhood? (c) What is the probability that there will be no more than one burglary in the Kohola Drive neighborhood? (d) What is the probability that there will be two or more burglaries in the Kohola Drive neighborhood?

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What does the random variable for a binomial experiment of \(n\) trials measure?

The Denver Post reported that a recent audit of Los Angeles 911 calls showed that \(85 \%\) were not emergencies. Suppose the 911 operators in Los Angeles have just received four calls. (a) What is the probability that all four calls are, in fact, emergencies? (b) What is the probability that three or more calls are not emergencies? (c) Quota Problem How many calls \(n\) would the 911 operators need to answer to be \(96 \%\) (or more) sure that at least one call is, in fact, an emergency?

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