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Public Safety: 911 Calls 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?

Short Answer

Expert verified
(a) 0.00051 (b) 0.00536 (c) At least 19 calls are needed.

Step by step solution

01

Identify key information and set up probabilities

Given that 85% of the 911 calls are not emergencies, the probability that a call *is* an emergency is 15%, or 0.15. We let the probability of an emergency be denoted as \( p = 0.15 \) and the probability of non-emergency calls as \( 1 - p = 0.85 \).
02

Calculate probability that all four calls are emergencies

To find the probability that all four calls are emergencies, calculate \( p^4 \) because each call is independent: \[ (0.15)^4 = 0.15 \times 0.15 \times 0.15 \times 0.15 = 0.00050625 \] Thus, the probability that all four calls are emergencies is 0.00050625.
03

Calculate probability that three or more calls are not emergencies

First, calculate the probability that zero, one, or two calls are emergencies and subtract from 1.- Zero emergencies: \( (0.85)^4 = 0.52200625 \)- One emergency: \( 4 \times (0.15)^1 \times (0.85)^3 = 0.3992625 \)- Two emergencies: \( \binom{4}{2} \times (0.15)^2 \times (0.85)^2 = 0.084375 \)The probability of three or more calls being non-emergencies is: \[ 1 - (0.52200625 + 0.3992625 + 0.084375) = 0.00535625 \]
04

Determine number of calls for 96% certainty of at least one emergency

We want to find \( n \) such that the probability of at least one call being an emergency is at least 0.96. The complementary probability that all calls are not emergencies is \( (0.85)^n \).Set \[ 1 - (0.85)^n \geq 0.96 \]Rearrange to \( (0.85)^n \leq 0.04 \).Solve for \( n \): - Use logarithms: - \( n \log(0.85) \leq \log(0.04) \), - \( n \geq \frac{\log(0.04)}{\log(0.85)} \), - approximating, \( n \geq 18.43 \)Thus, at least 19 calls are needed for a 96% certainty of at least one emergency.

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Key Concepts

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

Probability
Probability is the branch of mathematics that deals with the likelihood of different outcomes. When we say something is probable, we mean that there's a chance it can happen, but there's also a chance it might not. In our exercise, we look at 911 calls and use probability to figure out certain outcomes related to emergencies.

To understand the probability of an event, we divide the number of desired outcomes by the total number of possible outcomes. In the context of 911 calls, probability helps to determine the likelihood that any given call received is or isn't an emergency.
  • For example, if the probability of a call being an emergency is 15%, we denote this as: \( p = 0.15 \)
  • Conversely, the probability that the call is not an emergency is 85%, or \( 1 - p = 0.85 \)
Understanding these basics helps operators prepare for how many calls might be true emergencies.
Emergency Calls
Emergency calls, like those to 911, are crucial for ensuring public safety. In this exercise, the focus is on the characterization of incoming calls and determining which are actual emergencies versus non-emergencies.

It's important for operators to quickly and accurately assess these calls to provide appropriate responses. Based on statistical data, 85% of the 911 calls made in Los Angeles are not true emergencies. This statistic supports the need for effective call management systems that can prioritize genuine emergencies efficiently.
  • Operators rely on these statistics to make decisions about resource allocation.
  • Understanding the frequency and probability of emergency versus non-emergency calls helps in training and protocol development.
By examining probabilities, operators can anticipate call outcomes and respond more effectively.
Statistical Analysis
Statistical analysis is a critical tool in interpreting data about 911 calls, as seen in our problem. By analyzing statistics, we can make informed decisions about public safety and resource allocation. In this exercise, statistical analysis was used to solve several questions about emergency call probabilities.

  • The first analysis involved calculating the probability that all four 911 calls were emergencies, showing us how rare this possibility is, given the current statistics.
  • Secondly, we calculated the chances of three or more calls being non-emergencies. This analysis helps in anticipating the usual call load, which isn't pressing.
These statistical insights are powerful for developing strategies to improve efficiency and response times in emergency services. It involves using logical steps to derive useful information about the likelihood and frequency of different scenarios.
Binomial Distribution
The problem also explores the concept of binomial distribution, which is a probability distribution that summarizes the likelihood that a variable will take one of two independent values under a given set of parameters or conditions. In our 911 call scenario, each call is an independent event that is either an emergency or not.

The binomial distribution is applied here because it fits the characteristic of having two possible outcomes per each 911 call. To apply this:
  • We count the total number of calls, \( n \), as trials.
  • Each call is independently considered as either an emergency (success) or not (failure).
When assessing the situation where operators need a 96% certainty of having at least one emergency call, the formula for binomial distribution allows us to calculate how many calls, \( n \), are necessary. It's crucial for planning and ensures that resources are allocated efficiently based on likely outcomes.

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Most popular questions from this chapter

Law Enforcement: Officers Killed Chances: Risk and Odds in Everyday Life, by James Burke, reports that the probability a police officer will be killed in the line of duty is \(0.5 \%\) (or less). (a) In a police precinct with 175 officers, let \(r=\) number of police officers killed in the line of duty. 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 no officer in this precinct will be killed in the line of duty? (c) What is the probability that one or more officers in this precinct will be killed in the line of duty? (d) What is the probability that two or more officers in this precinct will be killed in the line of duty?

Critical Thinking Let \(r\) be a binomial random variable representing the number of successes out of \(n\) trials. (a) Explain why the sample space for \(r\) consists of the set \(\\{0,1,2, \ldots, n\\}\) and why the sum of the probabilities of all the entries in the entire sample space must be 1 . (b) Explain why \(P(r \geq 1)=1-P(0)\). (c) Explain why \(P(r \geq 2)=1-P(0)-P(1)\). (d) Explain why \(P(r \geq m)=1-P() 0-P(1)-\cdots-P(m-1)\) for \(1 \leq m \leq n\).

Statistical Literacy Consider each distribution. Determine if it is a valid probability distribution or not, and explain your answer. (a) \begin{tabular}{l|ccc} \hline\(x\) & 0 & 1 & 2 \\ \hline\(P(x)\) & \(0.25\) & \(0.60\) & \(0.15\) \\ \hline \end{tabular} (b) \begin{tabular}{c|ccc} \hline \(\mathbf{x}\) & 0 & 1 & 2 \\ \hline\(P(x)\) & \(0.25\) & \(0.60\) & \(0.20\) \\ \hline \end{tabular}

Statistical Literacy For a binomial experiment, how many outcomes are possible for each trial? What are the possible outcomes?

Basic Computation: Binomial Distribution Consider a binomial experiment with \(n=7\) trials where the probability of success on a single trial is \(p=0.60\). (a) Find \(P(r=7)\). (b) Find \(P(r \leq 6)\) by using the complement rule.

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