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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) 0.0005 (b) 0.8905 (c) 18 calls

Step by step solution

01

Define Variables for the Problem

Let the probability that a call is not an emergency be 0.85. Thus, the probability that a call is an emergency is \(1 - 0.85 = 0.15\). We will be using these probabilities for our calculations.
02

Calculate Probability for All Calls Being Emergencies (Part A)

To find the probability that all four calls are emergencies, we calculate \( (0.15)^4 \). This is because each call is independent, and we need all to be emergencies. Thus, \( P(4 \text{ emergencies}) = (0.15)^4 \).
03

Solve Part A Calculations

Calculate the value from Step 2: \((0.15)^4 = 0.15 \times 0.15 \times 0.15 \times 0.15 = 0.00050625.\) So, the probability is approximately 0.0005.
04

Calculate Probability for Three or More Non-emergencies (Part B)

To find the probability for three or more calls being non-emergencies, we consider two cases: exactly three non-emergencies and all four non-emergencies.
05

Probability of Exactly Three Non-emergencies

The probability of exactly three non-emergencies is given by the binomial formula: \( \binom{4}{3} (0.85)^3 (0.15) = 4 \times (0.85)^3 \times 0.15.\)
06

Solve for Exactly Three Non-emergencies

Calculate \((0.85)^3 \times 0.15 = 0.614125 \times 0.15 = 0.09211875.\) Multiply by 4, yielding \(4 \times 0.09211875 = 0.368475.\)
07

Probability of All Four Non-emergencies

The probability that all four calls are non-emergencies is \((0.85)^4 = 0.52200625.\)
08

Sum for Part B Total Probability

Add the probabilities of exactly three and all four non-emergencies: \(0.368475 + 0.52200625 = 0.89048125.\) Thus, the probability is approximately 0.8905.
09

Determine Calls Needed for Quota Problem (Part C)

We need to find \(n\) so that at least one call is an emergency with at least a 96% probability. This means we must solve \(1 - (0.85)^n \geq 0.96\).
10

Solve Part C Inequality

Rearrange to find \((0.85)^n \leq 0.04\). Use logarithms: \(n \log(0.85) \leq \log(0.04)\). Solving for \(n\), we get \(n \geq \frac{\log(0.04)}{\log(0.85)} \approx 17.782.\)
11

Round up to Solve Part C

Since \(n\) must be an integer, round up to 18. Therefore, 18 calls are needed to ensure there's a 96% probability at least one is an emergency.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a key concept in probability and statistics. It describes the number of successes in a fixed number of independent trials, each with the same probability of success. In this context, a "success" could be defined as an event of interest.
For instance, when handling 911 calls, a success might be identifying a call as an emergency. This situation fits well in a binomial distribution model since each call can either be an emergency or non-emergency and is statistically independent of others.
  • The total number of trials (calls, in this case) is fixed.
  • There are two possible outcomes for each trial: an emergency or not.
  • Each trial is independent, so the outcome of one does not affect the others.
  • The probability of success is constant (15% for an emergency, as given by the problem setup).
Understanding binomial distribution helps to solve problems like: calculating the probability that all calls in a set are emergencies or determining how many calls it takes to be confident that at least one is an emergency.
Statistics Problem
Statistics problems, like the one in the exercise, often involve calculating probabilities and making inferences based on data. These are essential skills in data analysis, helping us to make decisions under uncertainty.
In our exercise, there are several layers of probability calculations:
  • Calculating the probability of all calls being emergencies or non-emergencies.
  • Ensuring at least one call in a sequence is an emergency, which involves complex probability reasoning.
These probabilities help in understanding and predicting the behavior of random events. Grasping this statistical reasoning allows for better insight into real-life applications and enhances decision-making abilities. statistik problems also train analytical thinking by requiring step-by-step analysis and problem-solving skills.
Scenario Analysis
Scenario analysis involves exploring different possible outcomes in a given situation. It is a critical tool in risk assessment and decision making.
In the 911 call example:
  • Scenario 1: All four calls are emergencies. This is unlikely due to the low probability per call (15%), leading to a very low probability of 0.0005 for all being emergencies.
  • Scenario 2: Three or more calls are non-emergencies. This is more likely with a higher calculated probability of 0.8905.
  • Scenario 3: Ensuring at least one emergency call among many. Here, scenario analysis involves calculating how many calls need to be taken to achieve sufficient certainty (96%).
Scenario analysis like this involves comparing probabilities and consequences of different outcomes to guide strategic decisions or policies.
Random Variables
Random variables are a fundamental concept in probability that associate a numerical value with each outcome of a random process. In this exercise, the random variable is the number of emergency calls within a set of calls.
Random variables are important because they allow for:
  • Quantifying uncertain outcomes into measurable data.
  • Calculating probabilities for various outcomes in a statistic or probability problem.
In our example, we can define a random variable that represents whether each call is an emergency. The outcome (emergency or not) is then assigned a numerical value (1 for emergency and 0 for non-emergency).
The random variable hence helps in modeling and understanding the data, such as computing the probability of a certain number of emergency calls in a group of calls. Mastering random variables is crucial to effectively apply statistical methods in various contexts.

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Most popular questions from this chapter

USA Today reported that the U.S. (annual) birth rate is about 16 per 1000 people, and the death rate is about 8 per 1000 people. (a) Explain why the Poisson probability distribution would be a good choice for the random variable \(r=\) number of births (or deaths) for a community of a given population size.

Chances: Risk and Odds in Everyday Life, by James Burke, reports that only \(2 \%\) of all local franchises are business failures. A Colorado Springs shopping complex has 137 franchises (restaurants, print shops, convenience stores, hair salons, etc.). (a) Let \(r\) be the number of these franchises that are business failures. Explain why a Poisson approximation to the binomial would be appropriate for the random variable \(r\). What is \(n\) ? What is \(p ?\) What is \(\lambda\) (rounded to the nearest tenth)? (b) What is the probability that none of the franchises will be a business failure? (c) What is the probability that two or more franchises will be business failures? (d) What is the probability that four or more franchises will be business failures?

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?

A computer repair shop has two work centers. The first center examines the computer to see what is wrong, and the second center repairs the computer. Let \(x_{1}\) and \(x_{2}\) be random variables representing the lengths of time in minutes to examine a computer \(\left(x_{1}\right)\) and to repair a computer \(\left(x_{2}\right) .\) Assume \(x_{1}\) and \(x_{2}\) are independent random variables. Long-term history has shown the following times: Examine computer, \(x_{1}\) : \(\mu_{1}=28.1\) minutes; \(\sigma_{1}=8.2\) minutes Repair computer, \(x_{2}: \mu_{2}=90.5\) minutes; \(\sigma_{2}=15.2\) minutes (a) Let \(W=x_{1}+x_{2}\) be a random variable representing the total time to examine and repair the computer. Compute the mean, variance, and standard deviation of \(W\).

What does the expected value of a binomial distribution with \(n\) trials tell you?

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