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91Ó°ÊÓ

According to GMAC Insurance. \(20 \%\) of drivers aged \(60-65\) fail the written drivers' test. This is the lowest failure rate of any age group. (Source: http://www.gmacinsurance.com/SafeDriving/PressRelease.asp) If 200 people aged \(60-65\) independently take the exam, how many would you expect to pass? Give or take how many?

Short Answer

Expert verified
Expect approximately 160 people aged between 60 and 65 to pass the test, give or take around 8 people.

Step by step solution

01

Calculate Expected Pass

First, calculate the expected number of people who will pass the test. \(Expected \: value = Total \: number \: of \: people \: taking \: the \: exam \times Pass \: rate\). Substituting the given values, \(Expected \: value = 200 \times 80\% = 160\). Therefore, expect approximately 160 people to pass.
02

Calculate Standard Deviation

Calculate the standard deviation, which provides the 'give or take' value. The formula for the standard deviation is \(\sqrt{n \times p \times (1-p)}\). Substituting the values into the formula, \(Standard \: Deviation = \sqrt{200 \times 0.8 \times (1 - 0.8)} = 8\). Therefore, the standard deviation is approximately 8.
03

FInd the 'give or take' value

The 'give or take' value will be the calculated standard deviation. It's the measure of the amount of variation or dispersion there is in a set of values. Here, the 'give or take' value would be approximately 8.

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Key Concepts

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

Statistical Expectation
The concept of statistical expectation, also known as the expected value, is a fundamental idea in probability and statistics. It represents the average outcome you would expect from a random process if it were repeated many times. In the exercise provided, the expected number of people to pass a driving test is determined, given a certain probability of passing. To calculate this, we take the total number of individuals (200) and multiply it by the passing rate (80%), resulting in an expected 160 people passing the test.

Understanding expected value is crucial as it offers a prediction about the long-term behavior of the process being studied. In real-life scenarios, this concept could be used for a wide range of applications, from insurance to finance, to help make informed decisions based on probable outcomes.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values. It indicates how much individual data points in a statistical distribution deviate from the mean, or expected value. In the context of the exercise, the standard deviation tells us how much deviation we can expect from the 160 people predicted to pass the driving test. The calculation of a standard deviation involves the square root of the product of the sample size (200), the probability of success (0.8), and the probability of failure (0.2). The resulting standard deviation (8) indicates the 'give or take' value, which means we can reasonably expect the number of people passing to fall within 8 people above or below the expected value.

A low standard deviation would mean that data points are closely clustered around the mean, while a high standard deviation would indicate that data points are spread out over a wider range of values. Understanding standard deviation allows us to assess risks and make predictions with more confidence.
Probability Calculations
Probability calculations are integral to determining outcomes in statistics. These calculations involve determining the likelihood of various outcomes for an event. In our exercise, knowing that 20% of individuals fail the test enables us to deduce that there is an 80% chance of any given individual passing. Such calculations are the bedrock of expected value and standard deviation computations.

To calculate the probability of multiple independent events, like 200 people taking the test, statisticians multiply the individual probabilities. These principles underpin many areas of study and practical applications, including games of chance, predicting weather patterns, or estimating the success rates of medical procedures. By mastering probability calculations, students gain insight into how likely certain events are to occur, which is a powerful tool in decision-making processes.

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Most popular questions from this chapter

The homicide clearance rate in Florida is \(60 \%\). A crime is cleared when an arrest is made, a crime is charged, and the case is referred to a court. a. What is the probability that exactly 6 out of 10 independent homicides are cleared? b. Without doing a calculation, state whether the probability that 6 or more out of 10 homicides are cleared will be larger or smaller than the answer to part a? Why? c. What is the probability that 6 or more out of 10 independent homicides are cleared?

The average birth weight of elephants is 230 pounds. Assume that the distribution of birth weights is Normal with a standard deviation of 50 pounds. Find the birth weight of elephants at the 95 th percentile.

A study of U.S. births published on the website Medscape from WebMD reported that the average birth length of babies was \(20.5\) inches and the standard deviation was about \(0.90\) inch. Assume the distribution is approximately Normal. Find the percentage of babies with birth lengths of 22 inches or less.

The Normal model \(N(65,2.5)\) describes the distribution of heights of college women (inches). Which of the following questions asks for a probability and which asks for a measurement (and is thus an inverse Normal question)? a. What is the probability that a random college woman has a height of 68 inches or more? b. To be in the Tall Club, a woman must have a height such that only \(2 \%\) of women are taller. What is this height?

College women have heights with the following distribution (inches): \(N(65,2.5)\). a. Find the height at the 75 th percentile. b. Find the height at the 25 th percentile. c. Find the interquartile range for heights. d. Is the interquartile range larger or smaller than the standard deviation? Explain.

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