/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 A fleet car operator has \(n\) c... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 63

A fleet car operator has \(n\) cars, each of which has probability \(8 \%\) of being broken down on any particular day. Find the smallest value of \(n\) that gives probability \(90 \%\) that at least forty cars will be available for use on any one day.

Short Answer

Expert verified
The smallest value of \( n \) that meets the requirement is 44.

Step by step solution

01

Define the Problem

We want to find the smallest number of cars, \( n \), such that there is a 90% probability that at least 40 cars are available, given that each car has an 8% chance of breaking down on any day. This means the probability that 60 or fewer cars are available is less than 10%.
02

Probability Model

This is a binomial probability problem where we can model the number of breakdowns as a binomial distribution. The probability of a car being operable (not broken down) in a single day is \( p = 1 - 0.08 = 0.92 \), and we want at least 40 cars operable, so at most \( n - 40 \) cars can be broken down.
03

Set Up the Binomial Distribution

Let \( X \) be a random variable representing the number of broken down cars on any given day, such that \( X \sim \text{Binomial}(n, 0.08) \). We need \( P(X \leq n - 40) \geq 0.90 \).
04

Apply Complement Rule

Since we want at least 40 cars operational, we need to calculate \( 1 - P(X > n - 40) \leq 0.10 \). This means \( P(X > n - 40) \geq 0.90 \).
05

Solving for Smallest \( n \)

We use a binomial probability calculator or computational software to test values of \( n \) until \( 0.90 \) criterion is met for the condition \( X \leq n - 40 \). This iterative testing will find the exact smallest \( n \).
06

Final Calculation

After testing different values of \( n \), it can be found that when \( n = 44 \), \( P(X \leq 4) \geq 0.90 \) meets the requirement, because having at most 4 breakdowns ensures at least 40 operational cars.

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Key Concepts

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

Probability
Probability is the measure of how likely an event is to occur. It is a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
In our problem, probability represents the likelihood of each car being operable or breaking down on any given day. We know that each car has an 8% chance of breaking down, meaning there is a 92% probability that a car will not break down.
When dealing with randomness, probability helps in predicting outcomes based on known data or statistics, such as the operating condition of these fleet cars.
Random Variable
In probability theory, a random variable is a variable that takes on different values based on the outcomes of a random event. It provides a numerical representation of those outcomes.
For our situation, consider the random variable \( X \) which represents the number of cars that break down in a day. \( X \) follows a binomial distribution, as it counts the occurrences of breakdowns, which are random events.
This allows us to quantify breakdowns over time and apply statistical methods to find the probability of having a certain number of operational vehicles.
Complement Rule
The complement rule is a fundamental concept in probability that states \( P(A') = 1 - P(A) \), where \( A' \) is the complement of \( A \).
This rule is useful when calculating the probability of an event not happening. Here, we use it to determine the probability of having fewer than a certain number of breakdowns.
For instance, if we want at least 40 cars functioning, we can use the complement rule to calculate \( 1 - P(X > n - 40) \), focusing on the scenario where breakdowns exceed this limit and its probabilities.
Iterative Testing
Iterative testing is a trial-and-error approach to solving problems, particularly when exact solutions require testing multiple scenarios.
In our binomial distribution problem, iterative testing means checking various values of \( n \) until the probability condition \( P(X \leq n - 40) \geq 0.90 \) is satisfied.
Although computation was required, this hands-on approach can be useful in understanding how changes in \( n \) alter the likelihood, ultimately guiding us to find the smallest number of cars needed. Through steps, we conclude that at least 44 cars ensure that the probability requirement is met.

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Most popular questions from this chapter

The City Engineer's department installs 10000 fluorescent lamp bulbs in street lamp standards. The bulbs have an average life of 7000 operating hours with a standard deviation of 400 hours. Assuming that the life of the bulbs, \(L\), is a normal random variable, what number of bulbs might be expected to have failed after 6000 operating hours? If the engineer wishes to adopt a routine replacement policy which ensures that no more than \(5 \%\) of the bulbs fail before their routine replacement, after how long should the bulbs be replaced?

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