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Chapter 4: Problem 28

Fifty spotlights have just been installed in an outdoor security system. According to the manufacturer's specifications, these particular lights are expected to burn out at the rate of \(1.1\) per one hundred hours. What is the expected number of bulbs that will fail to last for at least seventy-five hours?

Short Answer

Expert verified
The expected number of bulbs that will fail to last for at least seventy-five hours is approximately 41.

Step by step solution

01

Calculate the Expected Rate of Failure for Seventy-Five Hours

The rate of failure given is for 100 hours. We need to find the rate for 75 hours. To accomplish this, we can use a proportion: if \(1.1\) bulbs fail per \(100\) hours, we can denote \(x\) as the number of bulbs that would fail per \(75\) hours. Hence, we get this proportion equation: \(1.1/100 = x/75\). Solving for \(x\), gives \(x = (1.1 * 75) / 100 = 0.825\). So the expected failure rate for 75 hours is \(0.825\) bulbs.
02

Apply the Poisson Distribution

The Poisson distribution is useful for calculating the probability of a number of events in a fixed interval of time or space. The formula for the Poisson distribution is: \( P(X = k) = (位^k * e^{-位}) / k! \), where 位 is the expected number of occurrences, k is the actual number of occurrences, and e is the base of the natural logarithm. Applying the Poisson distribution to 50 bulbs, we would have 位 = 50*0.825 = 41.25.
03

Calculate the Expected Number of Bulbs Failing

The expected value E(X) of a Poisson random variable X (which is the number of bulbs failing) is equal to the parameter 位. So, the expected number of bulbs that will fail in 75 hours is 41.25.

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Key Concepts

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

Expected Rate of Failure
In any statistical analysis involving probabilities of failures, the expected rate of failure is a crucial concept. For this specific exercise, we start with the manufacturer's failure rate, which is provided per 100 hours.To find out the expected rate of failure for a different time period, here 75 hours, we use a simple proportion equation.
  • We know that 1.1 bulbs are expected to fail per 100 hours.
  • We set up a proportion to find the failure rate for 75 hours: \( \frac{1.1}{100} = \frac{x}{75} \).
Solving for \(x\) involves basic arithmetic:\[x = \frac{1.1 \times 75}{100} = 0.825\]This means, approximately 0.825 bulbs are expected to fail every 75 hours.Using proportions allows us to easily determine expected rates over shorter or different time periods by scaling down the larger intervals provided by manufacturers.
Outdoor Security System
Outdoor security systems, such as the one described here, heavily rely on consistent performance from their components. The system mentioned consists of fifty spotlights. The longevity and reliability of each spotlight directly impact the overall effectiveness of the security system.
  • Having a grasp on how often these lights might fail helps in planning maintenance and replacements.
  • It enables security system managers to predict when and how often to check the systems to ensure they're always operational.
  • This minimizes unexpected dark spots that might compromise the security.
Understanding the expected rate of failure helps in the logistical management of such systems, ensuring that they serve their intended purpose without frequent outages.
Proportion Equation
The use of proportion equations is an essential mathematical tool in calculating expected outcomes based on known rates. In this exercise, proportions help bridge the manufacturer's failure rate over 100 hours to our required time span of 75 hours.The process involves:
  • Setting up a fraction representing the known rate of failure (per 100 hours),
  • Equaling it to a fraction representing the unknown failure rate for 75 hours.
For example:\[\frac{1.1}{100} = \frac{x}{75}\]Solving for \(x\) involves cross-multiplying and dividing, which yields:\[x = \frac{1.1 \times 75}{100} = 0.825\]This scenario illustrates how versatile proportion equations can be, as they allow for flexible adjustments in expected rates across various timescales.

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Most popular questions from this chapter

A door-to-door encyclopedia salesperson is required to document five in-home visits each day. Suppose that she has a \(30 \%\) chance of being invited into any given home, with each address representing an independent trial. What is the probability that she requires fewer than eight houses to achieve her fifth success?

Suppose that commercial airplane crashes in a certain country occur at the rate of \(2.5\) per year. (a) Is it reasonable to assume that such crashes are Poisson events? Explain. (b) What is the probability that four or more crashes will occur next year? (c) What is the probability that the next two crashes will occur within three months of one another?

Let \(X\) be the amount won or lost in betting \(\$ 5\) on red in roulette. Then \(p_{x}(5)=\frac{18}{38}\) and \(p_{x}(-5)=\frac{20}{38}\). If a gambler bets on red one hundred times, use the Central Limit Theorem to estimate the probability that those wagers result in less than \(\$ 50\) in losses.

Suppose \(X_{1}, X_{2}, X_{3}\), and \(X_{4}\) are independent Poisson random variables, each with parameter \(\lambda=3\). Let \(S=X_{1}+X_{2}+X_{3}+X_{4}\) (a) Use the Central Limit Theorem to approximate the probability that \(13 \leq S \leq 14\). (b) Calculate the exact probability that \(13 \leq S \leq 14\).

The IQs of nine randomly selected people are recorded. Let \(\bar{Y}\) denote their average. Assuming the distribution from which the \(Y_{i}\) 's were drawn is normal with a mean of 100 and a standard deviation of 16 , what is the probability that \(\bar{Y}\) will exceed 103 ? What is the probability that any arbitrary \(Y_{i}\) will exceed 103 ? What is the probability that exactly three of the \(Y_{i}\) 's will exceed 103 ?
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