/*! 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 19 A contractor is required by a co... [FREE SOLUTION] | 91Ó°ÊÓ

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A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) when applying for a building permit. Let \(y\) be the number of forms required of the next applicant. Suppose the probability that \(y\) forms are required is known to be proportional to \(y ;\) that is, \(p(y)=\) \(k y\) for \(y=1, \ldots, 5\) a. What is the value of \(k ?\) (Hint: \(\Sigma p(y)=1 .)\) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required?

Short Answer

Expert verified
The value of \(k\) is \(\frac{1}{15}\). The probability that at most three forms are required is \(\frac{6}{15}\), and the probability that between two and four forms are required is \(\frac{9}{15}\).

Step by step solution

01

Calculate the total probability

From the problem, we know that the probability of \(y\) forms is proportional to \(y\), so we can write the pmf as: \[p(y) = ky\] for \(y = 1, 2, 3, 4,\) or \(5\) and \(k\) is a constant. Since the sum of all probabilities must equal 1, we have: \[\sum_{y=1}^5 p(y) = 1\] Substituting \(p(y) = ky\), we obtain: \[k \sum_{y=1}^5 y = 1\]
02

Find the value of \(k\)

Now, we need to find the value of \(k\). We first evaluate the sum: \[\sum_{y=1}^5 y = 1+2+3+4+5= 15\] Using the formula, we have: \[k(15) = 1\] So, \(k=\frac{1}{15}\). This means our pmf is: \[p(y) = \frac{y}{15}\] for \(y=1, 2, 3, 4\) or \(5\).
03

Probability at most three forms are required

We are asked to find the probability that at most three forms are required, which is the sum of the probabilities for \(y=1, 2\) or \(3\). Using our pmf, we have: \[p(y\leq3) = p(1) + p(2) + p(3) = \frac{1}{15} + \frac{2}{15} + \frac{3}{15} = \frac{6}{15}\] So, the probability that at most three forms are required is \(\frac{6}{15}\).
04

Probability between two and four forms are required

We need to find the probability that between 2 and 4 forms are required, which is the sum of the probabilities for \(y=2, 3\) or \(4\). Using our pmf, we have: \[p(2\leq y\leq4) = p(2) + p(3) + p(4) = \frac{2}{15} + \frac{3}{15} + \frac{4}{15} = \frac{9}{15}\] So, the probability that between two and four forms are required is \(\frac{9}{15}\). To summarize, the value of \(k\) is \(\frac{1}{15}\), the probability that at most three forms are required is \(\frac{6}{15}\), and the probability that between two and four forms are required is \(\frac{9}{15}\).

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Key Concepts

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

Discrete Probability Distribution
When it comes to understanding probability, it's essential to start with the fundamentals, one of which is the discrete probability distribution. This concept describes a scenario where the outcomes are discrete, quantifiable, and countable, just like the number of forms required by a county planning department in our exercise. A discrete probability distribution is mathematically represented by the probability mass function (pmf), which assigns probabilities to each possible outcome of a discrete random variable.For instance, if we know that a certain discrete event can happen in five different ways, and the likeliness of each way is proportional to a constant multiple, we have a scenario described by a probability mass function. In the building permit example, the pmf is expressed as \(p(y) = ky\), where \(y\) represents the number of forms, and \(k\) is the constant of proportionality.
  • We ensure the total probability across all possible outcomes sums to 1 (a fundamental principle of probability).
  • From this, we can calculate the constant \(k\), giving us the specific pmf for the situation at hand.
Discrete probability distributions help us understand and forecast the likeliness of various outcomes in a quantified way, which is crucial in fields such as statistics, economics, and engineering.
Conditional Probability
Moving deeper into the probability terrain, we encounter conditional probability—a concept illustrating how the probability of an event can be affected by the occurrence of a previous event. Think of it as updating our expectations based on new evidence or information. Conditional probability is written as \(P(A|B)\), meaning 'the probability of event A occurring given that event B has occurred'.

Application in Real-Life Scenarios

In real life, you might use conditional probability to adapt your expectations. For example, if you know that it rains 30% of the time and that whenever it rains, your bus is late 50% of the time, you could calculate the probability of your bus being late on a rainy day (that's conditional probability in action).

Connection to the Exercise

Although our building permit exercise doesn't explicitly involve conditional probability, understanding this concept is crucial when we deal with more complex probability questions, where outcomes are dependent on prior events. As you delve into more advanced probability problems, you'll frequently need to figure out the chances of something happening based on other variables or conditions that have already taken place.
Random Variable
Central to the study of probability is the concept of a random variable. This is a numerical description of the outcome of a statistical experiment. Random variables can be classified into two types: discrete and continuous. In our exercise example, the number of forms (\(y\)) required by a contractor is a discrete random variable because it can take on a limited number of distinct and separate values (1, 2, 3, 4, or 5).
  • Discrete random variables are those that have countable outcomes, like the flip of a coin (Heads or Tails), or the roll of a die (1 to 6).
  • Continuous random variables, on the other hand, can take on an infinite number of possibilities, such as the exact time it takes for someone to run a mile.
The random variable is a cornerstone in probability and statistics because it connects real-world phenomena to the mathematical framework of probability theory, allowing us to quantify and model uncertainty. Thus, random variables serve as the bridge between theory and practice, enabling statisticians and data scientists to create predictions and make informed decisions based on probable outcomes.

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Most popular questions from this chapter

The light bulbs used to provide exterior lighting for a large office building have an average lifetime of 700 hours. If lifetime is approximately normally distributed with a standard deviation of 50 hours, how often should all the bulbs be replaced so that no more than \(20 \%\) of the bulbs will have already burned out?

A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well approximated by a normal distribution with mean \(3 \mathrm{~cm}\) and standard deviation \(0.1 \mathrm{~cm} .\) The specifications call for corks with diameters between 2.9 and \(3.1 \mathrm{~cm}\). A cork not meeting the specifications is considered defective. (A cork that is too small leaks and causes the wine to deteriorate; a cork that is too large doesn't fit in the bottle.) What proportion of corks produced by this machine are defective?

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Only \(20 \%\) of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the approximate probability that at most 70 mufflers are replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were replaced under warranty, would you question the \(20 \%\) figure? Explain.

Determine the following standard normal (z) curve areas: a. The area under the \(z\) curve to the left of 1.75 b. The area under the \(z\) curve to the left of -0.68 c. The area under the \(z\) curve to the right of 1.20 d. The area under the \(z\) curve to the right of -2.82 e. The area under the \(z\) curve between -2.22 and 0.53 f. The area under the \(z\) curve between -1 and 1 g. The area under the \(z\) curve between -4 and 4

Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1\). Determine each of the following probabilities: a. \(P(z<0.10)\) b. \(P(z<-0.10)\) c. \(P(0.40-1.25)\) g. \(P(z<-1.50\) or \(z>2.50)\)

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