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A cereal manufacturer is aware that the weight of the product in the box varies slightly from box to box. In fact, considerable historical data has allowed the determination of the density function that describes the probability structure for the weight (in ounces). In fact, letting \(X\) be the random variable weight, in ounces, the density function can be described as $$ f(x)=\left\\{\begin{array}{ll} \frac{2}{5}, & 23.75 \leq x \leq 26.25, \\ 0, & \text { elsewhere.} \end{array}\right. $$ (a) Verify that this is a valid density function. (b) Determine the probability that the weight is smaller than 24 ounces. (c) The company desires that the weight exceeding 26 ounces is an extremely rare occurrence. What is the probability that this "rare occurrence" does actually occur?

Short Answer

Expert verified
The given function is indeed a valid density function. The probability that the weight is less than 24 ounces is \(P(X < 24) = \frac{2}{5} \times (24 - 23.75)\). The probability that a 'rare occurrence' (i.e., the weight exceeding 26 ounces) happens is \(P(X > 26) = \frac{2}{5} \times (26.25 - 26)\).

Step by step solution

01

Verify the Density Function

Verification of a probability density function (pdf) involves confirming that the area under the curve equals 1. Mathematically, this verification of the given density function \(f(x)\) can be achieved with the following integral: \(\int_{-\infty}^{\infty} f(x) dx = 1\). Given that \(f(x) = \frac{2}{5}\) within the interval [23.75,26.25] and 0 elsewhere, this definite integral can be simplified into: \(\int_{23.75}^{26.25} \frac{2}{5} dx\). Upon calculating this integral, if the result is indeed 1, the function \(f(x)\) is a valid pdf.
02

Calculate the Probability of Weight being Less Than 24 Ounces

The probability that the weight is less than 24 ounces is given by the integral from the lower limit, 23.75, to 24 of the density function. This can be computed as follows: \(P(X < 24) = \int_{23.75}^{24} f(x) dx = \int_{23.75}^{24} \frac{2}{5} dx\). The result gives the probability of the weight being less than 24 ounces.
03

Determine Probability of a 'Rare Occurrence'

The 'rare occurrence' represents the event of the weight exceeding 26 ounces. The probability of this can be computed with the integral from 26 to the upper limit, 26.25. Calculated as follows: \(P(X > 26) = \int_{26}^{26.25} f(x) dx = \int_{26}^{26.25} \frac{2}{5} dx\). The result gives the probability of the weight exceeding 26 ounces.

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Key Concepts

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

Random Variable
In probability theory, a random variable is a fundamental concept used to describe outcomes of a random phenomenon. In simpler terms, it's a variable that can take on various values, each with a certain probability.

Random variables can be classified into two types:
  • Discrete Random Variables: These take on a countable number of values, like the roll of a dice which can be any integer between 1 and 6.
  • Continuous Random Variables: These can take on any value within a given interval, such as the weight of a cereal box, which can vary continuously between certain limits.
In the context of our exercise, the weight of the product, denoted as \(X\), is an example of a continuous random variable. This means that \(X\) can take any value within a continuous range, specified by the given probability density function (pdf). The pdf, in our case, helps us understand how likely it is to observe a specific weight value within the range of 23.75 to 26.25 ounces. It's an essential part of modeling real-world situations in probability theory.
Integral Calculus
Integral calculus is a branch of mathematics concerned with the concept of integration, which is essential for solving area-related problems. In probability and statistics, integration is used to compute probabilities for continuous random variables.

Given a probability density function (pdf), the integral of the pdf over a specified interval yields the probability that the random variable falls within that interval.

Here’s how it works:
  • The integral of the pdf over its entire range must equal 1. This confirms that the pdf represents a valid probability distribution.
  • The integral over a specific interval gives the probability of finding the random variable within that range.
In our exercise, verifying a valid density function involves integrating \(f(x) = \frac{2}{5}\) over the interval [23.75, 26.25] and checking if it equals 1. To find specific probabilities, like being less than 24 ounces, we change the limits of integration, using \(\int_{23.75}^{24} \frac{2}{5} \ dx\). Integral calculus thus aids us in quantifying these probabilities mathematically.
Probability Theory
Probability theory is the branch of mathematics that deals with the analysis of random variables and events. It provides the foundation for predicting the outcomes of random processes and understanding their likelihood quantitatively.

Key components of probability theory include:
  • Probability Density Function (pdf): It describes the likelihood of a continuous random variable assuming particular values. Our given pdf for weight showcases how the likelihood of obtaining weights in the range of 23.75 to 26.25 ounces is distributed.
  • Cumulative Distribution Function (CDF): It represents the probability that a random variable is less than or equal to a certain value and is derived by integrating the pdf from the lower bound up to this value.
  • Rare Events: Probability theory allows us to determine how unlikely a certain event is, such as a cereal box weighing over 26 ounces in our example. Such calculations help in understanding and managing risk in practical applications.
By utilizing these probability concepts, we can accurately predict and manage expectations in scenarios ranging from everyday activities to complex scientific experiments. In the exercise, probability theory is crucial in analyzing and solving for the likelihood of various weight outcomes of cereal boxes.

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Most popular questions from this chapter

Suppose a special type of small data processing firm is so specialized that some have difficulty making a profit in their first year of operation. The pdf that characterizes the proportion \(Y\) that make a profit is given by $$ f(x)=\left\\{\begin{array}{ll} k y^{4}(1-y)^{3}+ & 0 \leq y \leq 1, \\ 0_{1} & \text { elsewhere. } \end{array}\right. $$ (a) What is the value of \(k\) that renders the above a valid density function? (b) Find the probability that at most \(50 \%\) of the firms make a profit in the first year. (c) Find the probability that at least \(80 \%\) of the firms make a profit in the first year.

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