/*! 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 52 The thickness of wood paneling (... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The thickness of wood paneling (in inches) that a customer orders is a random variable with the following cumulative distribution function: $$ F(x)=\left\\{\begin{array}{lr} 0 & x<1 / 8 \\ 0.2 & 1 / 8 \leq x<1 / 4 \\ 0.9 & 1 / 4 \leq x<3 / 8 \\ 1 & 3 / 8 \leq x \end{array}\right. $$ Determine the following probabilities: (a) \(P(X \leq 1 / 18)\) (b) \(P(X \leq 1 / 4)\) (c) \(P(X \leq 5 / 16)\) (d) \(P(X>1 / 4)\) (e) \(P(X \leq 1 / 2)\)

Short Answer

Expert verified
(a) 0, (b) 0.9, (c) 0.9, (d) 0.1, (e) 1

Step by step solution

01

Understanding the Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) describes the probability that a random variable, such as the thickness of wood paneling in this problem, is less than or equal to a certain value. By examining how the CDF is defined in intervals, we understand which probabilities go with which thickness. For each of these questions, we will determine the appropriate CDF value based on the intervals given.
02

Evaluating Probability for (a) P(X \leq 1/18)

For the value \(x = 1/18\), which is less than 1/8, the CDF \(F(x) = 0\). Therefore, we have \(P(X \leq 1/18) = 0\).
03

Evaluating Probability for (b) P(X \leq 1/4)

At \(x = 1/4\), the CDF increases to 0.9, as this lies in the interval \(1/4 \leq x < 3/8\). Hence, \(P(X \leq 1/4) = 0.9\).
04

Evaluating Probability for (c) P(X \leq 5/16)

The value \(x = 5/16\) is contained within the interval \(1/4 \leq x < 3/8\), and the CDF defines it as \(F(x) = 0.9\). Therefore, \(P(X \leq 5/16) = 0.9\).
05

Calculating Complementary Probability for (d) P(X > 1/4)

To find \(P(X > 1/4)\), note that \(P(X > 1/4) = 1 - P(X \leq 1/4)\). Since \(P(X \leq 1/4) = 0.9\), we have \(P(X > 1/4) = 1 - 0.9 = 0.1\).
06

Evaluating Probability for (e) P(X \leq 1/2)

For \(x = 1/2\), \(x\) is greater than 3/8. Hence, according to the CDF definition \(F(x) = 1\) for \(x \geq 3/8\). Therefore, \(P(X \leq 1/2) = 1\).

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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. Imagine your random variable as a numerical label assigned to every outcome you can think of. In the context of the example, the thickness of the wood paneling is a random variable.

It's called "random" because each thickness has a certain probability of occurring. Consider it like rolling a dice: each face (or number) represents a possible outcome, just as each thickness is a possible measurement. Random variables can be discrete, taking on specific, distinct values, or continuous, with values in an interval. The key value of a random variable is that it helps us calculate probabilities. In this exercise, the random variable allowed us to use the cumulative distribution function (CDF) to find probabilities of different thicknesses. Understanding how these variables interact within a probability distribution is crucial to grasp how likely certain outcomes are, such as the thickness being less than or equal to a specified measurement.
Probability Calculation
Probability calculation involves determining how likely an event is to occur. Here, we use the cumulative distribution function (CDF) to calculate such probabilities. The CDF tells us the probability that a random variable takes on a value less than or equal to a specific point. Think of it as a running total of probabilities up to that point.
  • In this exercise, determining probabilities for values like \(P(X \leq 1/18)\), requires reading the CDF intervals. For \(x = 1/18\) which is less than \(1/8\), \(F(x)=0\), so the probability is zero.
  • For values within intervals, such as \(x = 1/4\), which corresponds with \(0.9\), you simply match the intervals. \(P(X \leq 1/4)\) is thus 0.9.
  • Complementary probabilities like \(P(X > 1/4)\) require a slight twist: subtracting the CDF value from one, since \(P(X > 1/4) = 1 - P(X \leq 1/4)\).
  • Using a CDF is particularly useful because it's clear and straightforward to follow. By understanding the relevant CDF intervals, you can easily calculate probabilities for any needed values.
Step-by-Step Solution
Breaking down complex problems into manageable steps is a hallmark of probability exercises. To better understand how to approach probability questions using the CDF, we'll take an exploratory tour through a structured process:1. **Understanding the CDF:** Recognize that the CDF provides probability values linked to specific intervals of the random variable's potential outcomes. In this case, it tells us the inbound probability of wood panel thickness being at or below any given boundary.2. **Evaluating Specific Probabilities:** As you see with \(P(X \leq 1/4)\), determine which segment of the CDF applies and apply its probability value directly.3. **Applying Complementary Rules:** Further, when calculating \(P(X > 1/4)\), leverage the rule that \(P(X > a) = 1 - P(X \leq a)\). It's a neat trick that simplifies flipping the perspective.4. **Adapting to New Queries:** Each new probability query might require checking a different interval or even stitching together results, as seen with \(P(X \leq 1/2)\), consolidating across all values up to a certain point.Following these clear steps, you can enhance your understanding and tackle similar problems more independently. Mastering this technique doesn’t just solve particular exercises; it builds a versatile skill set for probability analysis.

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Most popular questions from this chapter

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