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The thickness of wood paneling (in inches) that a customer orders is a random variable with the following cumulative distribution function: $$ F(x)=\left\\{\begin{array}{ll} 0 & x<1 / 8 \\ 0.2 & 1 / 8 \leq x<1 / 4 \\ 0.9 & 1 / 4 \leq x<3 / 8 \\ 1 & 3 / 8 \leq x \end{array}\right. $$ 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

The Cumulative Distribution Function (CDF) provided describes the probability that a random variable, the thickness of the wood paneling, takes on a value less than or equal to a certain value. The CDF is defined in piecewise sections, with specific probabilities for different ranges of thickness.
02

Calculating P(X ≤ 1/18)

Using the CDF, for any thickness less than 1/8 inch, the probability is given as 0. Therefore, \(P(X \leq 1/18) = 0\).
03

Calculating P(X ≤ 1/4)

For the range \(1/4 \leq x < 3/8\), the cumulative distribution function value is 0.9. Therefore, \(P(X \leq 1/4)\) corresponds to the lower boundary of this range, and the CDF value is \(0.9\).
04

Calculating P(X ≤ 5/16)

Since \(5/16\) is between \(1/4\) and \(3/8\), we use the range \(1/4 \leq x < 3/8\) where the CDF value is 0.9. Therefore, \(P(X \leq 5/16) = 0.9\).
05

Calculating P(X > 1/4)

The probability that \(X > 1/4\) can be found by calculating \(1 - P(X \leq 1/4)\). Since \(P(X \leq 1/4) = 0.9\), thus \(P(X > 1/4) = 1 - 0.9 = 0.1\).
06

Calculating P(X ≤ 1/2)

For any thickness \(x \geq 3/8\), the CDF value is 1. Since 1/2 is greater than 3/8, \(P(X \leq 1/2) = 1\).

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Key Concepts

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

Probability Calculations
Probability calculations are fundamental in understanding how likely an event is to happen. In the context of cumulative distribution functions, this often means determining the likelihood that a random variable falls within a certain range. Let's take event (a) from our exercise, where we calculate the probability that the thickness of wood paneling is less than or equal to 1/18 of an inch. Using the cumulative distribution function (CDF), since 1/18 is less than 1/8, we find that:
  • For all values less than 1/8, the probability is 0 according to the CDF.
Therefore, the probability, or likelihood, is 0.

Probability calculations, like those shown in steps (b) through (e), are essential in scenarios where conditions and ranges define potential outcomes. By using the CDF values directly from our piecewise function, we can quickly determine the probabilities over various intervals.

Key steps in such calculations include identifying the correct range and using the corresponding CDF value. This provides a swift method to calculate even complex probabilities in real world scenarios.
Random Variables
Random variables are central in probability theory and statistics. They are variables whose values result from the outcomes of a random phenomenon. In our problem, the thickness of the wood paneling is a random variable.

To fully understand this concept, imagine measuring the thickness each time an order is placed. Each measurement could vary slightly, representing different outcomes:
  • When we say the thickness is a random variable, we mean that it can take on different values, mostly influenced by random processes or inherent variability in production.
  • In probability, we express these variations using a distribution function, such as the CDF, which tells us the likelihood of the thickness being less than or equal to a specified value.
The use of a CDF illustrates how likely it is that the thickness falls within a certain range at any given time. This framework allows businesses, manufacturers, and researchers to account for variability and uncertainty during production and decision making.
Piecewise Functions
Piecewise functions are mathematical expressions defined by multiple sub-functions, each applying to a specific interval of the main function's domain. Our cumulative distribution function (CDF) is a clear example of a piecewise function.

Here's how it works:
  • The CDF switches between different constant values based on the range of the random variable, in our case, the thickness of the wood paneling.
  • For instance, if the thickness is between 1/8 and 1/4 inch, the CDF offers a probability of 0.2. This changes to 0.9 when the thickness ranges from 1/4 to 3/8 inches.
The importance of piecewise functions lies in their ability to model real-life situations with varying behaviors over different intervals effectively. They are particularly useful in representing scenarios where observations change abruptly or have distinct stages, as often seen in engineering, economics, and statistics.

Mastering piecewise functions is instrumental in understanding various complex phenomena that cannot be represented by a single mathematical expression across their entire domain.

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Most popular questions from this chapter

Messages that arrive at a service center for an information systems manufacturer have been classified on the basis of the number of keywords (used to help route messages) and the type of message, either e-mail or voice. Also, \(70 \%\) of the messages arrive via e-mail and the rest are voice. $$ \begin{array}{llllll} \text { number of keywords } & 0 & 1 & 2 & 3 & 4 \\ \text { e-mail } & 0.1 & 0.1 & 0.2 & 0.4 & 0.2 \\ \text { voice } & 0.3 & 0.4 & 0.2 & 0.1 & 0 \end{array} $$ Determine the probability mass function of the number of keywords in a message.

Each main bearing cap in an engine contains four bolts. The bolts are selected at random, without replacement, from a parts bin that contains 30 bolts from one supplier and 70 bolts from another. (a) What is the probability that a main bearing cap contains all bolts from the same supplier? (b) What is the probability that exactly three bolts are from the same supplier?

The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8 . Assume the trials are independent. (a) What is the probability that the first successful alignment requires exactly four trials? (b) What is the probability that the first successful alignment requires at most four trials? (c) What is the probability that the first successful alignment requires at least four trials?

Printed circuit cards are placed in a functional test after being populated with semiconductor chips. A lot contains 140 cards, and 20 are selected without replacement for functional testing. (a) If 20 cards are defective, what is the probability that at least 1 defective card is in the sample? (b) If 5 cards are defective, what is the probability that at least one defective card appears in the sample?

Derive the formula for the mean and standard deviation of a discrete uniform random variable over the range of integers \(a, a+1, \ldots, b\).

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