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$$ F(x)=\left\\{\begin{array}{lr} 0 & x<-10 \\ 0.25 & -10 \leq x<30 \\ 0.75 & 30 \leq x<50 \\ 1 & 50 \leq x \end{array}\right. $$ (a) \(P(X \leq 50)\) (b) \(P(X \leq 40)\) (c) \(P(40 \leq X \leq 60)\) (d) \(P(X<0)\) (e) \(P(0 \leq X<10)\) (f) \(P(-10

Short Answer

Expert verified
(a) 1, (b) 0.75, (c) 0.25, (d) 0, (e) 0, (f) 0.25.

Step by step solution

01

Understanding the CDF Function

The given function \( F(x) \) is a cumulative distribution function (CDF). A CDF shows the probability that a random variable \( X \) is less than or equal to a certain value \( x \). It is always non-decreasing and ranges from 0 to 1.
02

Solving for (a) \( P(X \leq 50) \)

Look at the CDF. For \( x = 50 \), we have \( F(x) = 1 \). Therefore, \( P(X \leq 50) = 1 \).
03

Solving for (b) \( P(X \leq 40) \)

For \( x = 40 \), we find \( F(x) \) in the interval \( 30 \leq x < 50 \), where \( F(x) = 0.75 \). Thus, \( P(X \leq 40) = 0.75 \).
04

Solving for (c) \( P(40 \leq X \leq 60) \)

Calculate \( F(60) - F(40) \). Since \( 60 \geq 50 \), \( F(60) = 1 \). For \( x = 40 \), as found earlier, \( F(40) = 0.75 \). Thus, \( P(40 \leq X \leq 60) = 1 - 0.75 = 0.25 \).
05

Solving for (d) \( P(X < 0) \)

For \( x < 0 \), we look at the intervals that cover these values. Since the CDF does not increase until \( x = -10 \), \( F(x) = 0 \) for all \( x < 0 \). Hence, \( P(X < 0) = 0 \).
06

Solving for (e) \( P(0 \leq X < 10) \)

For both \( 0 \leq X < 10 \), \( F(x) \) still falls within the range defined by \( -10 \leq x < 30 \), so \( F(x) = 0.25 \). As \( P(X < 10) - F(0) = F(10) - F(0) = 0.25 - 0.25 = 0 \).
07

Solving for (f) \( P(-10 < X < 10) \)

Find the probability from \( -10 < x < 10 \). Since \( F(x) \) remains \( 0.25 \) over this interval, calculate \( F(10) - F(-10) = 0.25 - 0 = 0.25 \).

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Key Concepts

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

Probability Calculations
Probability calculations help us determine the likelihood of certain events occurring. They revolve around understanding how a random variable behaves within a defined framework. In our case, we're dealing with a cumulative distribution function (CDF), which provides the probability that a random variable \(X\) is less than or equal to a certain value.

For example, to find \(P(X \leq 50)\), the CDF directly provides this information. When \(x = 50\), \(F(x)\) is given as 1, meaning there's a 100% chance that a value \(X\) will be less than or equal to 50.

Understanding how to read and interpret these values from the CDF is crucial for solving probability calculation exercises accurately. You take the value from the table and look at which interval it falls into to find the probability. This practice will sharpen your skills in measuring probabilities efficiently.
Step-by-Step Solutions
Breaking down problems into step-by-step solutions can make complex probability calculations more manageable. This approach is beneficial not only for internalizing methods but also for ensuring all important details are considered.

Here's a snapshot of how you can tackle a problem using these steps:
  • Identify the range of interest from the cumulative distribution function (CDF).
  • Match the given values to the correct interval in the CDF.
  • Apply the relevant logic or subtract adjacent probabilities to find your answer. For instance, you subtract \(F(40)\) from \(F(60)\) to find \(P(40 \leq X \leq 60)\).
This method demystifies the process and equips you with the precision needed to solve similar problems with ease.
Random Variables
Random variables are foundational elements in probability and statistics. They represent outcomes that result from random phenomena. We often describe them using probability distributions to predict outcomes based on possible values.

In this exercise, random variable \(X\) is defined through the CDF, which depicts probabilities aligned with certain intervals. This linkage helps bring structure to unpredictable events, allowing for more focused predictions and analyses. With random variables, understanding the role and behavior of each variable enhances interpretation skills in probability tasks.
Probability Distributions
Probability distributions describe how probabilities are assigned to different outcomes of a random variable. The cumulative distribution function (CDF) is one type of probability distribution that provides a cumulative probability for a random variable getting a value less than or equal to \(x\).

For our particular exercise, the CDF segments different intervals with specific probabilities, offering a structured overview of where a random variable \(X\) falls along the number line.
  • \(P(X \leq x)\) reflects the cumulative probability from the CDF.
  • The CDF guarantees a non-decreasing function, meaning probabilities do not drop as you move along the graph.
Understanding these distributions helps in the broader analysis of data and probability, offering tools to forecast and interpret events in varied contexts.

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Most popular questions from this chapter

The number of cracks in a section of interstate highway that are significant enough to require repair is assumed to follow a Poisson distribution with a mean of two cracks per mile. (a) What is the probability that there are no cracks that require repair in 5 miles of highway? (b) What is the probability that at least one crack requires repair in \(1 / 2\) mile of highway? (c) If the number of cracks is related to the vehicle load on the highway and some sections of the highway have a heavy load of vehicles whereas other sections carry a light load, how do you feel about the assumption of a Poisson distribution for the number of cracks that require repair?

An automated egg carton loader has a \(1 \%\) probability of cracking an egg, and a customer will complain if more than one egg per dozen is cracked. Assume each egg load is an independent event. (a) What is the distribution of cracked eggs per dozen? Include parameter values. (b) What are the probability that a carton of a dozen eggs results in a complaint? c) What are the mean and standard deviation of the number of cracked eggs in a carton of one dozen?

The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a mean of 0.02 failure per hour. (a) What is the probability that the instrument does not fail in an eight-hour shift? (b) What is the probability of at least one failure in a 24 -hour day?

Samples of 20 parts from a metal punching process are selected every hour. Typically, \(1 \%\) of the parts require rework. Let \(X\) denote the number of parts in the sample of 20 that require rework. A process problem is suspected if \(X\) exceeds its mean by more than three standard deviations. (a) If the percentage of parts that require rework remains at \(1 \%,\) what is the probability that \(X\) exceeds its mean by more than three standard deviations? (b) If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds \(1 ?\) (c) If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds 1 in at least one of the next five hours of samples?

Suppose that the number of customers who enter a bank in an hour is a Poisson random variable, and suppose that \(P(X=0)=0.05 .\) Determine the mean and variance of \(X .\)

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