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Suppose that the cumulative distribution function of the random variable \(X\) is \(F(x)=\left\\{\begin{array}{lc}0 & x<-2 \\ 0.25 x+0.5 & -2 \leq x<2 \\ 1 & 2 \leq x\end{array}\right.\) Determine the following: (a) \(P(X<1.8)\) (b) \(P(X>-1.5)\) (c) \(P(X<-2)\) (d) \(P(-1

Short Answer

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(a) 0.95; (b) 0.875; (c) 0; (d) 0.5.

Step by step solution

01

Determine Given Information

We are provided with the cumulative distribution function (CDF) of the random variable \(X\), defined piecewise as:\\[F(x) = \begin{cases}0, & x < -2 \0.25x + 0.5, & -2 \leq x < 2 \1, & x \geq 2\end{cases}\]We need to calculate probabilities for different ranges using this function.
02

Calculate P(X < 1.8)

Since \(1.8\) lies in the range \(-2 \leq x < 2\), we use the second piece of the CDF function: \(F(x) = 0.25x + 0.5\).\[\begin{align*}F(1.8) & = 0.25(1.8) + 0.5 \& = 0.45 + 0.5 \& = 0.95\end{align*}\] Thus, \(P(X < 1.8) = 0.95\).
03

Calculate P(X > -1.5)

To find \(P(X > -1.5)\), we need \(1 - P(X \leq -1.5)\). \(-1.5\) lies in the range \(-2 \leq x < 2\). \[\begin{align*}F(-1.5) & = 0.25(-1.5) + 0.5 \& = -0.375 + 0.5 \& = 0.125\end{align*}\]Thus, \(P(X > -1.5) = 1 - 0.125 = 0.875\).
04

Calculate P(X < -2)

For \(x < -2\), the cumulative distribution function \(F(x) = 0\), thus: \(P(X < -2) = 0\).
05

Calculate P(-1 < X < 1)

This probability requires \(F(1) - F(-1)\). Both values lie in the range \(-2 \leq x < 2\).\[\begin{align*}F(1) & = 0.25(1) + 0.5 = 0.75,\F(-1) & = 0.25(-1) + 0.5 = 0.25\end{align*}\]Hence, \(P(-1 < X < 1) = 0.75 - 0.25 = 0.5\).

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Key Concepts

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

Probability Calculation
Probability calculation often involves determining the likelihood of certain events occurring based on the known data and statistical models. In the context of cumulative distribution functions (CDF), calculating probabilities requires understanding the relationship between the CDF and the probabilities of events involving a random variable. For a random variable \(X\), the CDF \(F(x)\) gives the probability that \(X\) takes a value less than or equal to \(x\). Hence, the probability \(P(X < x)\) can be directly derived from \(F(x)\) when \(x\) is within the domain of the function.
To calculate the probability of a range, such as \(P(a < X < b)\), we find \(F(b)\) and \(F(a)\), and then subtract the two: \(P(a < X < b) = F(b) - F(a)\).
This method allows for straightforward computation using the CDF, as seen in the solutions for the problems like finding \(P(X < 1.8)\) and \(P(-1 < X < 1)\). This illustrates how understanding the role of the CDF facilitates the computation of probabilities over both individual values and intervals.
Random Variables
Random variables are a fundamental concept in probability and statistics. They represent numerical outcomes of a random phenomenon. When dealing with random variables, it's crucial to understand their distribution, which describes how probabilities are assigned to each possible value of the random variable.
In the problem provided, the random variable \(X\) has a distribution defined by a piecewise CDF. This distribution tells us how \(X\) behaves over different intervals. A cumulative distribution function, like the one given in the exercise, sums probabilities up to a certain point, helping us understand which values \(X\) might realistically take.
Random variables can be discrete or continuous. In this case, the random variable \(X\) is continuous, as indicated by the continuous range of values it can take within specified intervals. Understanding the behavior and characteristics of random variables is essential for accurate probability calculations and statistical analysis.
Piecewise Function Analysis
Piecewise functions are functions defined by multiple sub-functions, each applying to a certain interval of the domain. In the context of a cumulative distribution function, a piecewise function allows for modeling of complex systems where the behavior varies over different ranges.
In the original exercise, the piecewise CDF for the random variable \(X\) is described as follows: \(F(x) = 0\) for \(x < -2\), \(F(x) = 0.25x + 0.5\) for \(-2 \leq x < 2\), and \(F(x) = 1\) for \(x \geq 2\). This allows the probability distribution to capture different scenarios that \(X\) may encounter.
When analyzing such piecewise functions, it's important to determine which part of the function applies to a given evaluation point. This requires understanding both the mathematical structure of the function and the contextual meaning of each section. Successfully using piecewise functions involves calculating probabilities or expected values by leveraging the defined sub-functions correctly, ensuring precise solutions, like determining \(P(X < 1.8)\), which falls under the interval \(-2 \leq x < 2\).
By dissecting each segment of the piecewise function, one can fully grasp its overall impact on the probability calculations.

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Most popular questions from this chapter

The time between calls is exponentially distributed with a mean time between calls of 10 minutes. (a) What is the probability that the time until the first call is less than 5 minutes? (b) What is the probability that the time until the first call is between 5 and 15 minutes? (c) Determine the length of an interval of time such that the probability of at least one call in the interval is \(0.90 .\) (d) If there has not been a call in 10 minutes, what is the probability that the time until the next call is less than 5 minutes? (e) What is the probability that there are no calls in the intervals from 10: 00 to \(10: 05,\) from 11: 30 to \(11: 35,\) and from 2: 00 to \(2: 05 ?\) (f) What is the probability that the time until the third call is greater than 30 minutes? (g) What is the mean time until the fifth call?

The length of time (in seconds) that a user views a page on a Web site before moving to another page is a lognormal random variable with parameters \(\theta=0.5\) and \(\omega^{2}=1\). (a) What is the probability that a page is viewed for more than 10 seconds? (b) By what length of time have \(50 \%\) of the users moved to another page? (c) What are the mean and standard deviation of the time until a user moves from the page?

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch. (a) Suppose that the specifications require the dot diameter to be between 0.0014 and 0.0026 inch. If the probability that a dot meets specifications is to be \(0.9973,\) what standard deviation is needed? (b) Assume that the standard deviation of the size of a dot is 0.0004 inch. If the probability that a dot meets specifications is to be \(0.9973,\) what specifications are needed? Assume that the specifications are to be chosen symmetrically around the mean of 0.002 .

The time between calls to a corporate office is exponentially distributed with a mean of 10 minutes. (a) What is the probability that there are more than three calls in one-half hour? (b) What is the probability that there are no calls within onehalf hour? (c) Determine \(x\) such that the probability that there are no calls within \(x\) hours is 0.01 . (d) What is the probability that there are no calls within a twohour interval? (e) If four nonoverlapping one-half-hour intervals are selected, what is the probability that none of these intervals contains any call? (f) Explain the relationship between the results in part (a) and (b).

Use integration by parts to show that \(\Gamma(r)=(r-1)\) \(\Gamma(r-1).\)

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