/*! 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 3 For each of the following distri... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 3

For each of the following distributions, compute \(P(\mu-2 \sigma

Short Answer

Expert verified
The computations involve first calculating the means and standard deviations of the given distributions, and then evaluating the probabilities by integrating or summing the distribution from \( \mu - 2 \sigma \) to \( \mu + 2 \sigma \). Due to the complex nature of these calculations, numerical or software-assisted methods might be required to obtain the final result.

Step by step solution

01

Compute the Mean and Standard Deviation for the First Distribution

For the first distribution \(f(x)=6 x(1-x), 0<x<1\), zero elsewhere, the mean \( \mu \) is given by \( \mu = \int_{0}^{1} x f(x) dx = \int_{0}^{1} 6 x^2 (1-x) dx \). The standard deviation \( \sigma \) is the square root of the variance, which is given by \( Var(X) = \int_{0}^{1} (x-\mu)^2 f(x) dx = \int_{0}^{1} 6 (x^2 - 2 \mu x + \mu^2)(x-x^2) dx \). After calculating \( \mu \) and \( \sigma \), we can evaluate \( P(\mu-2 \sigma<X<\mu+2 \sigma) \) by integrating the probability density function \(f(x)\) from \(\mu-2 \sigma\) to \( \mu+2 \sigma\).
02

Compute the Mean and Standard Deviation for the Second Distribution

For the second distribution \(p(x)=(\frac{1}{2})^{x}, x=1,2,3,\ldots \), zero elsewhere, the mean \( \mu \) is given by \( \mu = \sum_{x=1}^{\infty} x p(x) = \sum_{x=1}^{\infty} x (\frac{1}{2})^{x} \). The standard deviation \( \sigma \) is the square root of the variance, which is given by \( Var(X) = \sum_{x=1}^{\infty} (x-\mu)^2 p(x) = \sum_{x=1}^{\infty} ((x^2-2 \mu x + \mu^2)(\frac{1}{2})^{x}) \). After calculating \( \mu \) and \( \sigma \), we can evaluate \( P(\mu-2 \sigma<X<\mu+2 \sigma) \) by summing the probability mass function \(p(x)\) from \(\mu-2 \sigma\) to \( \mu+2 \sigma\).
03

Evaluate the Probabilities

In both cases, after the mean and standard deviation are calculated, the probability \( P(\mu-2 \sigma<X<\mu+2 \sigma) \) can be evaluated as described in the previous steps, using the appropriate methods of integration for a continuous distribution and summation for a discrete distribution.

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Most popular questions from this chapter

Let \(X\) denote a random variable such that \(K(t)=E\left(t^{X}\right)\) exists for all real values of \(t\) in a certain open interval that includes the point \(t=1 .\) Show that \(K^{(m)}(1)\) is equal to the \(m\) th factorial moment \(E[X(X-1) \cdots(X-m+1)]\).

Suppose \(C_{1}, C_{2}, C_{3}, \ldots\) is a nondecreasing sequence of sets, i.e., \(C_{k} \subset C_{k+1}\), for \(k=1,2,3, \ldots\) Then \(\lim _{k \rightarrow \infty} C_{k}\) is defined as the union \(C_{1} \cup C_{2} \cup C_{3} \cup \cdots\). Find \(\lim _{k \rightarrow \infty} C_{k}\) if (a) \(C_{k}=\\{x: 1 / k \leq x \leq 3-1 / k\\}, k=1,2,3, \ldots\). (b) \(C_{k}=\left\\{(x, y): 1 / k \leq x^{2}+y^{2} \leq 4-1 / k\right\\}, k=1,2,3, \ldots\)

Find the mean and variance, if they exist, of each of the following distributions. (a) \(p(x)=\frac{3 !}{x !(3-x) !}\left(\frac{1}{2}\right)^{3}, x=0,1,2,3\), zero elsewhere. (b) \(f(x)=6 x(1-x), 0

If the pdf of \(X\) is \(f(x)=2 x e^{-x^{2}}, 0

Let a random variable \(X\) of the continuous type have a pdf \(f(x)\) whose graph is symmetric with respect to \(x=c .\) If the mean value of \(X\) exists, show that \(E(X)=c\) Hint: Show that \(E(X-c)\) equals zero by writing \(E(X-c)\) as the sum of two integrals: one from \(-\infty\) to \(c\) and the other from \(c\) to \(\infty\). In the first, let \(y=c-x\); and, in the second, \(z=x-c\). Finally, use the symmetry condition \(f(c-y)=f(c+y)\) in the first.
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