i.c P(x - 14 ≥ ko) ≤ 2 < And now we proof our main result. Proof: Letz be a continuous random variable 0² E[(x-μ)²] - = (x-μ)² f(x)dr 88 μ-ko - ko μ + ko rutko (-) (a)de + f (x-4)²(a) dx + (x − p)³. f(x)dr f - -μ)² -ka µ+ko
i.c P(x - 14 ≥ ko) ≤ 2 < And now we proof our main result. Proof: Letz be a continuous random variable 0² E[(x-μ)²] - = (x-μ)² f(x)dr 88 μ-ko - ko μ + ko rutko (-) (a)de + f (x-4)²(a) dx + (x − p)³. f(x)dr f - -μ)² -ka µ+ko
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Please write a detailed preliminary explaining the first 2 lines. Explain E, the integral , and variance
![The text presented is an excerpt from a mathematical proof regarding probabilities and random variables. It focuses on the idea that for a given continuous random variable \(x\) with mean \(\mu\) and standard deviation \(\sigma\), the probability that \(x\) deviates from its mean \(\mu\) by more than \(k\sigma\) is bounded by \( \frac{1}{k^2} \). This is related to Chebyshev's inequality.
### Transcription:
The statement begins with the inequality:
\[
i.e., \; P(|x - \mu| \geq k\sigma) \leq \frac{1}{k^2}
\]
This is an expression of Chebyshev’s inequality which provides a bound for the probability.
#### Proof:
Let \(x\) be a continuous random variable.
The proof begins by defining the variance \(\sigma^2\) as the expected value of the squared deviations from the mean:
\[
\sigma^2 = E[(x - \mu)^2] = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) \, dx
\]
A number line diagram is used to split the integration limits into three segments at \( \mu - k\sigma \) and \( \mu + k\sigma \).
The proof then segments the integral into three parts:
\[
= \int_{-\infty}^{\mu-k\sigma} (x - \mu)^2 f(x) \, dx + \int_{\mu-k\sigma}^{\mu+k\sigma} (x - \mu)^2 f(x) \, dx + \int_{\mu+k\sigma}^{\infty} (x - \mu)^2 f(x) \, dx
\]
### Explanation of the Diagram:
- **Number Line:** It divides the \(-\infty\) to \(+\infty\) interval into three regions:
- \( (-\infty, \mu-k\sigma) \)
- \( (\mu-k\sigma, \mu+k\sigma) \)
- \( (\mu+k\sigma, \infty) \)
These intervals correspond to the three integrals in the variance expression, demonstrating how the probability and variance are distributed across these regions.
This type of analysis is useful for understanding the distribution of random variables and applying probabil](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2F014512ca-a89d-4524-8175-2fff06f04697%2Fxwb9zp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The text presented is an excerpt from a mathematical proof regarding probabilities and random variables. It focuses on the idea that for a given continuous random variable \(x\) with mean \(\mu\) and standard deviation \(\sigma\), the probability that \(x\) deviates from its mean \(\mu\) by more than \(k\sigma\) is bounded by \( \frac{1}{k^2} \). This is related to Chebyshev's inequality.
### Transcription:
The statement begins with the inequality:
\[
i.e., \; P(|x - \mu| \geq k\sigma) \leq \frac{1}{k^2}
\]
This is an expression of Chebyshev’s inequality which provides a bound for the probability.
#### Proof:
Let \(x\) be a continuous random variable.
The proof begins by defining the variance \(\sigma^2\) as the expected value of the squared deviations from the mean:
\[
\sigma^2 = E[(x - \mu)^2] = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) \, dx
\]
A number line diagram is used to split the integration limits into three segments at \( \mu - k\sigma \) and \( \mu + k\sigma \).
The proof then segments the integral into three parts:
\[
= \int_{-\infty}^{\mu-k\sigma} (x - \mu)^2 f(x) \, dx + \int_{\mu-k\sigma}^{\mu+k\sigma} (x - \mu)^2 f(x) \, dx + \int_{\mu+k\sigma}^{\infty} (x - \mu)^2 f(x) \, dx
\]
### Explanation of the Diagram:
- **Number Line:** It divides the \(-\infty\) to \(+\infty\) interval into three regions:
- \( (-\infty, \mu-k\sigma) \)
- \( (\mu-k\sigma, \mu+k\sigma) \)
- \( (\mu+k\sigma, \infty) \)
These intervals correspond to the three integrals in the variance expression, demonstrating how the probability and variance are distributed across these regions.
This type of analysis is useful for understanding the distribution of random variables and applying probabil
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