i.c P(x-μ ≥ko) ≤ 1/1/2 < And now we proof our main result. Proof: Let z be a continuous random variable 0² E[(x-μ)²] [(x − μ)² f(x) dx - = ∞ "-ko 00 FL μ + ko 8X μ-ko rutko (x-μ)²) (x − 1)² f(x)}dx + [lth (x − 1)²f(x)dx + -ka Suv (2 - 11) ² µ+ko
i.c P(x-μ ≥ko) ≤ 1/1/2 < And now we proof our main result. Proof: Let z be a continuous random variable 0² E[(x-μ)²] [(x − μ)² f(x) dx - = ∞ "-ko 00 FL μ + ko 8X μ-ko rutko (x-μ)²) (x − 1)² f(x)}dx + [lth (x − 1)²f(x)dx + -ka Suv (2 - 11) ² µ+ko
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
How did the E turn to the integral ? Please explain in detail not understanding the first two lines
![Here is the transcribed content suitable for an educational website:
---
**Main Result: Chebyshev's Inequality**
i.e. \( P(|x - \mu| \geq k\sigma) \leq \frac{1}{k^2} \)
And now we prove our main result.
**Proof:** Let \( x \) be a continuous random variable.
\[
\sigma^2 = E[(x - \mu)^2]
\]
\[
= \int_{-\infty}^{\infty} (x - \mu)^2 f(x) \, dx
\]
Visual representation with the range:
\[
-\infty \quad | \quad \mu - k\sigma \quad | \quad \mu \quad | \quad \mu + k\sigma \quad | \quad \infty
\]
Further expansion:
\[
= \int_{-\infty}^{\mu-k\sigma} (x - \mu)^2 f(x) \, dx + \int_{\mu-k\sigma}^{\mu+k\sigma} (x - \mu)^2 f(x) \, dx + \int_{\mu+k\sigma}^{\infty} (x - \mu)^2 f(x) \, dx
\]
---
This text explains the derivation of the Chebyshev’s inequality and breaks down the variance calculation into integrals over specific intervals. This could serve as part of a lesson on probability theory or statistics, specifically focusing on variance and Chebyshev's inequality.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2Facee21c4-c7bd-4b85-99ea-9beeb6032dfc%2F8a6f8ar_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Here is the transcribed content suitable for an educational website:
---
**Main Result: Chebyshev's Inequality**
i.e. \( P(|x - \mu| \geq k\sigma) \leq \frac{1}{k^2} \)
And now we prove our main result.
**Proof:** Let \( x \) be a continuous random variable.
\[
\sigma^2 = E[(x - \mu)^2]
\]
\[
= \int_{-\infty}^{\infty} (x - \mu)^2 f(x) \, dx
\]
Visual representation with the range:
\[
-\infty \quad | \quad \mu - k\sigma \quad | \quad \mu \quad | \quad \mu + k\sigma \quad | \quad \infty
\]
Further expansion:
\[
= \int_{-\infty}^{\mu-k\sigma} (x - \mu)^2 f(x) \, dx + \int_{\mu-k\sigma}^{\mu+k\sigma} (x - \mu)^2 f(x) \, dx + \int_{\mu+k\sigma}^{\infty} (x - \mu)^2 f(x) \, dx
\]
---
This text explains the derivation of the Chebyshev’s inequality and breaks down the variance calculation into integrals over specific intervals. This could serve as part of a lesson on probability theory or statistics, specifically focusing on variance and Chebyshev's inequality.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 5 steps with 4 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
