Solution Found!
Chebyshev's inequality .(Section 2.4) states that for any
Chapter 4, Problem 26E(choose chapter or problem)
Chebyshev’s inequality (Section 2.4) states that for any random variable X with mean μ and variance σ2, and for any positive number k, \(P(|X=\mu| \geq k \sigma) \leq 1 / k^{2}\). Let X ∼ N\(\left(\mu, \sigma^{2}\right)\). Compute \(P(|X-\mu| \geq k \sigma)\) for the values k = 1, 2, and 3. Are the actual probabilities close to the Chebyshev bound of \(1 / k^{2}\), or are they much smaller?
Equation transcription:
Text transcription:
P(|X=mu| \geq k sigma) \leq 1 / k^{2}
(mu, sigma^{2})
P(|X-mu| geq k sigma)
1 / k^{2}
Questions & Answers
QUESTION:
Chebyshev’s inequality (Section 2.4) states that for any random variable X with mean μ and variance σ2, and for any positive number k, \(P(|X=\mu| \geq k \sigma) \leq 1 / k^{2}\). Let X ∼ N\(\left(\mu, \sigma^{2}\right)\). Compute \(P(|X-\mu| \geq k \sigma)\) for the values k = 1, 2, and 3. Are the actual probabilities close to the Chebyshev bound of \(1 / k^{2}\), or are they much smaller?
Equation transcription:
Text transcription:
P(|X=mu| \geq k sigma) \leq 1 / k^{2}
(mu, sigma^{2})
P(|X-mu| geq k sigma)
1 / k^{2}
ANSWER:
Solution 26E
Step1 of 2:
We have chebyshev’s inequality and it states that for any random variable X with mean and variance and for any number of k,
Also we have a random variable X it follows normal distribution with mean “n” and
variance “” .
We need to compute for the values for k = 1, 2 and 3 also we have to check whether the actual probabilities are close to chebyshev’s bound or are they much smaller.
Step2 of 2:
From the given informa we have chebyshev’s inequality
There are 68% of the population is in the interval therefore
We need to compute for k = 1, 2, 3.