Chebyshev’s theorem provides that for a random variable X with any probability distribution and a positive constant k greater than or equal to 1, the inequality P (|X − μ| ≥ kσ) ≤ 1 k2 holds. This signifies that the probability of X deviating from its mean by at least k standard deviations is at most 1 k2 . a. Determine the upper limit for k being 2, 3, 4, 5, and 10. b. Calculate the mean μ and standard deviation σ for the following distribution. Then, assess P (|X − μ| ≥ kσ) for the given k values and compare these to the upper bounds found previously. x 0 1 2 3 4 5 6 p(x) .1 .15 .2 .25 .06 .04 2 c. Assume X can take on the values -1, 0, and 1 with probabilities 1 18 , 8 9 , and 1 18 respectively. Estimate P (|X − μ| ≥ 3σ), and relate it to the corresponding upper limit. d. Propose a probability distribution where P (|X − μ| ≥ 5σ) is 0.04.
Chebyshev’s theorem provides that for a random variable X with any probability distribution and a positive constant k greater than or equal to 1, the inequality P (|X − μ| ≥ kσ) ≤ 1 k2 holds. This signifies that the probability of X deviating from its mean by at least k standard deviations is at most 1 k2 . a. Determine the upper limit for k being 2, 3, 4, 5, and 10. b. Calculate the mean μ and standard deviation σ for the following distribution. Then, assess P (|X − μ| ≥ kσ) for the given k values and compare these to the upper bounds found previously. x 0 1 2 3 4 5 6 p(x) .1 .15 .2 .25 .06 .04 2 c. Assume X can take on the values -1, 0, and 1 with probabilities 1 18 , 8 9 , and 1 18 respectively. Estimate P (|X − μ| ≥ 3σ), and relate it to the corresponding upper limit. d. Propose a probability distribution where P (|X − μ| ≥ 5σ) is 0.04.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
Chebyshev’s theorem provides that for a random variable X with any
probability distribution and a positive constant k greater than or equal
to 1, the inequality P (|X − μ| ≥ kσ) ≤ 1
k2 holds. This signifies that
the probability of X deviating from its mean by at least k standard
deviations is at most 1
k2 .
a. Determine the upper limit for k being 2, 3, 4, 5, and 10.
b. Calculate the mean μ and standard deviation σ for the following
distribution. Then, assess P (|X − μ| ≥ kσ) for the given k values
and compare these to the upper bounds found previously.
x 0 1 2 3 4 5 6
p(x) .1 .15 .2 .25 .06 .04
2
probability distribution and a positive constant k greater than or equal
to 1, the inequality P (|X − μ| ≥ kσ) ≤ 1
k2 holds. This signifies that
the probability of X deviating from its mean by at least k standard
deviations is at most 1
k2 .
a. Determine the upper limit for k being 2, 3, 4, 5, and 10.
b. Calculate the mean μ and standard deviation σ for the following
distribution. Then, assess P (|X − μ| ≥ kσ) for the given k values
and compare these to the upper bounds found previously.
x 0 1 2 3 4 5 6
p(x) .1 .15 .2 .25 .06 .04
2
c. Assume X can take on the values -1, 0, and 1 with probabilities
1
18 , 8
9 , and 1
18 respectively. Estimate P (|X − μ| ≥ 3σ), and relate
it to the corresponding upper limit.
d. Propose a probability distribution where P (|X − μ| ≥ 5σ) is 0.04.
1
18 , 8
9 , and 1
18 respectively. Estimate P (|X − μ| ≥ 3σ), and relate
it to the corresponding upper limit.
d. Propose a probability distribution where P (|X − μ| ≥ 5σ) is 0.04.
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