A scheme of random number generator is described as follows. First, one generates a normal random number with mean 0 and variance 16, a uniform random number on [0,24], then a geometric random number of parameter p = 0.25 (each independently). Then we define the random number to be the average of these three numbers. 30 random numbers are generated with the scheme above, denoted as {X}30₁. Estimate/approximate the probability that the average 1 X is deviated from μ = E[X₁] by a distance of more than 0.2 using: (a) Chebyshev's inequality; (b) central limit theorem. k=1*
A scheme of random number generator is described as follows. First, one generates a normal random number with mean 0 and variance 16, a uniform random number on [0,24], then a geometric random number of parameter p = 0.25 (each independently). Then we define the random number to be the average of these three numbers. 30 random numbers are generated with the scheme above, denoted as {X}30₁. Estimate/approximate the probability that the average 1 X is deviated from μ = E[X₁] by a distance of more than 0.2 using: (a) Chebyshev's inequality; (b) central limit theorem. k=1*
A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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![A scheme of random number generator is described as follows. First, one
generates a normal random number with mean 0 and variance 16, a uniform random number
on [0,24], then a geometric random number of parameter p = 0.25 (each independently).
Then we define the random number to be the average of these three numbers. 30 random
numbers are generated with the scheme above, denoted as {X}31. Estimate/approximate
the probability that the average 1 X is deviated from μ = E[X₁] by a distance of more
than 0.2 using: (a) Chebyshev's inequality; (b) central limit theorem.
k=1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd94c6fe5-0a79-4b34-8240-29323f6fd175%2Fc61a2a6c-4ec1-420e-8e45-5b29bac2e7ec%2Fq6k4tfo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A scheme of random number generator is described as follows. First, one
generates a normal random number with mean 0 and variance 16, a uniform random number
on [0,24], then a geometric random number of parameter p = 0.25 (each independently).
Then we define the random number to be the average of these three numbers. 30 random
numbers are generated with the scheme above, denoted as {X}31. Estimate/approximate
the probability that the average 1 X is deviated from μ = E[X₁] by a distance of more
than 0.2 using: (a) Chebyshev's inequality; (b) central limit theorem.
k=1
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