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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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
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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