Question

Use Chebyshev's theorem to determine at least what percentage of data values fall between 13 and 99 for a distribution with a mean of 56 and a standard deviation of 24.

Answer 1

Chebychevs inequality is

P(|x-mu|>=k sigma)>=1/k^2

P((x-mu)<-ksigma or x-mu>k sigma)>=1/k^2

P(x<56 -24k or x> 56+24k)>=1/k^2

when k=43/24,

P(x<13 or x>99)>=1/(43/24)^2=24^2/43^2=0.3116

Therefore P(13<X<99)>= 1-0.3116=0.6884

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