Tchebysheff’s Theorem (Chebyshev's Theorem)

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Tchebysheff’s Theorem (Chebyshev's Theorem)

Given a number t>1t > 1t>1 and a population with n measurements, then at least 1−1t2\boxed{\large1-\frac{1}{t^2}}1−t21​​ of the measures will lie within t\bcfi tt standard deviations of their mean.

Example
If the interval is (μ−2σ,μ+2σ)(\mu-2\sigma,\mu+2\sigma)(μ−2σ,μ+2σ), then t=2: t=2:t=2: 

1−1t2=1−1(2)2=1−14=34\displaystyle{1-\frac{1}{t^2}=1-\frac{1}{(2)^2}=1-\frac{1}{4}=\frac{3}{4}}1−t21​=1−(2)21​=1−41​=43​ 

 
At least 3/4 or 75% of the measurements lie in the (μ−2σ,μ+2σ)(\mu-2\sigma,\mu+2\sigma)(μ−2σ,μ+2σ) interval.
Notice that the Empirical Rule states that 95% of the measurements lie within the (μ−2σ,μ+2σ)(\mu-2\sigma,\mu+2\sigma)(μ−2σ,μ+2σ) interval. 
Tchebysheff’s Theorem is therefore much more conservative, and it applies to any shape of relative frequency histogram. This includes data that is skewed or not normally distributed.

You are given that:
1−1t2=0.65\bm 1\bm-\frac{\bm 1}{\bf t^2}\bm =\bm0.\bm6\bm51−t21​=0.65

Fill in the blanks:

“At least 
% of observations fall within t=t=t=
 standard deviations of the mean.”

I don't know

Extra Practice

Tchebysheff’s Theorem Question

You are given that:
1−1t2=0.65\bm 1\bm-\frac{\bm 1}{\bf t^2}\bm =\bm0.\bm6\bm51−t21​=0.65
Fill in the blanks:

Tchebysheff's Theorem

Tchebysheff’s Theorem: Which of the following is a correct statement?
1−1(1.25)2=0.361-\frac{1}{\left(1.25\right)^2}=0.361−(1.25)21​=0.36

Wize University Statistics Textbook > Continuous Probability Distributions

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Tchebysheff’s Theorem (Chebyshev's Theorem)

Extra Practice

Tchebysheff’s Theorem Question

Tchebysheff's Theorem