The Empirical Rule and Other Rules in Statistics


Empirical Rule For the Normal Distribution

{μσXμ+σ}={σXμσ}={1Xμσ1}\left\{ \mu -\sigma \le X\le \mu +\sigma \right\}=\left\{ -\sigma \le X-\mu \le \sigma \right\}=\left\{ -1\le \frac{X-\mu }{\sigma }\le 1 \right\}

{μσXμ+σ}={σXμσ}={1Xμσ1}={1Z1}\left\{ \mu -\sigma \le X\le \mu +\sigma \right\}=\left\{ -\sigma \le X-\mu \le \sigma \right\}=\left\{ -1\le \frac{X-\mu }{\sigma }\le 1 \right\}=\left\{ -1\le Z\le 1 \right\} Pr(μσXμ+σ)=Pr(1Xμσ1)=Pr(1Z1)Pr \left( \mu -\sigma \le X\le \mu +\sigma \right)=\Pr \left( -1\le \frac{X-\mu }{\sigma }\le 1 \right)=\Pr \left( -1\le Z\le 1 \right) =Pr(Z1)Pr(Z1)0.8413450.1586550.682689=\Pr \left( Z\le 1 \right)-\Pr \left( Z\le -1 \right)\approx 0.\text{841345}-0.\text{158655}\approx 0.\text{682689} Pr(μ2σXμ+2σ)=Pr(2Xμσ2)=Pr(2Z2)\Pr \left( \mu -2\sigma \le X\le \mu +2\sigma \right)=\Pr \left( -2\le \frac{X-\mu }{\sigma }\le 2 \right)=\Pr \left( -2\le Z\le 2 \right) =Pr(Z2)Pr(Z2)0.9772498680.0227501320.9544997=\Pr \left( Z\le 2 \right)-\Pr \left( Z\le -2 \right)\approx 0.\text{977249868}-0.0\text{2275}0\text{132}\approx 0.\text{9544997} Pr(μ3σXμ+3σ)=Pr(3Xμσ3)=Pr(3Z3)\Pr \left( \mu -3\sigma \le X\le \mu +3\sigma \right)=\Pr \left( -3\le \frac{X-\mu }{\sigma }\le 3 \right)=\Pr \left( -3\le Z\le 3 \right) =Pr(Z3)Pr(Z3)0.9986501020.0013498980.9973002=\Pr \left( Z\le 3 \right)-\Pr \left( Z\le -3 \right)\approx 0.\text{99865}0\text{1}0\text{2}-0.00\text{1349898}\approx 0.\text{9973}00\text{2}

The Rule of Thumb for the Standard Deviation

sRange4s\approx \frac{Range}{4}

Chebyshev's Rule

Pr(μkσXμ+kσ)11k2\Pr \left( \mu -k\sigma \le X\le \mu +k\sigma \right)\ge 1-\frac{1}{{{k}^{2}}} Pr(μ2σXμ+2σ)1122=0.75\Pr \left( \mu -2\sigma \le X\le \mu +2\sigma \right)\ge 1-\frac{1}{{{2}^{2}}}=0.75
This tutorial is brought to you courtesy of MyGeekyTutor.com

log in to your account

Don't have a membership account?
REGISTER

reset password

Back to
log in

sign up

Back to
log in