Let X be uniformly distributed over (-2,2). Use Chebyshev's inequality to estimate P(abs(x) >= 1) and compare to the exact value.(adsbygoogle = window.adsbygoogle || []).push({});

The answer in the book got something different than what i got so i wanted to see if i was right

EX of a uniformly distrubuted RV is (a+b)/2 so -2+2/0 =0

So i plugged into the formula.

P(abs(x)>=1 <= sigma^2

the variance of a uniformly RV is (b-a)^2/12 so (2-(-2))^2=14 and 14/12 is 4/3.

my first answer ======= P(abs(x)>=1 <= 4/3

For the exact part I used the P(X<=x) cdf of a unform RV and got F(1)=3/4. So 1-F(1) = 0.25 which is the probability we are looking for.

Did i do this right? Thanks!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Chebyshev's Inequality Problem!

**Physics Forums | Science Articles, Homework Help, Discussion**