scorpiomanoj
New Member
Hi David,
Probability that a random variable is 5 deviations from mean in a NORMAL DISTRIBUTION is 0.1350% (as calculated from excel function, ie, 1-normsdist(5) ).
For any unknown distribution, applying Chebyshev's inequality, prob. that the r.v. is 5 sigma from mean is LESS THAN OR EQUAL TO 1/25, which is 4%.
1) Does the prob. of k-sigma dev. in a normal distribution and that of any unknown dist. (applying Chebyshev's ineq) vary significantly. Or am I making any mistakes in the above two calculations?
2) I have so far understood that Chebyshev's inequality provides only a maximum boundary. Is my understanding correct?
Thanks in anticipation,
Halan Manoj Kumar.
Probability that a random variable is 5 deviations from mean in a NORMAL DISTRIBUTION is 0.1350% (as calculated from excel function, ie, 1-normsdist(5) ).
For any unknown distribution, applying Chebyshev's inequality, prob. that the r.v. is 5 sigma from mean is LESS THAN OR EQUAL TO 1/25, which is 4%.
1) Does the prob. of k-sigma dev. in a normal distribution and that of any unknown dist. (applying Chebyshev's ineq) vary significantly. Or am I making any mistakes in the above two calculations?
2) I have so far understood that Chebyshev's inequality provides only a maximum boundary. Is my understanding correct?
Thanks in anticipation,
Halan Manoj Kumar.
