# chebyshev inequality

• Oct 4th 2010, 06:17 PM
mathworld
chebyshev inequality
Hi,

A product is sold by box of 5000 products. The mean and variance for the number of defective products in one box is respectivly 10 and 5.76. Use chebyshev inequality to find a boundary for the probability that a box contains 8 to 12 defective products.

What i did:

P(8 <= X <= 12) = P(|X - 10| <= 2) = 1 - P(|X -10| > 2), and with chebyshev inequality i find P(|X - 10| > 2) <= 5.76/2^2 = 1.44?? What is my mistake? Thank you so much
• Oct 4th 2010, 08:14 PM
harish21
If X is a real valued random variable with mean $\displaystyle \mu$
and variance $\displaystyle \sigma^{2}$, then your chebyshef's inequality is given by:

$\displaystyle P\{|X - \mu| < k \sigma\} \geq 1 - k^{-2}$

• Oct 4th 2010, 10:42 PM
CaptainBlack
Quote:

Originally Posted by harish21
If X is a real valued random variable with mean $\displaystyle \mu$
and variance $\displaystyle \sigma^{2}$, then your chebyshef's inequality is given by:

$\displaystyle P\{|X - \mu| < k \sigma\} \geq 1 - k^{-2}$

Not a real criticism, but for other reasons I would prefer:

$\displaystyle P\left(\left|\dfrac{X - \mu}{\sigma}\right| < k \right) \geq 1 - k^{-2}$

CB
• Oct 4th 2010, 11:33 PM
mathworld
yes but the problem is that k < 1...

P(|X - u| < ka) = P(|X - 10| < 2) where ka=2 and a = 2.4 so k=5/6, then 1 - (5/6)^-2 = - 0.44 ...
• Oct 5th 2010, 05:44 AM
CaptainBlack
Quote:

Originally Posted by mathworld
yes but the problem is that k < 1...

P(|X - u| < ka) = P(|X - 10| < 2) where ka=2 and a = 2.4 so k=5/6, then 1 - (5/6)^-2 = - 0.44 ...

And the interpretation is that the bound is zero.

CB