1. ## 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

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

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

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

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

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

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

CB

4. 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 ...

5. 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