Maybe the next lesson is to evaluate how much quicker it would have been to have used the inequality.

(that's me trying to get inside the head of the teacher. Dangerous activity that! )

Bob

EDIT: Hey. It seems reality has got ahead of my thoughts.

]]>I understand now. Thanks.

]]>One more question. If you want the probability and this is a Chebychev problem why not just use the inequality?

]]>Isn't Chebychev's an inequality that bounds an unknown distribution? What does it have to do with this question since you know the distribution? And why are you integrating?

]]>I'm not fully following this.

For a probability density function the integral over all popssible values must be 1.

If I integrate from - ∞ to + ∞ that doesn't happen, so I'm assuming it should be from 0 to + ∞. then it works.

That fits with

http://en.wikipedia.org/wiki/Exponential_distribution

and my calculations for mean and variance are 1/3 and 1/9 as you have. So far so good.

But k ?

According to

http://en.wikipedia.org/wiki/Chebyshev%27s_inequality

k is an value indicating how many 'standard deviations from the mean. Your calculations for what x values to use follow ok.

But if x > 0 for the pdf what are we to do with x = -2/3 ?

The integral up to zero will be zero, so I'm thinking you should integrate from 0 to 4/3.

Hope that's correct.

Bob

]]>Expected value = 1/3. The variance is 1/9 and the standard deviation is 1/3 so for k=3 I got:

(1/3) - 3*(1/3) < X < (1/3) + 3*(1/3) (Wasn't sure about this, all I did was sub 2 with 3 from the problem where k=2.)

= -2/3 < X < 4/3

So to get the probability, you integrate f(x) from -2/3 to 4/3? Would this be correct?

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