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# Probability - continuous random variables watch

1. Ok, I have 2 questions:

1. Nicotine levels in smokers can be modelled by a normal random variable with mean 315 and variance 1312. What is the probability, if 20 smokers are tested, that at most one has a nicotine level higher than 500?

2. fX,Y (x,y) = xe-x-y 0<x<y<
Find c.
Find the marginal probability density functions.

1. I have worked out that each smoker individually has a 0.079 probability of having a nicotine level higher than 500, but I'm not sure about the at most one section. Would I need to work out 1-P(at least 2 have a nicotine level higher than 500) where P(at least 2 have a nicotine level higher than 500) = P(2 have a nicotine level higher than 500) + P(3 do) + P(4 do) + ... + P(20 do). In this case would it be a geometric sum with a = 0.0792, r = 0.079 and n = 19?

2. It is the limits of integration that I am finding confusing here. I know to find c I need to do the double integral equal to one, but is it fX,Y (x,y) dx dy (i.e. integrating x between 0 and y and y between 0 and infinity)?

Similarly for the marginal distributions would the limits when integrating with respect to y be 0 and infinity and x be 0 and y?

Thanks
2. (Original post by Kate2010)
Ok, I have 2 questions:

1. I have worked out that each smoker individually has a 0.079 probability of having a nicotine level higher than 500, but I'm not sure about the at most one section. Would I need to work out 1-P(at least 2 have a nicotine level higher than 500) where P(at least 2 have a nicotine level higher than 500) = P(2 have a nicotine level higher than 500) + P(3 do) + P(4 do) + ... + P(20 do). In this case would it be a geometric sum with a = 0.0792, r = 0.079 and n = 19?

Agree with 0.079.

Probability that at most one has a level > 500, means either none have or one has.

So you only need to work out P(X=0) and P(X=1).
Your method would work, but is rather long winded and going round the houses.
Distribution is not geometric however.
3. But X is measuring the nicotine level not the amount of people. Do I need to introduce another random variable to model the amount of people? Also X is a continuous random variable so you can't work it out for a specific point.
4. (Original post by Kate2010)
But X is measuring the nicotine level not the amount of people. Do I need to introduce another random variable to model the amount of people? Also X is a continuous random variable so you can't work it out for a specific point.
You hadn't mentioned X in question 1, so I was using it as an abbreviation whose meaning I thought would be clear.

I'll rephrase it as:

So you only need to work out P(0 have a nicotine level higher than 500) + P(1 has a nicotine level higher than 500).
5. Thanks think I've got it now. Sorry I didn't realise I hadn't mentioned about X in the question.
6. (Original post by Kate2010)
Thanks think I've got it now. Sorry I didn't realise I hadn't mentioned about X in the question.
No problem. But, do note my final comment in my first post; it's not a geometric distribution.

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