# Probability - continuous random variables

## Homework Statement

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<$$\infty$$
Find c.
Find the marginal probability density functions.

## The Attempt at a Solution

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 $$\int$$$$^{infinity}_{0}$$$$\int$$$$^{y}_{0}$$ 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?

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LCKurtz
Homework Helper
Gold Member

## Homework Statement

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<$$\infty$$
Find c.
Find the marginal probability density functions.

## The Attempt at a Solution

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?
P(at most 2) = P(0) + P(1). And if your p = .079 is correct and X is the number with nicotine over 500, then isn't X binomial(20,p)?

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 $$\int$$$$^{infinity}_{0}$$$$\int$$$$^{y}_{0}$$ 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?
Your limits are correct. Click on the expression below to see how to render it in tex:

$$\int_0^\infty \int_0^y f(x,y)dxdy$$

LCKurtz
Homework Helper
Gold Member
the limits when integrating with respect to y be 0 and infinity and x be 0 and y?

I didn't notice this final question. For the marginals your limits are determined by the non-zero region. So when integrating in the x direction you will go from 0 to y and when integrating in the y direction, it will go from x to infinity. Your nonzero domain is in the first quadrant above the line y = x.

Thanks a lot, really helpful :)

Ah ok, so fX(x) = $$\int_x^\infty f(x,y) dxdy$$ not fX(x) = $$\int_0^\infty f(x,y) dxdy$$?

vela
Staff Emeritus
Ah ok, so fX(x) = $$\int_x^\infty f(x,y) dxdy$$ not fX(x) = $$\int_0^\infty f(x,y) dxdy$$?