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3 years ago

11

For every person who has the flow there are six people who have when we feel like symptoms if a doctor sees 40 patients the time

and approximately how many patients you would expect to have only flu like symptoms

1 answer:

krok68 [10]3 years ago

8 0

The question doesn't seem to be written correctly, I'm not able to make much sense of it.

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4 years ago

Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I

Hatshy [7]

a. $f_{X,Y}$ is a joint density function if its integral over the given support is 1:

$\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1$

so the answer is yes.

b. We should first find the density of the marginal distribution, $f_Y(y)$ :

$f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy$

$f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}$

Then

$P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}$

or about 0.2019.

For the other probability, we can use the joint PDF directly:

$P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}$

which is about 0.7326.

c. We already know the PDF for $Y$ , so we just integrate:

$E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5$

5 0

3 years ago