# Let $X=$ amount of insurance reimbursement for dental expense. The pdf of $X$ is $F(x) = ce^{-.004x}$ for $x>0$. Find $c$.

Let $X$= amount of insurance reimbursement for dental expense. The pdf of $X$ is $F(x) = ce^{-.004x} \text{ for } x>0$.

a) Find c. Hint: You can do this several ways. One way is to notice that X is an exponential RV.

For this question I am aware of two things. One is that for an exponential distribution $g(x, \theta) = 1/e^{-x/\theta}$ for $x>0$. I also know that the integral for any pdf of the function equals $1$.

I'm just unsure how to combine this information to find $c$.

Part b asks to calculate the $80$th percentile. [Hint: The $r$th percentile is the $x$-value, say $x=R$, for which the areas under $f$ to the left of $x=R$ is equal to $r$ percent.]

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Hint: the total area under a pdf equals $1$. What is the area under the given pdf? (It will work out to be of the form $c\times a$ where $a = \int_0^\infty e^{-0.004x}\,dx$ is a number that you are expected to calculate for yourself by computing the value of the integral) Can you figure out what $c$ is from this? –  Dilip Sarwate Dec 12 '12 at 3:15

The density function of an exponential has shape $\lambda^{-\lambda x}$ if $x\ge 0$, and $0$ elsewhere. So $c=0.004$.
Alternately, find $\int_0^\infty ce^{-0.004 x}\,dx$. We get $\dfrac{c}{0.004}$. To make the integral $1$, we need $c=0.004$.
For the $80$-th percentile, argue thus. The probability that $X\gt x$ is (by integration or otherwise) equal to $e^{-0.004x}$. We want this to be $0.80$. So we get the equation $$e^{-0.004x}=0.80.$$ Take the logarithm of both sides. We get $$-0.004 x=\ln(0.80).$$ Now with a little help from the calculator we can find $x$.