# Thread: cdf for funcion including E(X) and V(X)

1. ## cdf for funcion including E(X) and V(X)

I seem to have totally messed up this problem because my results do not make sense.
Could anyone please show me where I went wrong? Many thanks.

The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv X with pdf

$f(x)=2(1-\frac{1}{x^{2}})$ for $1\leq x\leq2$ and 0 otherwise.

a) Compute the cdf of X. $F(X)=\intop_{1}^{x}2(1-\frac{1}{y^{2}})dy,over 1\leq x\leq2$ $F(X)=2x+\frac{2}{x}\mid_{1}^{2}$=4+1-2-2=1.

b) Obtain an expression for the (100p)th percentile. What is the value of $\tilde{\mu}$?

$p=F(\eta(p))=\intop_{1}^{\eta(p)}2(1-\frac{1}{y^{2}})dy=2y+\frac{2}{y}\mid_{1}^{\eta(p) }=2\eta(p)+\frac{2}{\eta(p)}-4$ So $2\eta(p)^{2}+\frac{2}{\eta(p)}-4-p=0$

$\tilde{\mu}:=2\eta(p)+\frac{2}{\eta(p)}-4.5=0$. So $2\eta(p)^{2}-4.5\eta(p)+2=0$, $4\eta(p)^{2}-9\eta(p)+4=0$ So $\tilde{\mu}=.60961$

c) Compute E(X) and V(X).

$E(X)=\intop_{1}^{2}(2x-\frac{1}{x})dx=x^{2}-ln(x)\mid_{1}^{2}=4-ln(2)-1+ln(1)=2.3069
$

$E(X^{2})=\intop_{1}^{2}(2x^{2}-1)dx=\frac{2}{3}x^{3}-x\mid_{1}^{2}=\frac{16}{3}-2-\frac{2}{3}+1=\frac{11}{3}=3\frac{2}{3}$ $V(X)=E(X^{2})-[E(X)]^{2}=-1.6551$

d) If 1.5 thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the 1.5 thousand gallons is expected to be left at the end of the week? [Hint: Let h(x) = amount left when demand = x.]
Since my other answers do not make sense, I haven't attempted this one yet.

2. EditCant help on the percentile one, sorry.

c) Compute E(X) and V(X).

$E(X)=\intop_{1}^{2}(2x-\frac{1}{x})dx=x^{2}-ln(x)\mid_{1}^{2}=4-ln(2)-1+ln(1)=2.3069
$

$E(X^{2})=\intop_{1}^{2}(2x^{2}-1)dx=\frac{2}{3}x^{3}-x\mid_{1}^{2}=\frac{16}{3}-2-\frac{2}{3}+1=\frac{11}{3}=3\frac{2}{3}$
I think you mean
$E(X})=\intop_{1}^{2}(2x-\frac{2}{x})dx=....$
$E(X^{2})=\intop_{1}^{2}(2x^{2}-2)dx=....$

will that solve the problem?