Question

The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv X with the following pdf f(x)-21 - 0 1 sxs 2 2 otherwise (a) Compute the cdf of X. 0 F(x)2 2 > X (b) Obtain an expression for the (100p)th percentile n(p) 2 What is the value of μ? (Round your answer to three decimal places.) 2 (c) Compute E(X) and V(X). (Round your answers to four decimal places.) E(X) = 2 V(X) = 2 x thousand gallons x thousand gallons squared (d) If 1.4 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 1.4 thousand gallons is expected to be left at the end of the week? [Hint: Let h(x) = amount left when demand = (Round your answer to three decimal places.) 2 x thousand gallons

Answer 1

(a)

From the given pdf, the CDF of X is

- 1 1 f (x)dx F(x) P(X ) 2 1 dr 1 2

Hence, the CDF is

0. 1 F()2 -2, 1 2 1, 2 r

(b)

Let $\eta (p)$ shows the 100pth percentile. So we have

PX np)= p

1 - 2 P η(p) 2 | η (p) +

P n(P)np +2 (p) 2 IG

2(p)1 2n(p)

n2(p) 2 7(p ) = 0 2np)

Let b=-(p/2+2) so

(p) 2) ± /+2}? _ 4 (p+4)t Vp2 8p 2 2 4

Now the root lies between 1 and 2 so pth percentile is

(p4) p2 +8p 4

$\eta (p)=\frac{(p+4)+\sqrt {p^{2}+8p}}{4}$

For median we have p=0.5 so

(0.5 4)0.58.0.5 1.64 4

(c)

The expected value of X is

$E(X)=\int_{1}^{2}xf(x)dx=\int_{1}^{2}2\left [x-\frac{1}{x} \right ]dx=2\left [ \frac{x^{2}}{2}-ln(x) \right ]_{1}^{2}=2\left [2-ln(2)-\frac{1}{2}+0 \right ]=3-2ln(2)=1.6137$Now

$E(X^{2})=\int_{1}^{2}x^{2}f(x)dx=\int_{1}^{2}2\left [x^{2}-1 \right ]dx=2\left [ \frac{x^{3}}{3}-x \right ]_{1}^{2}=2\left [\frac{8}{3}-2-\frac{1}{3}+1 \right ]=\frac{8}{3}=2.6667$

So variance will be

$Var(X)=E(X^{2})-[E(X)]^{2}=0.0627$

(d)

Let random variable Y shows the amount left. So we have

Y = 1.4 -x

Therefore

E(Y) = 1.4-E(X) = 1.4 - 1.6137 = -0.2137