Answer to Question #199143 in Statistics and Probability for mandona

Question

A manufacturer has observed that the time that elapses between the placement of an order with a just-in-time supplier and the delivery of parts is uniformly distributed between 100 and 180 minutes. a) What proportion of orders takes between 2 and 2.5 hours to be delivered? b) What proportion of orders takes between 2 and 3 hours to be delivered? c) What is the expected delivery time? d) What is the standard deviation of the delivery time? e) The contract between the manufacturer and supplier stipulates that the cost of the order will be $4,000 minus ten dollars per minute that it takes between the order and the delivery. What is the expected cost of an order? What is the standard deviation of the order cost? What is the probability that a continuous random variable X takes on any

Answer 1

For uniform distribution:

$P(x_1<x<x_2)=\frac{x_2-x_1}{b-a}$

We have:

$b-a=180-100=80\ min$

a)

$x_1=2\ hours=120\ min$

$x_2=2.5\ hours=150\ min$

$x_2-x_1=2.5-2=0.5\ hour=30\ min$

$P(120<x<150)=\frac{30}{80}=\frac{3}{8}$

b)

$x_1=2\ hours=120\ min$

$x_2=3\ hours=180\ min$

$P(120<x<180)=\frac{180-120}{80}=\frac{60}{80}=\frac{3}{4}$

c)

$E(X)=\frac{a+b}{2}=\frac{100+180}{2}=140$

d)

$\sigma=\sqrt{V(X)}=\frac{b-a}{\sqrt{12}}=\frac{180-100}{\sqrt{12}}=23.09$

e)

For cost distribution:

$a=4000-10\cdot180=2200$

$b=4000-10\cdot100=3000$

$E(X)=\frac{2200+3000}{2}=2600$

$\sigma=\frac{3000-2200}{\sqrt{12}}=230.94$