Question #199143

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


Expert's answer

For uniform distribution:

P(x1<x<x2)=x2x1baP(x_1<x<x_2)=\frac{x_2-x_1}{b-a}

We have:

ba=180100=80 minb-a=180-100=80\ min


a)

x1=2 hours=120 minx_1=2\ hours=120\ min

x2=2.5 hours=150 minx_2=2.5\ hours=150\ min

x2x1=2.52=0.5 hour=30 minx_2-x_1=2.5-2=0.5\ hour=30\ min

P(120<x<150)=3080=38P(120<x<150)=\frac{30}{80}=\frac{3}{8}


b)

x1=2 hours=120 minx_1=2\ hours=120\ min

x2=3 hours=180 minx_2=3\ hours=180\ min

P(120<x<180)=18012080=6080=34P(120<x<180)=\frac{180-120}{80}=\frac{60}{80}=\frac{3}{4}


c)

E(X)=a+b2=100+1802=140E(X)=\frac{a+b}{2}=\frac{100+180}{2}=140


d)

σ=V(X)=ba12=18010012=23.09\sigma=\sqrt{V(X)}=\frac{b-a}{\sqrt{12}}=\frac{180-100}{\sqrt{12}}=23.09


e)

For cost distribution:

a=400010180=2200a=4000-10\cdot180=2200

b=400010100=3000b=4000-10\cdot100=3000


E(X)=2200+30002=2600E(X)=\frac{2200+3000}{2}=2600


σ=3000220012=230.94\sigma=\frac{3000-2200}{\sqrt{12}}=230.94


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