Question

Ship-to-shore transfer times. Lack of port facilities or shallow water may require cargo on a large ship to be transferred to a pier in a smaller craft. The smaller craft may often have to cycle back and forth from ship to shore. Researchers developed models of this transfer process that provide estimates of ship-to-shore transfer times (Naval Research Logistics, Vol. 41, 1994). They used an exponential distribution to model the time between arrivals of the smaller craft at the pier.

a. Assume that the mean time between arrivals at the pier is 17 minutes. Give the value of u for this exponential distribution. Graph the distribution.

b. Suppose there is only one unloading zone at the pier available for the small craft to use. If the first craft docks at 10:00 a.m. and doesn’t finish unloading until 10:15 a.m., what is the probability that the second craft will arrive at the unloading zone and have to wait before docking? Applying the Concepts—Advanced

Short answer

1. The value of is 17, and the graph for the exponential distribution will look like

b. The probability that the second craft will arrive at the unloading zone and have to wait before docking is 0.5862.

Step-by-step solution

Given information

Researchers used an exponential distribution to model the time between arrivals of the smaller craft at the pier.

Calculating the value of μ

Given$\mu$=17

Since p.d.f of the exponential distribution is given by,

localid="1660754292801" $f\left(x\right)=\frac{1}{\mu }{e}^{-\frac{x}{\mu }}$

Also,

localid="1660754297655" $f\left(x\right)=\frac{1}{17}{e}^{-\frac{x}{17}}$

Now,

Put x=0,

localid="1660754325516" $\begin{array}{rcl}f\left(0\right)& =& \frac{1}{17}\\ & =& 0.0588\\ & & \end{array}$

 Step 3: Computing the probability that the second craft will arrive at the unloading zone and have to wait before docking

The second craft will have to wait if it arrives between 10:00 a.m. and 10:15 a.m.

Therefore,

$\begin{array}{rcl}P\left(x<15\right)& =& 1-e^{-\frac{15}{17}}\\ & =& 1-e^{-0.88235}\\ & =& 1-0.4138\\ & & \end{array}$

Hence,

$P\left(x<15\right)=0.5862$

The probability that the second craft will arrive at the unloading zone and have to wait before docking is 0.5862.