Question

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest bulk cargo port in the world. The U.S. Army Corps of Engineers reports that the port handles a mean of 4.5 million tons of cargo per week (USA Today, September 25, 2012). Assume that the number of tons of cargo handled per week is normally distributed with a standard deviation of .82 million tons.
a. What is the probability that the port handles less than 5 million tons of cargo per week (to 4 decimals)?
b. What is the probability that the port handles 3 or more million tons of cargo per week (to 4 decimals)?
c. What is the probability that the port handles between 3 million and 4 million tons of cargo per week (to 4 decimals)?
d. Assume that 85% of the time the port can handle the weekly cargo volume without extending operating hours. What is the number of tons of cargo per week that will require the port to extend its operating hours (to 2 decimals)?

Answer 1

Answer:

(a) The probability that the port handles less than 5 million tons of cargo per week is 0.7291.

(b) The probability that the port handles 3 or more million tons of cargo per week is 0.9664.

(c) The probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.

(d) The number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.

Step-by-step explanation:

Let X = amount of cargo the port handles per week.

The random variable X is Normally distributed with parameters,

μ = 4.5 million

σ = 0.82 million

(a)

Compute the probability that the port handles less than 5 million tons of cargo per week as follows:

$P(X$

*Use a z-table for the probability.

Thus, the probability that the port handles less than 5 million tons of cargo per week is 0.7291.

(b)

Compute the probability that the port handles 3 or more million tons of cargo per week as follows:

$P(X\geq 3)=P(\frac{X-\mu}{\sigma}\geq \frac{3-4.5}{0.82})\\=P(Z>-1.83)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that the port handles 3 or more million tons of cargo per week is 0.9664.

(c)

Compute the probability that the port handles between 3 million and 4 million tons of cargo per week as follows:

$P(3$

Thus, the probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.

(d)

It is provided that P (X < x) = 0.85.

Then, P (Z < z) = 0.85.

The value of z for this probability is:

z = 1.04.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\1.04=\frac{x-4.5}{0.82}\\x=4.5-+1.04\times 0.82)\\x=5.3528\approx5.35$

Thus, the number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.