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. Assume that the number of tons of cargo handled per week is normally distributed with a standard deviation of .82 million tons.
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)?
To find the number of tons of cargo per week that will require the port to extend its operating hours, we need to find the value that represents the cutoff for the top 85% of the distribution.
To do this, we can use the z-score formula:
z = (x - mean) / standard deviation
Since we want to find the value that represents the top 85% of the distribution, we need to find the z-score that corresponds to the cumulative probability of 0.85. We can use a standard normal distribution table or a calculator to find this value.
The z-score that corresponds to a cumulative probability of 0.85 is approximately 1.04.
Now, we can use the z-score formula to find the corresponding tonnage:
1.04 = (x - 4.5) / 0.82
Solving for x:
x - 4.5 = 1.04 * 0.82
x - 4.5 = 0.8528
x = 4.5 + 0.8528
x = 5.3528
Therefore, the number of tons of cargo per week that will require the port to extend its operating hours is approximately 5.35 million tons.