Question

You may need to use the appropriate appendix table or technology to answer this 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.† Assume that the number of tons of cargo handled per week is normally distributed with a standard deviation of 0.84 million tons.

(a) What is the probability that the port handles less than 5 million tons of cargo per week? (Round your answer to four decimal places.)

(b) What is the probability that the port handles 3 or more million tons of cargo per week? (Round your answer to four decimal places.)

(c) What is the probability that the port handles between 3 million and 4 million tons of cargo per week? (Round your answer to four decimal places.)

(d) Assume that 83% 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? (Round your answer to one decimal places.) tons of cargo per week

Answer 1

Solution:

Let X be a random variable which represents the weight of corgo handled by the US army corps.

Given that, X ~ N(4.5, 0.84²)

i.e.  μ = 4.5 million tons and σ = 0.84 million tons

a) We have to obtain P(X < 5 million tons).

We know that if X ~ N(μ ,σ^{​​​​​​2}) then

Using "pnorm" function of R we get, P(Z < 0.5952) = 0.7241

Hence, the probability that the port handles less than 5 million tons of cargo per week is 0.7241.

b) We have to obtain P(X ≥ 3 million tons).

We know that if X ~ N(μ ,σ^{​​​​​​2}) then

Using "pnorm" function of R we get, P(Z ≥ -1.7857) = 0.9629

Hence, the probability that the port handles 3 or more million tons of cargo per week is 0.9629.

c) We have to obtain P(3 < X < 4 ).

We know that if X ~ N(μ ,σ^{​​​​​​2}) then

Using "pnorm" function of R we get,

P(Z < -0.5952) = 0.2759 and P(Z < -1.7857) = 0.0371

Hence, the probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2388.

d) The value corresponding to 83th percentile will be the required number of million tons cargo so that they need to extend their operating hours.

Let 83th percentile be k.

P(X < k) = 0.83

We know that if X ~ N(μ ,σ^{​​​​​​2}) then

......................(1)

Using "qnorm" function of R we get, P(Z < 0.9542) = 0.83

Comparing, P(Z < 0.9542) = 0.83 and (1) we get,

The 95th percentile is 5.3

Hence, 5.3 million tons of cargo per week that will require the port to extend its operating hours.

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