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he port of South Louisiana, located along 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 million tons of cargo per week (USA Today, September , ). Assume that the number of tons of cargo handled per week is normally distributed with a standard deviation of million tons. a. What is the probability that the port handles less than million tons of cargo per week (to 4 decimals)? b. What is the probability that the port handles or more million tons of cargo per week (to 4 decimals)? c. What is the probability that the port handles between million and 4 million tons of cargo per week (to 4 decimals)? d. Assume that 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)?

Respuesta :

Answer:

a) P(x < 5) = 0.7291

b) P(x ≥ 3) = 0.9664

c) P(3 < x < 4) = 0.2373

d) 5.35 million tons of cargo in a week will require the port to extend operating hours.

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 4.5 million tons of cargo per week

Standard deviation = σ = 0.82 million

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

= P(x < 5)

We first standardize/normalize 5.

The standardized score of any value is the value minus the mean divided by the standard deviation.

z = (x - μ)/σ = (5 - 4.5)/0.82 = 0.61

To determine the probability that the port handles less than 5 million tons of cargo per week

P(x < 5) = P(z < 0.61)

We'll use data from the normal probability table for these probabilities

P(x < 5) = P(z < 0.61) = 0.72907 = 0.7291 to 4 d.p

b) The probability that the port handles 3 or more million tons of cargo per week?

P(x ≥ 3)

We first standardize/normalize 3.

z = (x - μ)/σ = (3 - 4.5)/0.82 = -1.83

To determine the probability that the port handles less than 3 or more million tons of cargo per week

P(x ≥ 3) = P(z ≥ -1.83)

We'll use data from the normal probability table for these probabilities

P(x ≥ 3) = P(z ≥ -1.83)

= 1 - P(z < -1.83)

= 1 - 0.03362

= 0.96638 = 0.9664

c) The probability that the port handles between 3 million and 4 million tons of cargo per week = P(3 < x < 4)

We first standardize/normalize 3 and 4.

For 3 million

z = (x - μ)/σ = (3 - 4.5)/0.82 = -1.83

For 4 million

z = (x - μ)/σ = (4 - 4.5)/0.82 = -0.61

To determine the probability that the port handles between 3 million and 4 million tons of cargo per week

P(3 < x < 4) = P(-1.83 < z < -0.61)

We'll use data from the normal probability table for these probabilities

P(3 < x < 4) = P(-1.83 < z < -0.61)

= P(z < -0.61) - P(z < -1.83)

= 0.27093 - 0.03362

= 0.23731 = 0.2373 to 4 d.p

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?

Let x' represent the required number of tons of cargo thay will require the port to extend its operating hours.

Let its z-score be z'

P(x < x') = P(z < z') = 85% = 0.85

Using the normal distribution table,

z' = 1.036

z' = (x' - μ)/σ

1.036 = (x' - 4.5)/0.82

x' - 4.5 = (0.82 × 1.036) = 0.84952

x' = 0.84952 + 4.5 = 5.34952 = 5.35 to 2 d.p

Therefore, 5.35 million tons of cargo in a week will require the port to extend its operating hours.

Hope this Helps!!!

lhmarianateixeira

In this exercise we have to use probability knowledge, so this will result in:

a) [tex]P(X < 5) = 0.7291[/tex]

b) [tex]P(X \geq 3) = 0.9664[/tex]

c) [tex]P(3 < x < 4) = 0.2373[/tex]

d) 5.35 million tons of cargo in a week will require the port to extend operating hours.

This is a normal distribution problem with: Mean = μ = 4.5 million tons of cargo per week Standard deviation = σ = 0.82 million

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

[tex]= P(X < 5)[/tex]

We first standardize/normalize 5. The standardized score of any value is the value minus the mean divided by the standard deviation:  

[tex]z = (X - \mu )/\sigma = (5 - 4.5)/0.82 = 0.61[/tex]

To determine the probability that the port handles less than 5 million tons of cargo per week:

[tex]P(X < 5) = P(Z < 0.61)[/tex]

We'll use data from the normal probability table for these probabilities:  

[tex]P(X < 5) = P(Z < 0.61) = 0.72907 = 0.7291[/tex]

b) The probability that the port handles 3 or more million tons of cargo per week?

[tex]P(x \geq 3)[/tex]

We first standardize/normalize 3:

[tex]Z = (X - \mu )/\sigma = (3 - 4.5)/0.82 = -1.83[/tex]

To determine the probability that the port handles less than 3 or more million tons of cargo per week:

[tex]P(X \geq 3) = P(Z \geq -1.83)[/tex]

We'll use data from the normal probability table for these probabilities

[tex]P(X \geq 3) = P(Z \geq -1.83)\\= 1 - P(Z \leq (-1.83))\\= 1 - 0.03362\\= 0.96638 = 0.9664[/tex]

c) The probability that the port handles between 3 million and 4 million tons of cargo per week  [tex]= P(3 < x < 4)[/tex] .

We first standardize/normalize 3 and 4. For 3 million:

[tex]Z = (X - \mu )/\sigma = (3 - 4.5)/0.82 = -1.83[/tex]

For 4 million:

[tex]z = (X - \mu)/\sigma= (4 - 4.5)/0.82 = -0.61[/tex]

To determine the probability that the port handles between 3 million and 4 million tons of cargo per week:

[tex]P(3 < x < 4) = P(-1.83 < z < -0.61)[/tex]

We'll use data from the normal probability table:

[tex]P(3 < x < 4) = P(-1.83 < z < -0.61)\\= P(z \leq -0.61) - P(z < -1.83)\\= 0.27093 - 0.03362\\= 0.23731 = 0.2373[/tex]

d) Let x' represent the required number of tons of cargo thay will require the port to extend its operating hours. Let its z-score be z':

[tex]P(x < x') = P(z < z') = 85%[/tex]

Using the normal distribution table:

[tex]z' = 1.036\\z' = (x' - \mu)/\sigma\\1.036 = (x' - 4.5)/0.82\\x' - 4.5 = (0.82 × 1.036) = 0.84952\\x' = 0.84952 + 4.5 = 5.34952 = 5.35[/tex]

Therefore, 5.35 million tons of cargo in a week will require the port to extend its operating hours.

See more about probability at brainly.com/question/795909

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