Respuesta :
Answer:
(A) Probability that the port handles less than 5 million tons of cargo per week is 0.7291.
(B) Probability that the port handles 3 or more million tons of cargo per week is 0.9664.
(C) 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 3.65 million tons.
Step-by-step explanation:
We are given that the port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge.
The U.S. Army Corps of Engineers reports that the port handles a mean of 4.5 million tons of cargo per week with a standard deviation of 0.82 million tons.
Let X = Number of tons of cargo handled per week
SO, X ~ Normal([tex]\mu=4.5,\sigma^{2}=0.82^{2}[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 4.5 million tons
[tex]\sigma[/tex] = standard deviation = 0.82 million tons
(A) Probability that the port handles less than 5 million tons of cargo per week is given by = P(X < 5 million tons)
P(X < 5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{5-4.5}{0.82}[/tex] ) = P(Z < 0.61) = 0.7291
The above probability is calculated by looking at the value of x = 0.61 in the z table which has an area of 0.7291.
(B) Probability that the port handles 3 or more million tons of cargo per week is given by = P(X [tex]\geq[/tex] 3 million tons)
P(X [tex]\geq[/tex] 3) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{3-4.5}{0.82}[/tex] ) = P(Z [tex]\geq[/tex] -1.83) = P(Z [tex]\leq[/tex] 1.83)
= 0.9664
The above probability is calculated by looking at the value of x = 1.83 in the z table which has an area of 0.9664.
(C) Probability that the port handles between 3 million and 4 million tons of cargo per week is given by = P(3 million tons < X < 4 million tons)
P(3 million tons < X < 4 million tons) = P(X < 4) - P(X [tex]\leq[/tex] 3)
P(X < 4) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{4-4.5}{0.82}[/tex] ) = P(Z < -0.61) = 1 - P(Z [tex]\leq[/tex] 0.61)
= 1 - 0.7291 = 0.2709
P(X [tex]\leq[/tex] 3) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{3-4.5}{0.82}[/tex] ) = P(Z [tex]\leq[/tex] -1.83) = 1 - P(Z < 1.83)
= 1 - 0.9664 = 0.0336
The above probability is calculated by looking at the value of x = 0.61 and x = 1.83 in the z table which has an area of 0.7291 and 0.9664 respectively.
Therefore, P(3 < X < 4) = 0.2709 - 0.0336 = 0.2373
(D) Now, it is given that 85% of the time the port can handle the weekly cargo volume without extending operating hours, that means;
P(X > x) = 0.85 {where x is the required no. of tons of cargo}
P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-4.5}{0.82}[/tex] ) = 0.85
P(Z > [tex]\frac{x-4.5}{0.82}[/tex] ) = 0.85
In the z table, the critical value of x which represents top 85% area is given as -1.036, that is;
[tex]\frac{x-4.5}{0.82} = -1.036[/tex]
[tex]{x-4.5} = -1.036\times 0.82[/tex]
x = 4.5 - 0.85 = 3.65 million tons
Hence, the number of tons of cargo per week that will require the port to extend its operating hours is 3.65 million tons.
