Answer:
a) 0.7287
b) 0.9663
c) 0.237
d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4.5 million tons of cargo per week
Standard Deviation, σ = 0 .82 million tons
We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.
Formula:
![z_{score} = \displaystyle\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z_%7Bscore%7D%20%3D%20%5Cdisplaystyle%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
a) P( port handles less than 5 million tons of cargo per week)
P(x < 5)
![P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)](https://tex.z-dn.net/?f=P%28%20x%20%3C%205%29%20%3D%20P%28%20z%20%3C%20%5Cdisplaystyle%5Cfrac%7B5%20-%204.5%7D%7B0.82%7D%29%20%3D%20P%28z%20%3C%200.609%29)
Calculation the value from standard normal z table, we have,
![P(x < 5) =0.7287= 72.87\%](https://tex.z-dn.net/?f=P%28x%20%3C%205%29%20%3D0.7287%3D%2072.87%5C%25)
b) P( port handles 3 or more million tons of cargo per week)
![P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)](https://tex.z-dn.net/?f=P%28x%20%5Cgeq%203%29%20%3D%20P%28z%20%5Cgeq%20%5Cdisplaystyle%5Cfrac%7B3-4.5%7D%7B0.82%7D%29%20%3D%20P%28z%20%5Cgeq%20%E2%88%921.82926%29%5C%5C%5C%5CP%28%20z%20%5Cgeq%20%E2%88%921.82926%29%20%3D%201%20-%20P%28z%20%3C%20-1.829%29)
Calculating the value from the standard normal table we have,
![1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%](https://tex.z-dn.net/?f=1%20-%200.0337%20%3D%200.9663%20%3D%2096.63%5C%25%5C%5CP%28%20x%20%5Cgeq%203%29%20%3D%2096.63%5C%25)
c)P( port handles between 3 million and 4 million tons of cargo per week)
![P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%](https://tex.z-dn.net/?f=P%283%20%5Cleq%20x%20%5Cleq%204%29%20%3D%20P%28%5Cdisplaystyle%5Cfrac%7B3%20-%204.5%7D%7B0.82%7D%20%5Cleq%20z%20%5Cleq%20%5Cdisplaystyle%5Cfrac%7B4-4.5%7D%7B0.82%7D%29%20%3D%20P%28-1.829%20%5Cleq%20z%20%5Cleq%20-0.609%29%5C%5C%5C%5C%3D%20P%28z%20%5Cleq%20-0.609%29%20-%20P%28z%20%3C%20-1.829%29%5C%5C%3D%200.271-0.034%20%3D%200.237%3D%2023.7%5C%25)
![P(3 \leq x \leq 4) = 23.7\%](https://tex.z-dn.net/?f=P%283%20%5Cleq%20x%20%5Cleq%204%29%20%3D%2023.7%5C%25)
d) P(X=x) = 0.85
We have to find the value of x such that the probability is 0.85.
P(X > x)
Calculation the value from standard normal z table, we have,
![P( z \leq -1.036) = 0.15](https://tex.z-dn.net/?f=P%28%20z%20%5Cleq%20-1.036%29%20%3D%200.15)
Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.