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2 answers:
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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:
a) P( port handles less than 5 million tons of cargo per week)
P(x < 5)
Calculation the value from standard normal z table, we have,
b) P( port handles 3 or more million tons of cargo per week)
Calculating the value from the standard normal table we have,
c)P( port handles between 3 million and 4 million tons of cargo per week)
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,
Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.