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!!!
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