Respuesta :
Answer:
(a) The probability that the port handles less than 5 million tons of cargo per week is 0.7291.
(b) The probability that the port handles 3 or more million tons of cargo per week is 0.9664.
(c) The 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 5.35 million tons.
Step-by-step explanation:
Let X = amount of cargo the port handles per week.
The random variable X is Normally distributed with parameters,
μ = 4.5 million
σ = 0.82 million
(a)
Compute the probability that the port handles less than 5 million tons of cargo per week as follows:
[tex]P(X<5)=P(\frac{X-\mu}{\sigma}<\frac{5-4.5}{0.82})\\=P(Z<0.61)\\=0.7291[/tex]
*Use a z-table for the probability.
Thus, the probability that the port handles less than 5 million tons of cargo per week is 0.7291.
(b)
Compute the probability that the port handles 3 or more million tons of cargo per week as follows:
[tex]P(X\geq 3)=P(\frac{X-\mu}{\sigma}\geq \frac{3-4.5}{0.82})\\=P(Z>-1.83)\\=P(Z<1.83)\\=0.9664[/tex]
*Use a z-table for the probability.
Thus, the probability that the port handles 3 or more million tons of cargo per week is 0.9664.
(c)
Compute the probability that the port handles between 3 million and 4 million tons of cargo per week as follows:
[tex]P(3<X<4)=P(\frac{3-4.5}{0.82}<\frac{X-\mu}{\sigma}<\frac{4-4.5}{0.82})\\=P(-1.83<Z<-0.61)\\=P(Z<-0.61)-P(Z<-1.83)\\=0.2709-0.0336\\=0.2373[/tex]
Thus, the probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.
(d)
It is provided that P (X < x) = 0.85.
Then, P (Z < z) = 0.85.
The value of z for this probability is:
z = 1.04.
Compute the value of x as follows:
[tex]z=\frac{x-\mu}{\sigma}\\1.04=\frac{x-4.5}{0.82}\\x=4.5-+1.04\times 0.82)\\x=5.3528\approx5.35[/tex]
Thus, the number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.
