Answer:
The expected number of days per week the captain can leave port is 4.05.
Step-by-step explanation:
We are given the following distribution:
[tex]P(X = 0) = 0.2[/tex]
[tex]P(X = 1) = 0.05[/tex]
[tex]P(X = 2) = 0.05[/tex]
[tex]P(X = 3) = 0.05[/tex]
[tex]P(X = 4) = 0.15[/tex]
[tex]P(X = 5) = 0.1[/tex]
[tex]P(X = 6) = 0.15[/tex]
[tex]P(X = 7) = 0.25[/tex]
What is the expected number of days per week the captain can leave port.
We multiply each outcome by its probability. So
[tex]E = 0*0.2 + 1*0.05 + 2*0.05 + 3*0.05 + 4*0.15 + 5*0.1 + 6*0.15 + 7*0.25 = 4.05[/tex]
The expected number of days per week the captain can leave port is 4.05.