Answer:
a) Attached
b) P=0.60
c) P=0.80
d) The expected flight time is E(t)=122.5
Step-by-step explanation:
The distribution is uniform between 1 hour and 50 minutes (110 min) and 135 min.
The height of the probability function will be:
[tex]h=\frac{1}{Max-Min}=\frac{1}{135-110} =\frac{1}{25}[/tex]
Then the probability distribution can be defined as:
[tex]f(t)=\frac{1}{25}=0.04 \,\,\,\,\\\\t\in[110,135][/tex]
b) No more than 5 minutes late means the flight time is 125 or less.
The probability of having a flight time of 125 or less is P=0.60:
[tex]F(T<t)=0.04(t-min)\\\\F(T<125)=0.04*(125-110)=0.04*15=0.60[/tex]
c) More than 10 minutes late means 130 minutes or more
The probability of having a flight time of 130 or more is P=0.80:
[tex]F(T>t)=1-0.04(t-110)\\\\F(T>130)=1-0.04*(130-110)=1-0.04*20=1-0.8=0.2[/tex]
d) The expected flight time is E(t)=122.5
[tex]E(t)=\frac{1}{2}(max+min)= \frac{1}{2}(135+110)=\frac{1}{2}*245=122.5[/tex]