Answer :
Answer:
a) Figure attached
b) [tex] P(X<130 min)= F(130) = \frac{130-120}{20}=0.5[/tex]
c) [tex] P(X>135) =1-P(X<135) = 1- F(135) = 1- \frac{135-120}{20}= 0.25[/tex]
d) [tex] E(X) =\frac{a+b}{2}= \frac{120+140}{2}= 130min[/tex]
Step-by-step explanation:
2 hr = 120 min
2hr 20 min = 120+20 = 140 min
Let X the random variable that represent "flight time". And we know that the distribution of X is given by:
[tex]X\sim Uniform(120 ,140)[/tex]
For this case the density function for A is given by:
[tex] f(X) = \frac{1}{b-a}= \frac{1}{20} , 120\leq X \leq 140 [/tex]
And [tex] f(X)= 0[/tex] for other case
Part a
For this case we can see the figure attached. We see that the probability density function is defined between 120 and 140 minutes.
Part b
The arrival time is 2hr 5 min = 125 min. So then 5 minutes leate means 130 min
For this case we want to find this probability:
[tex] P(X<130 min)[/tex]
And we can use the cumulative distribution function for X given by:
[tex] F(X) = \frac{x-120}{140-120} = \frac{x-120}{20} , 120\leq X \leq 140 [/tex]
And if we use this formula we got:
[tex] P(X<130 min)= F(130) = \frac{130-120}{20}=0.5[/tex]
Part c
For this case more than 10 minutes late means 125 +10 = 135 min or more, so we want this probability:
[tex] P(X>135)[/tex]
And using the complement rule we have:
[tex] P(X>135) =1-P(X<135) = 1- F(135) = 1- \frac{135-120}{20}= 0.25[/tex]
Part d
The expected value for the uniform distribution is given by:
[tex] E(X) =\frac{a+b}{2}= \frac{120+140}{2}= 130min[/tex]
