Answer:
0.3888
Step-by-step explanation:
To solve the exercise we need a neutral time, so we define at 12:00 as time 0.
If the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.
If the woman arrives in a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.
Assuming that each one arrives at 12: X and 12: Y for men and women, then the space of our time is defined as
Man [-20,10]
Woman [-15,20]
So,
[tex]f_x(x)=\frac{1}{10-(-20)}=\frac{1}{30}\\f_y(y)=\frac{1}{20-(-15)}=\frac{1}{35}[/tex]
Then,
[tex]f_{xy}(x,y)=f_x(x)f_y(y)=\frac{1}{30} \frac{1}{35} = \frac{1}{1050}[/tex]
The probability of finding an arrival less than 5 minutes is
[tex]P(|X-Y|\leq 5) = \int\limit^20_{-15} \int\limit^{x+5}_{x-5} f(x,y)dydx[/tex]
[tex]P(|X-Y|\leq 5)= \frac{1}{1050}\int\limit^{20}_{-15} y|^{x+5}_{x-5}dx[/tex]
[tex]P(|X-Y|\leq 5)= \frac{35*10}{900}=0.3888[/tex]
The probability that the first to arrive waits no longer than 5 minutes is 0.3888.
We would make use of a neutral time so we define 12:00 as time 0.
Hence, if the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.
Also, if the woman arrives at a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.
Therefore, if we are to assume that each one arrives at 12: X and 12: Y= for men and women, then the space of our time is defined as
fx(x)= [tex]1/10-(-20)= 1/30[/tex]
fy(y)= [tex]1/20-(-15)= 1/35[/tex]
Then,
[tex]fxy(x,y) = fx(x) fy(y)= 1/30.1/35 = 1/1050[/tex]
The probability of finding an arrival less than 5 minutes is
P(|X-Y|≤ 5) = 35 x 10/900
=0.3888.
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