Question
please can anyone help with question 1
Question 1. [12 Marks] Two airplanes are lying in the same direction in adjacent parallel corridors. At time t-0, the first airplane is 15 km behind the second one. Suppose the speed of the first plane (km/hr) is normally distributed with mean 510 and standard deviation 14 and the second plane's speed is also normally distributed with mean and standard deviation 498 and 14, respectively. a) [3 marks] What is the probability that after 1 hour of flying, the first plane has not caught up to the second plane? b) [3 marks] Determine the probability that the planes are apart by at least 10 km after 1 hour. c) 13 marks] What is the probability that after 2 hours of flying, the first plane has caught up to the second plane? d) (3 marks] Determine the probability that the planes are apart by at most 10 km after 2 hours.
Explanation / Answer
X1~N1(510,14) mu=510,S1=14
X2~N2(498,14). Mu=498,s1=14
A)distance between second and first aeroplane is given by
y=-X1+X2+15
E(y) =E(-X1+X2+15)
=-E(X1)+E(X2)+15
= -Mu1+Mu2+15
=-510+498+15
= 3
V(y) =V(-X1+X2+15)
=V(X1)+V(X2)
=S1^2 +S2^2
=14^2+14^2
=392
S(t)=?392. =19.80
A
(y>0) =p(z>0-3/19.80)
=P(z>-0.15)
=0.0596+0.5
=0.5596
B)p(y>10)=p(y-E(y)/s(t)>10-3/19.8)
=. P(z>0.36)
=0.5-0.1406
=0.3594