Answer: 0.5507
Step-by-step explanation:
Given : The waiting time, in hours, between successive speeders spotted by a radar unit is a continuous random variable X with a cumulative distribution function
[tex]F(x)= \begin{cases}0,& x<0 \\ 1-e^{-4x}, & x\geq0\end{cases}\end{document}[/tex]
Since , the waiting time is in hours , then we can write 12 minutes as [tex]\dfrac{12}{60}[/tex] hour i.e.0.2 hour.
Now, the probability of waiting fewer than 12 minutes between successive speeders is given by :-
[tex]P(X<0.2)=1-e^{-4(0.2)}=1-e^{-0.8}\\\\=1- 0.449328964117=0.550671035883\approx0.5507[/tex]
Hence, the required probability = 0.5507