Peter and Paul agree to meet at a restaurant at noon. Peter arrives at a time normally distributed with mean 12:00 noon, and standard deviation 5 minutes. Paul arrives at a time normally distributed with mean 12:02 P.M., and standard deviation 3 minutes. Assuming the two arrival times are independent, find the chance that (a) Peter arrives before Paul; (b) both men arrive within 3 minutes of noon; (c) the two men arrive within 3 minutes of each other.

Answer:0.5137,0.3082,0.3955Step-by-step explanation:Let X be the arriving time of Peter and Y of Paul.X is N(12,5) Y Is N(12.02, 3)Hence X-Y is normal with mean [tex]=12-12.02 =-0.02[/tex]Variance = [tex]5^2+3^2 =34[/tex]Std dev = 5.831X-Y is N(-0.02, 5.831)a) Prob Peter arrives before Paul;= [tex]P((X-Y)\leq 0)=P(Z\leq \frac{0.2}{5.831} ) \\=0.5137[/tex]b) both men arrive within 3 minutes of noon;=[tex]P(|x-12|<3) *P(|Y-12.02|<3\\= P(|Z|<0.6)*P(|z|<1)\\= 0.4514*0.6826\\=0.3082[/tex]c) both men arrive within 3 minutes of noon;=[tex]P(|x-y|<3)= P(|z|<\frac{3.02}{5.831} \\=P(|z|<0.52)\\=0.3955[/tex]