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.

0.5137,0.3082,0.3955

Step-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

Variance =

Std dev = 5.831

X-Y is N(-0.02, 5.831)

a) Prob Peter arrives before Paul;

=

b) both men arrive within 3 minutes of noon;

=

c) both men arrive within 3 minutes of noon;

=

to have more of what you want like money

step-by-step explanation:

save it by not spending on unnecessary stuff and have more

step-by-step explanation:

use context clues to figure out the meaning of the bolded word:  

the sage gardener knew when she had to water her plants.  

question 8 options:

mint plant

wise

happy

lazy