There are bus and train services between Town A to Town B. A student living in Town A travels to his school in Town B by bus or train. The journey times are independent and normally distributed with mean and standard deviations, in minutes as shown below.
BUS: mean is 35 and standard deviation is 10 Train: mean is 40 and standard deviation is 2
Question 1: two students who live in Town A set out at the same time to travel to school, one by bus and the other by train. Find the probability that the student who travelled by train will arrive in school first.
Let the time taken by the student who travel by bus be X and the time by train be Y.
Finding the expected value for Y - X : 40 - 35 = 5
Find the variance value for Y - X : $10^2$ + $2^2$ = 104
Y-x is normally distributed with mean of 5 and variance of 104
Using graphic calculator
1 - P(Y-X>0) = 0.312
Question 2: one of the student will arrive more than 10 minutes after the other.
X-Y is normally distributed with mean of -5 and variance of 104
Using graphic calculator:
P(X-Y >10) = 0.070663
Are both my answers or approach/ steps correct? If not what should it be?
Your first answer is correct.
The second answer is not, because the event that one of the students arrives more than 10 minutes after the other, implies that $$|X - Y| > 10;$$ that is to say, the train arrives more than $10$ minutes after the bus, or the bus could arrive more than $10$ minutes after the train. You have the second case but not the first.
To compute this probability, you could write $$\Pr[|X - Y| > 10] = \Pr[X - Y > 10] + \Pr[X - Y < -10].$$ The first term you already calculated; all you need to do now is compute the second and add the two together.