Confused on how to approach this

Question

have three errands to take care of in the Administration Building. Let Xi = the time that it takes for the ith errand (i = 1, 2, 3), and let X4 = the total time in minutes that I spend walking to and from the building and between each errand. Suppose the Xi's are independent, and normally distributed, with the following means and standard deviations: μ1 = 15, σ1 = 4, μ2 = 5, σ2 = 1, μ3 = 8, σ3 = 2, μ4 = 12, σ4 = 3. I plan to leave my office at precisely 10:00 A.M. and wish to post a note on my door that reads, "I will return by t A.M." What time t should I write down if I want the probability of my arriving after t to be 0.05? (Round your answer to two decimal places.)

Answer 1

To start, we're going to let

Y=total time to complete errands and return to office

Given that X_i=time to run the i-th errand, we can then rewrite Y=Σ⁴_i=1(X_i) or Y=X₁+X₂+X₃+X₄

Because Y is made of normal, independently distributed random variables, we can infer that Y is also independent and normally distributed, and we can then calculate the mean and variance of Y

μ_Y=Σ⁴_i=1(μ_Xi)=μ₁+μ₂+μ₃+μ₄. We have these values given to us above, so we can calculate this as 15+5+8+12=40.

Similarly, Var(Y)=Σ⁴_i=1(Var(X_i)). Here, we have been given not variance values, but standard deviation values. However, all we need to do is square the standard deviation values to make them variance values. So then Var(Y) can be calculated again using the information above to get (σ₁²)+(σ₂²)+(σ₃²)+(σ₄²)

=4²+1²+2²+3²=16+1+4+9=30. Standard Deviation of Y is then √30 = 5.4772 = σ_Y

So now that we have the mean and the st.dev of Y, we can figure out what we need to get Y, and we can get Y at a particular percentile as well using the formula

Y=μ_Y+(σ_Y*Z()). Here, Z() is a percentile that we're looking for, and we can plug this value into a Z table to get the number that we're looking for. Specifically, since we're looking for p=.05, we're in the 95th percentile. This has a corresponding Z value of 1.645. So, putting all of this together is

Y = 40 + (5.4772 * 1.645) = 40 + 9.009994 = 49.01. This is how many minutes the professor will be gone from the office to ensure that they have returned, with all errands completed, in the 95th percentile of returning before the time wrote on the note. If the professor is leaving at 10 am and the time they would need is what we have just calculated out to be 49.01 minutes, then the professor should write that they will be back at 10:50 on the note, if it must be rounded to the best minute. If the professor is writing exactly, then they should write that they will return at 10:49:01 miliseconds (I believe).

I think this is the correct answer. I did my best to explain everything as clear as possible, though I'm unsure of how specific you're being asked to answer on the note. I hope that this is able to clear things up for you, though! Good Luck!