Probability

Question

I 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 = 16, σ1 = 4, μ2 = 7, σ2 = 1, μ3 = 9, σ3 = 2, μ4 = 13, 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." How long should I estimate my trip will take if I want the probability of the trip taking longer than my estimate to be 0.01? (Round your answer to two decimal places.)

Answer 1

We could work out the Z-score for each X_i giving us Z_i.

Z₁=(X₁-16)/4, Z₂=X₂-7, Z₃=(X₃-9)/2, Z₄=(X₄-13)/3. The total time T to complete the errands, including walking, = X₁+X₂+X₃+X₄, X_i=μ_i±σ_iZ_i. 

We need a critical value for Z such that the probability of exceeding this value is 1%. When Z=2.3263 approx, p=99%.

We are considering X_i>μ_i+σ_iZ.

Therefore X₁>25.31, X₂>9.33, X₃>13.65, X₄>19.98, so T=68.26 min, that is, 1hr 8.26min. Therefore t=11:08.26. There is a 1% probability that she will be back later than 11:08.26 am.