Question

I have three errands to take care of in the Administration Building. Let  X_i = 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

X_i's are independent, and normally distributed, with the following means and standard deviations:

μ₁ = 16,

σ₁ = 4,

μ₂ = 6,

σ₂ = 1,

μ₃ = 8,

σ₃ = 2,

μ₄ = 14,

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

Given,

Xi = the time that it takes for the ith errand (i=1,2,3)

X4 = the total time in minutes that I spend walking to and from the building and between each errand

Let Total time taken for the trip is X

Then X =  X1+X2+X3+X4

Given,

Xi's are independent, and normally distributed, with the following means and standard deviations:

μ1 = 16,

σ1 = 4,

μ2 = 6,

σ2 = 1,

μ3 = 8,

σ3 = 2,

μ4 = 14,

σ4 = 3.

Then

X : (X1+X2+X3+X4) follows a normal distribution with mean = () and standard deviation :

=   = 16+6+8+14 = 44

Therefore,

Total time taken for the trip is X follows a normal distribution with mean = 44 nd standard deviation : = 5.4772

Let 'mx' be the estimate for the time it takes for the trip.

Probability of the trip taking longer than estimate: 'mx' to be 0.01 i.e P(X>mx) = 0.01

P(X > mx ) = 0.01

So we need to find the value of mx such that P(X>mx) = 0.01

P(X>mx) = 1-P(Xmx)

1-P(Xmx) = 0.01 ; P(Xmx) = 1-0.01 = 0.99

P(Xmx) = 0.99

Zx : Zscore of mx = (mx - )/ = (mx - 44)/ 5.4772

P(Xmx) = P(Z<Zx) = 0.99

From standard normal tables ,

Zx = 2.33

Zx = (mx - 44)/ 5.4772;

mx = 44 + 5.4772 Zx =44 + 5.4772 x 2.33 = 44+ 12.761876 = 56.761876 56.76

Estimate time of the trip is 56.76 minutes.