Question

Problem:

A university administrator is interested in whether a new building can be planned and built on campus within a four-year time frame. He considers the process in two phases. Phase I: Phase I involves lobbying the state legislature and governor for permission and funds, issuing bonds to obtain funds, and obtaining all the appropriate legal documents. Past experience indicates that the time required to complete phase I is approximately normally distributed with a mean of 16 months and standard deviation of 4 months. If X = phase I time, then X ~ N(μ = 16 months, σ = 4 months). Phase II: Phase II involves creation of blue prints, obtaining building permits, hiring contractors, and, finally, the actual construction of the building. Past data indicates that the time required to complete these tasks is approximately normally distributed with a mean of 18 months and a standard deviation of 12 months. If Y = phase II time, then Y ~ N(μ = 18 months, σ = 12 months). a) A new random variable, T = total time for completing the entire project, is defined as T = X + Y. What is the probability distribution of T? (Give both the name of the distribution and its parameters.) b) Find the probability that the total time for the project is less than four years. (In symbols, calculate P(T < 48 months).) c) Find the 95th percentile of the distribution of T.

Answer 1

X denotes phase 1 time and X ~ N(μ = 16 months, σ = 4 months)

Y denotes phase 2 time and Y ~ N(μ = 18 months, σ = 12 months)

a) T denotes total time for the project ==> T = X + Y

** Sum of two normal distributions also follows a normal distribution with mean equal to the sum of the individual means and variance equal to the sum of individual variances

Therefore T ~ N(μ = 18+16 = 34 months, σ = = 12.65 months)

b) P(T < 48) =

where Z follows standard normal distribution

From Z table P(Z < 1.11) = 0.8665

Therefore the probability that the total time for the project is less than four years is 0.8665

c) Let be the 95th percentile of distribution of T

P(T < ) = =0.95

where Z follows standard normal distribution

From Z table P(Z < 1.645) = 0.95

Therefore the 95th percentile of the distribution of T is 1.645*12.65+34 = 54.81 months