On his way to work every morning, Bob first takes the bus from his house, exits

Question

On his way to work every morning, Bob first takes the bus from his house, exits near his workplace, and walks the remaining distance. His time spent on the bus (X) is a random variable that follows a normal distribution, with mean mu = 20 minutes, and standard deviation sigma = 2 minutes, i.e., X N(20, 2). Likewise, his walking time (K) is also a random variable that follows a normal distribution, with mean mu = 10 minutes, and standard deviation sigma = 1.5 minutes, i.e., Y N(10, 1.5). Find the probability that Bob arrives at his workplace in 35 minutes or less.

Explanation / Answer

X~ N(20 2), Y~ N( 10, l.5)
X + Y ~ N( 20+10, 2+1.5)
X + Y ~ N( 30, 3.5 )

Mean ( u ) =30
Standard Deviation ( sd )=3.5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

P(X > 35) = (35-30)/3.5
= 5/3.5 = 1.4286
= P ( Z >1.429) From Standard Normal Table
= 0.0766                  

P(X < = 35) = (1 - P(X > 35)
= 1 - 0.0766 = 0.9234  

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