Topic: Normal Distribution A survey reveals that the waiting time for the bus se

Question

Topic: Normal Distribution

A survey reveals that the waiting time for the bus service 179 during off-peak hours is approximated by a normal distribution, whose mean and standard deviation are respectively 7 and 2 minutes.

(a) Find the probability of waiting for less than 6 minutes before boarding (4 decimals).

(b) Find the probability of waiting for more than 8 minutes before boarding (4 decimals).

(c) Assume that you have waited 5 minutes at the bus stop. What is the probability that you wait for a total of more than 8 minutes before boarding (4 decimals)?

Explanation / Answer

Mean ( u ) =7
Standard Deviation ( sd )=2
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 6) = (6-7)/2
= -1/2= -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.3085                  
b.
P(X > 8) = (8-7)/2
= 1/2 = 0.5
= P ( Z >0.5) From Standard Normal Table
= 0.3085                  
c.
P(X>8|X>5) = P(X>8 and X>5) / P(X>5)
P(X > 5) = (5-7)/2
= -2/2 = -1
= P ( Z >-1) From Standard Normal Table
= 0.8413                  
P(X>8|X>5)=P(X>8 and X>5) / P(X>5)=P(X>8)/ P(X>5) = .3085 / 0.8413 = 0.36669

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