Question 1 Discrete Distribution-8 marks

There are two routes that John goes to work (Route A and Route B). Route A has a shorter distance to work, but occasionally John has to stop at the railway crossing. Route B is farther away from work, but there is no railway crossing to navigate around. Eighty (80) percent of the time, John would take Route A to work, and he would take Route B to work 20% of the time. But when taking Route A, 40% of the time a train will cross at the railway crossing and it takes him 20 minutes to get to work. While taking Route A to work, 60% of the time, there will not be any train passing by, hence he only takes 10 minutes to work. If he takes Route B, it takes him 18 minutes to work.

a) Define X as the time (in minutes) it takes for John to get to work. Construct a probability distribution table of X. [3 marks]

Route. A (with stop) A (without stop) B

Time (minute) 20 10 18

Probability. 0.32. 0.48 0.20

b) Calculate the expected value of X and provide an interpretation of its value. [1+2 mark]

E(X) = 20*0.32 + 10*0.48 + 18*0.2 = 6.4 + 4.8 + 3.6 = 14.8

When the experiment of "recording the time when taking one of the three routes to work" is repeated infinite number of times, the average amount of time John takes to work is 14.8 minutes.

Please explain how this 0.32, 0.48 and 0.20 was found out?