A total of 4 buses carrying 148 students from the same school arrive at a football stadium. The buses carry, respectively, 40, 33, 25, and 50 students. One of the students is randomly selected. Let X denote the number of students that were on the bus carrying this randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on her bus. What is E[X+Y]?

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Answer:

76.2838

Step-by-step explanation:

E[X+Y] = E[X] + E[Y]

E[X] is the average of X  or expectancy of X

E[Y] is the average or expectancy f Y

E[X] = 40P(student is in bus 1)+33P(student is in bus 2)+25P(student is in bus 3)+50P(student is in bus 4)=

40(40/148) + 33(33/148) + 25(25/148) + 50(50/148) = 39.2838

E[Y] = (number of students)/(number of drivers) = 148/4 = 37

So E[X+Y] = 39.2838 + 37 = 76.2838

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