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]?

Question

Gill

Mathematics

16 September, 17:50

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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Alyssa Newman · Answer 1

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