Four 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 the randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on her bus. (a) Which of E[X] or E[Y] do you think is larger? Why? (b) Compute E[X] and E[Y].

Answer:E[X] is larger than  E[Y]E[X]  = 39.283784  and E[Y] = 37Step-by-step explanation:Given data total students = 148bus 1 students = 40bus 2 students = 33bus 3 students = 25bus 4 students = 50to find out E[X] and E[Y]solutionwe know bus have total 148 students and 4 bus so E[X] is larger than  E[Y] because maximum no of students are likely to chosen to bus and probability of bus is 1/4  as chosen studentsand probability  of 40 i.e. P(40)  students = 40/148P(33) = 33/148 P(25) = 25 / 148P(50) = 50 / 148 first we find out i.eE[X]  = 40 P(40) + 33 P(33)+  25 P(25)+  50 P(50)E[X]  = 40  (40/148) + 33 (33/148)+  25 (25/148)+  50 (50/148) E[X]  = 39.283784 and y is bus chosenE[Y] = 1/4 (40+ 33 + 25 + 50)so E[Y] = 1/4 (40+ 33 + 25 + 50)E[Y] = 1/4 (148)E[Y] = 37so E[X]  = 39.283784  and E[Y] = 37