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].

E[X] is larger than E[Y]

E[X] = 39.283784 and E[Y] = 37

Step-by-step explanation:

Given data

total students = 148

bus 1 students = 40

bus 2 students = 33

bus 3 students = 25

bus 4 students = 50

to find out

E[X] and E[Y]

solution

we 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 students

and probability of 40 i. e. P (40) students = 40/148

P (33) = 33/148

P (25) = 25 / 148

P (50) = 50 / 148

first we find out i. e

E[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 chosen

E[Y] = 1/4 (40 + 33 + 25 + 50)

so E[Y] = 1/4 (40 + 33 + 25 + 50)

E[Y] = 1/4 (148)

E[Y] = 37

so E[X] = 39.283784 and E[Y] = 37