Question

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 1

Answer:

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