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].

Four buses

hansel

Answered by Aaron 2 years ago

**part (a)**

E[X] is larger, since the driver selected is equally likely to be from any of the 4 buses, but the student selected is more likely to have come from a bus carrying a large number of students.

**part (b)**

$$E(X)=\sum_{i=1}^4 x_i p(x_i)=40\bigg(\frac{40}{148}\bigg)+33\bigg(\frac{33}{148}\bigg)+25\bigg(\frac{25}{148}\bigg)+50\bigg(\frac{50}{148}\bigg)=39.28$$

$$E(Y)=\sum_{i=1}^4 y_i p(y_i)=40\bigg(\frac{1}{4}\bigg)+33\bigg(\frac{1}{4}\bigg)+25\bigg(\frac{1}{4}\bigg)+50\bigg(\frac{1}{4}\bigg)=37$$

which justifies our answer in part (a)

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Asked: 2 years ago