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 who 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].

Short answer

1. $E\left(X\right)islargerthanE\left(Y\right)$Because the number of students is more than the number of buses
2. $E\left(X\right)=39.3\text{and}E\left(Y\right)=37$
   

Step-by-step solution

Given Information (Part-a)

From the information, 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 who 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.

We need to find which is larger$E\left[X\right]orE\left[Y\right]$

Explanation (Part-a)

Here, the buses carrying total 148 students.

The buses carry, respectively, 40, 33, 25, and 50 students.

Therefore, we need to find which one is larger.

So we need to apply part b first.

from that we get, $E\left(X\right)$is larger than $E\left(Y\right)$.

Because the number of students is more than the number of buses.

:Final Answer (Part-a)

$E\left(X\right)=$$39.3$.Therefore$E\left(X\right)$is larger than$E\left(Y\right)$

Given Information (Part-b)

From the information, 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 who 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.

We need to compute the$E\left[X\right]andE\left[Y\right]$

Calculation (Part-b)

To find $E\left(X\right)$

$\begin{array}{r}E\left(X\right)=40\times \frac{40}{148}+33\times \frac{33}{148}+25\times \frac{25}{148}+50\times \frac{50}{148}\end{array}$

$\begin{array}{c}=\frac{5814}{148}\\ =39.3\end{array}$

To find $E\left(Y\right)$

localid="1647082915151" $E\left(Y\right)=40\times \frac{1}{4}+33\times \frac{1}{4}+25\times \frac{1}{4}+50\times \frac{1}{4}$

$=\frac{148}{4}$

$=37$

Final Answer (Part-b)

The value of$E\left(X\right)=39.3andE\left(Y\right)=37$