Four buses carrying 153 high school students arrive to Montreal. The buses carry, respectively, 36, 46, 32, and 39 students. One of the students is randomly selected. Let X denote the number of students that were on the bus carrying this randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on his bus. Compute the expectations and variances of X and Y: E(X)

Given - Four buses carrying 153 high school students arrive to Montreal. The buses carry, respectively, 36, 46, 32, and 39 students. One of the students is randomly selected.One of the 4 bus drivers is also randomly selected.

To find - Compute the expectations and variances of X and Y.

Proof -

Let us assume that,

X be The number of students that were on the bus carrying this randomly selected student.

Y denote the number of students on his bus.

Now,

E(X) = [tex]\frac{36(36) + 46(46) + 32(32) + 39(39)}{153}[/tex]

= [tex]\frac{5957}{153}[/tex]

= 38.9346

⇒E(X) = 38.9346

Now,

E(Y) = [tex]\frac{36 + 46 + 32 + 39}{4}[/tex]

= [tex]\frac{153}{4}[/tex]

= 38.25

⇒E(Y) = 38.25

Now,

Var(X) = [tex]\frac{36(36 - 38.9346)^{2} + 46(46 - 38.9346)^{2} + 32(32 - 38.9346)^{2} + 39(39 - 38.9346)^{2}}{153}[/tex]

= [tex]\frac{310.0276 + 2296.3144 + 1308.5091 + 0.1668}{153}[/tex]

= [tex]\frac{3915.0179}{153}[/tex]

= 25.588

⇒Var(X) = 25.588

Now,

Var(Y) = E(Y²) - [E(Y)]²

= [tex]\frac{36^{2} + 46^{2} + 32^{2} + 39^{2} }{4} - (38.25)^{2}[/tex]

= [tex]\frac{5957}{4} - 1463.0625[/tex]

= 1489.25 - 1463.0625

= 26.1875

⇒Var(Y) = 26.1875