Question

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)

Answer 1

E(X) = 38.9346

Var(X) = 25.588

E(Y) = 38.25

Var(Y) = 26.1875

Explanation:

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) =
`(36(36) + 46(46) + 32(32) + 39(39))/(153)`

=
`(5957)/(153)`

= 38.9346

⇒E(X) = 38.9346

Now,

E(Y) =
`(36 + 46 + 32 + 39)/(4)`

=
`(153)/(4)`

= 38.25

⇒E(Y) = 38.25

Now,

Var(X) =
`(36(36 - 38.9346)^(2) + 46(46 - 38.9346)^(2) + 32(32 - 38.9346)^(2) + 39(39 - 38.9346)^(2))/(153)`

=
`(310.0276 + 2296.3144 + 1308.5091 + 0.1668)/(153)`

=
`(3915.0179)/(153)`

= 25.588

⇒Var(X) = 25.588

Now,

Var(Y) = E(Y²) - [E(Y)]²

=
`(36^(2) + 46^(2) + 32^(2) + 39^(2) )/(4) - (38.25)^(2)`

=
`(5957)/(4) - 1463.0625`

= 1489.25 - 1463.0625

= 26.1875

⇒Var(Y) = 26.1875