Four buses carrying 159 high school students arrive to Montreal. The buses carry, respectively, 31, 46, 33, and 49 students. One of the studetns 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:

Let us name the buses as follows. Bus A has 31 students, bus B has 46 students, Bus C has 33 and Bus D has 49. Define the events A: The student is from bus A, B: the student is from bus B, C: the student is from bus C, D: The student is from bus D. Then, by simply dividing the number of students per bus by the total number of students, we get the probabilities

[tex]P(A) = \frac{31}{159}, P(B) = \frac{46}{159}, P(C) = \frac{33}{159}, P(D) = \frac{49}{159}[/tex]

In this case, note that X=31 if the student is from bus A. Then, P(X=31) = P(A). Then

[tex]E[X] = 31P(X=31)+46P(X=46)+33P(X=33)+49P(X=49)= 31P(A)+46P(B)+33P(C)+49P(D)=41.301[/tex].

For the random variable Y, note that Y = 31 if we pick the driver from bus A. In this case, each driver is equally likely to be picked, so the probability of picking the drive from bus A is 1/4 (the same for each the drivers), then

[tex]E[Y] =31P(Y=31)+46P(Y=46)+33P(Y=33)+49P(Y=49)=\frac{159}{4} = 39.75[/tex].

We will calculate the expected value of X^2 and Y^2. That is

[tex]E[X^2] = 31^2 P(A) + 46^2P(B) + 33^2P(C) + 49^2P(D) = 1765.49[/tex]

[tex]E[Y^2] = 31^2 \frac{1}{4} + 46^2\frac{1}{4} + 33^2\frac{1}{4} + 49^2\frac{1}{4} = 1641.49[/tex]

For the variance, we use the formula [tex]Var(X) = E[X^2]-(E[X])^2[/tex].

Then,

[tex]Var[X] =E[X^2]-(E[X])^2 = 1765.49- (41.301)^2 = 59.6447[/tex]

[tex]Var[Y] = E[Y^2]-(E[Y])^2 = 1641.49- (39.75)^2 = 61.6875[/tex]

Omm2 Omm2 · Answer 2 · 2022-03-16T12:39:52+01:00

The required variance is [tex]E(X)=41.30189[/tex]

Variance:

Variance is the expected value of the squared variation of a random variable from its mean value, in probability and statistics. Informally, variance estimates how far a set of numbers (random) are spread out from their mean value.

The formula for calculating the variance is,

[tex]E(X)=xP(X=x)[/tex]

Now, substituting the given values into the above formula we get,

[tex]E(X)=xP(X=x)\\=31(\frac{31}{159} )+46(\frac{46}{159} )+33(\frac{33}{159} )+49(\frac{49}{159} )\\=41.30189[/tex]

Learn more about the topic Variance:

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