Q:

Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 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

Accepted Solution

A:
Answer:The expected value of X is [tex]E(X)=\frac{2754}{73} \approx 37.73[/tex] and the variance of X is [tex]Var(X)=\frac{226192}{5329} \approx 42.45[/tex]The expected value of Y is [tex]E(Y)=\frac{73}{2} \approx 36.5[/tex] and the  variance of Y is [tex]Var(Y)=\frac{179}{4} \approx 44.75[/tex]Step-by-step explanation:(a) Let X be a discrete random variable with set of possible values D and  probability mass function p(x). The expected value, denoted by E(X) or [tex]\mu_x[/tex], is[tex]E(X)=\sum_{x\in D} x\cdot p(x)[/tex]The probability mass function [tex]p_{X}(x)[/tex] of X is given by[tex]p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}[/tex]Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function [tex]p_{Y}(x)[/tex] of Y is given by[tex]p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}[/tex]The expected value of X is[tex]E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)[/tex][tex]E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73[/tex]The expected value of Y is[tex]E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)[/tex][tex]E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5[/tex](b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is[tex]V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2[/tex]The variance of X is[tex]E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)[/tex][tex]E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}[/tex][tex]Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45[/tex]The variance of Y is[tex]E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)[/tex][tex]E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377[/tex][tex]Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75[/tex]