Answer to Question #323839 in Statistics and Probability for Bless

Question

 Each of the 50 students in class belongs to exactly one of the four groups A, B, C, or D. The membership numbers for the four groups are as follows: A: 5, B: 10, C: 15, D: 20. First, choose one of the 50 students at random and let X be the size of that student’s group. Next, choose one of the four groups at random and let Y be its size. (Recall: all random choices are with equal probability, unless otherwise specified.)

(a) Write down the probability mass functions for X and Y . (b) Compute E[X] and E[Y ] . (c) Compute V (X) and V (Y ). (d) Assume you have s students divided into n groups with membership numbers s1,s2,··· ,sn and again X is the size of the group of a randomly chosen student, while Y is the size of the randomly chosen group. Let E[Y ] = µ and V (Y ) = σ2. Express E[X] with s,n,µ, and σ.

Answer 1

$a)\space P(X=5)=\frac{5}{5+10+15+20}=0.1\\ P(X=10)=\frac{10}{5+10+15+20}=0.2\\ P(X=15)=\frac{15}{5+10+15+20}=0.3\\ P(X=20)=\frac{20}{5+10+15+20}=0.4$

For all other values of X P(X) = 0

Probability of selecting one group out of 4 is 1 / 4 = 0.25

$P(Y=5)=0.25\\ P(Y=10)=0.25\\ P(Y=15)=0.25\\ P(Y=20)=0.25$

For all other values of Y P(Y) = 0

$b)\space\Epsilon[X]=0.1\cdot5+0.2\cdot10+\\ +0.3\cdot15+0.4\cdot20=15\\ \Epsilon[Y]=0.25\cdot(5+10+15+20)=12.5$

$c)\space V(X)=\frac{1}{4-1}(0.1\cdot(5-15)^2+0.2\cdot(10-15)^2+\\ 0.3\cdot(15-15)^2+0.4\cdot(20-15)^2)]\approx\\ \approx8.333\\ V(Y)=\frac{1}{3}\cdot0.25\cdot(5+10+15+20)\approx4.167$

$d)\space \Epsilon[X]=\sum_{k=1}^{n}\frac{s_k}{s}s_k=\frac{1}{s}\sum_{k=1}^{n}s_k^2\\ \Epsilon[Y]=\sum_{k=1}^{n}\frac{1}{n}s_k=\frac{1}{n}\sum_{k=1}^{n}s_k=\frac{s}{n}\\ V[Y]=\frac{1}{n-1}\sum_{k=1}^{n}\frac{1}{n}(s_k-\frac{s}{n})^2=\\ =\frac{1}{n(n-1)}\sum_{k=1}^{n}(s_k^2-\frac{2s_ks}{n}+\frac{s^2}{n^2})=\\ =\frac{1}{n(n-1)}(s\Epsilon[X]-\frac{2s^2}{n}+\frac{s^2}{n})=\\ =\frac{1}{n(n-1)}(s\Epsilon[X]-\frac{s^2}{n})=\sigma^2\Rarr\\ \Rarr\Epsilon[X]=\frac{1}{s}(n(n-1)\sigma^2+\frac{s^2}{n})=\\ =\frac{n(n-1)}{s}\sigma^2+\mu$