Please solve the following probabilities

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 ] = u and V (Y ) = standard deviation^2. Express E[X] with s; n; u, and variance.

Answer 1

part a&b pmf(X) = p(5)=.1 , p(10) = .2, p(15) = .3 p(20) = .4

pmf(Y) P(A)=P(B)=P(C)=P(D)=1/4

E(x)=5*.1+10*.2+15*.3+20*.4 = 15

E(Y) = (5+10+15+20)/4 = 12.5

part c definition of V = ∑(x_i-x¯)² P(x_i) x‾ is x bar is same as E(X)

V(X) = .1*(5-15)²+ .2*(10-15)²+.3*(15-15)²+.4*(20-15)²⁼ 25

V(Y) = .25*(5-12.5)²+ .25*(10-12.5)²+.25*(15-12.5)²+.25*(20-12.5)²⁼ 31.25

d. Here we have to make some simple assumptions. We can see that the selection process of the essentially Gaussian so we will assume that the distribution of the samples is normal.

more later: (hint: the E(X) is defined as the sum of the expected value of each group a student could belong to, times the probability that that group would be the one the student belongs to, which is related to the size of that group. )