Joint Distribution Functions/ part8

Statistics Marginal distributions and condition distribution?
Exaple
A refrigerator manufacturer subjects his finished products to
a final inspection. Of interest are two categories of defects: scratches or flaws in
the porcelain finish, and mechanical defects. The number of each type of defects
is a random variable. The results of inspecting 50 refrigerators are shown in the
following joint p.m.f. table, where X represents the occurrence of finish defects and
Y represents the occurrence of mechanical defects.

Y\X--0--------1---------2---------3---…
0---11/50--4/50----2/50-----1/50----1/…
1---8/50----3/50----2/50-----1/50---1/…
2----4/50---3/50----2/5-------1/50
3----3/50---1/50
4---1/50

(a) Find the marginal distributions of X and Y .
(b) Find the conditional distribution of mechanical defects, given that there are
no finish defects.
(c) Find the conditional distribution of finish defects, given that there are no
mechanical defects.

Solution

(a) marginal distributions are the sums of each row and column.
X --- 0 -- -- 1 -- -- 2 -- -- 3 -- -- 4 -- -- 5 --
-- 27/50 11/50 06/50 03/50 01/50 02/50

Y --- 0 -- -- 1 -- -- 2 -- -- 3 -- -- 4 --
-- 20/50 15/50 10/50 04/50 01/50

(b) conditional distributions are constructed by taking each row and each column and dividing all probabilities in that row/column by the marginal probability of that row/column. For example, the conditional distribution of Y given X=0 is

Y --- 0 -- -- 1 -- -- 2 -- -- 3 -- -- 4 --
-- 11/50 08/50 04/50 03/50 01/50
-- -------- -------- -------- -------- --------
-- 27/50 27/50 27/50 27/50 27/50

Y --- 0 -- -- 1 -- -- 2 -- -- 3 -- -- 4 --
-- 11/27 | 8/27 | 4/27 | 3/27 | 1/27

(c)
X --- 0 -- -- 1 -- -- 2 -- -- 3 -- -- 4 --
-- 11/50 04/50 02/50 01/50 01/50
-- -------- -------- -------- -------- --------
-- 19/50 19/50 19/50 19/50 19/50

X --- 0 -- -- 1 -- -- 2 -- -- 3 -- -- 4 --
-- 11/19 | 4/19 | 2/19 | 1/19 | 1/19