Download Problem 3.15

ENGR 323
BHW #4
Tien Tai
Problem 3-15
Many manufacturers have quality control programs that include inspection of incoming
materials for defects. Suppose a computer manufacturer receives computer boards in lots of
five. Two boards are selected from each lot for inspection. We can represent possible
outcomes of the selection process by pairs. For example, the pair (1,2) represents the
selection of boards 1 and 2 for inspection.
a. List 10 different possible outcomes
In part a, we are asked to find out all outcomes. So I list all the possible combinations below.
(1,2) (1,3) (1,4) (1,5)
(2,3) (2,4) (2,5)
(3,4) (3,5)
(4,5)
Figure 1. Total Outcomes
b. Suppose that board 1 and 2 are only defective boards in a lot of five. Two boards are to
be chosen at random. Define X to be the number of defective boards observed among those
inspected. Find the probability distribution for X.
In part b, we are asked to find the pmf of the function, and by definition from probability
textbook, we know that the probability mass function (pmf) of a discrete random variable
is defined for every number y by
p(y) = P(X = x) = P(all s ∈ζ: Y(s) = y). *
For every possible value x of X, P(X=x) specifies the probability of observing that value
when the experiment is performed.
So, Let X = No. of defective boards
The pmf is constructed below



p ( x ) = 



0 .3
x = 0
0 .6
x = 1
0 .1
x = 2
0
Otherwise
1
The result from the pmf is given from the result of part a.
Example.
Probability of none of the boards are defective
p(0)=0.3
The sum of blue highlighted = No. of defective boards picked for examination
Probability of Getting Defected Boards
After we calculate pmf, then we graph it. The graph is referred to figure 2.
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
No. of Defected Boards
Figure 2 A pmf Graph of Problem 3.15
c. Let F(x) denote the cdf of X. First determine F(0) = P(X≤0), F(1), F(2) then F(x) for all
other x.
We can now calculate cdf from the pmf we calculated from part b.
By definition on p.98 in the probability textbook, we know that the cumulative
distribution function (cdf) F(y) of a discrete rv variable Y with pmf p(x) is defined for
every number y by
F ( x ) = P ( X ≤ x) =
∑
p( y ) \
x: x≤ y
For any number y, F(y) is the probability that the observed value of Y will be at most y
2
Therefore, we can construct the cumulative distribution function from the pmf we just
calculated, and the cdf is listed on following page.
x < 0
1 < x ≤ 0
2 < x ≤ 1
x ≥ 2
 0
 0 .3

F ( x ) = 
 0 .9
 1
After we know the value of cdf, we can construct a graph of cdf (see figure 3)
Probability of Getting Defected
Boards
1.5
1
0.5
0
-1
0
1
2
-0.5
-1
No. of Defected Boards
Figure 3 Graph of cdf Problem 3.15
Compute E(X), V(X)
Beth also asked me to find the expected value E(X) and sample variance V(X) of the
function
d) To calculate E(X), we need to use the equation that’s listed on p.104
3
3
E (X ) = µ
x
=
∑ x ⋅ p(x) = (0 ⋅ 0.3 ) + (1 ⋅ 0.6 ) + (2 ⋅ 0.1 ) = 0.6 + 0.2 = 0.8
x ∈D
The expected value of defective boards = 0.8 boards
e) To calculate V(X), we need to use the equation that’s listed on p. 110
V(X) = s 2 = E(X 2 ) − [E(X)]2 = [(02 ⋅ 0.3) + (1 2 ⋅ 0.6) + (2 2 ⋅ 0.1)] − [0.8]2
= 0 + 0.6 + 0.4
= 1 − 0.64
= 0.36
Sample Variance of defective board =0.36 board 2
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