Download Beautiful Homework No. 4

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Beautiful Homework No. 4
Engineering 323
Problem 3-13
Bridges
Problem Statement 3-13:
A mail-order computer business has six telephone lines. Let X denote the number of
lines in use at a specified time. Suppose the probability mass function of X is as given
in the accompanying table.
x
0
1
2
3
4
5
6
P(x)
.10
.15
.20
.25
.20
.06
.04
Probability Mass Function
0.3
P(x) values
0.25
0.2
0.15
p(x)
0.1
0.05
0
0
1
2
3
4
5
6
7
x values
Getting Started:
The probability mass function (pmf) of a discrete random variable X, is defined for every
number x by
p(x) = P(X = x) = P(all s ε S : X(s) = x)
For every possible value x of X, P(X = x) specifies the probability of observing that value
when the experiment is performed.
Now using the table we get,
Beautiful Homework No. 4
2 of 3
Engineering 323
Problem 3-13
Bridges
P(X=0) = p(0) = .10
P(X=1) = p(1) = .15
P(X=2) = p(2) = .20
P(X=3) = p(3) = .25
P(X=4) = p(4) = .20
P(X=5) =p(5) = .06
P(X=6) = p(6) = .04
To calculate the probability of each of the following events, we need to find out what
probabilities are in the set given then add them up.
a. {At most 3 lines are in use}
This set includes the probability of 0, 1, 2, and 3. Therefore,
P(X ≤ 3) = p(0) +p(1) + p(2) + p(3) = .1 + .15 + .2 + .25 = .7
b. {Fewer than 3 lines are in use}
This question is similar to part (a), except 3 is not included.
P(X < 3) = p(0) + p(1) + p(2) = .1 + .15 + .2 = .45
c. {At least 3 lines are in use}
This is the compliment to part (b) or we can add the probabilities of 3, 4, 5, and 6.
P(X ≥3) = p(3) + p(4) + p(5) + p(6) = .25 + .2 + .06 + .04 = .55
Or P(X ≥ 3) = 1 – P(X < 2 ) = 1 - .45 = .55
d. {Between 2 and 5 lines, inclusive, are in use}
Again this is just adding up the probabilities we want, in this case 2, 3, 4, and 5.
Beautiful Homework No. 4
3 of 3
Engineering 323
Problem 3-13
Bridges
P(2 ≤ X ≤ 5) = p(2) + p(3) + p(4) + p(5) = .2 + .25 + .2 + .06 = .71
e. {Between 2 and 4 lines, inclusive, are not in use}
This problem has to be handled a bit differently. Because our random variable is
defined in terms of lines in use, we need to interpret our given set in those terms.
Notice if 2 lines are not in use then 4 are in use. If 3 lines are not in use then 3
lines are in use. If 4 lines are not in use, then 2 lines are in use. Looking at it
this way our new set becomes {Between 4 and 2 lines, inclusive, are in use}.
Now that we have our problem in terms of lines in use like our random variable
we can calculate like before in part (d).
P(4 ≥ X ≥ 2) = p(4) + p(3) + p(2) = .2 + .25 + .2 = .65
f. {At least 4 lines are not in use}
Again we need to define are set in terms of lines in use. This works out to be
{At most 2 lines are in use}.
P(X ≤ 2) = p(0) + p(1) + p(2) = .1 + .15 + .2 = .45