# Quiz Three Solutions

```May 1, 2009
Math 244 Spring 2009 Quiz 3
Name:
1. An appliance dealer advertises three different models of upright freezers having 13.5, 16.2 and
19.7 cubic feet of storage space, respectively. Let X = the advertised storage space purchased
by the next customer to buy a freezer. Suppose that X has the pmf given by:
x
p(x)
13.5
.3
16.2
.45
19.7
.25
(a) (Points) Compute E(X), and V (X).
E(X) =
X
E(X 2 ) =
xp(x) = (13.5)(0.3) + (16.2)(.45) + (19.7)(.25) = 16.625
X
x2 p(x) = (13.5)2 (0.3) + (16.2)2 (.45) + (19.7)2 (.25) =
V (X) = E(X 2 ) − (E(X))2 = −(16.265)2 ≈ 5.245
(b) ( Points) Suppose that while X is the advertised storage space, the actual storage space
is given by h(X) = X − 2 X1 . What is the expected actual capacity of the freezer purchased by the next customer?
E(h(x)) =
X
1
1
1
h(x)p(x) = 13.5 − 2
(.3) + 16.2 − 2
(.45) + 19.7 − 2
≈ 16.14f ft3
13.5
16.2
19.7
One might also think of doing this problem by calculating E(h(X)) in the following way:
2
1
E X−
= E(X) − 2E
= 16.265 − 2 ((1/13.3)(.3) + (1/16.2)(.45) + (1/19.7)(.25)) ≈ 16.14
X
X
(c) (2 Points) Compute the cumulative distribution function for X and write it as a piecewise function.
The cdf for this function is:


0
x &lt; 13.5



.3
13.5 ≤ x &lt; 16.2
F (x) =

.75 16.2 ≤ x &lt; 19.7



1
x ≥ 19.7
It might help to graph this function as well. The graph looks like:
1.0
0.8
0.6
0.4
0.2
5
10
15
20
2. An instructor who taught two sections of engineering statistics last term, the first with 23
students and the second with 27 students, decided to assign each student a term project.
After all projects had been turned in, the instructor randomly ordered them before grading.
Consider the first 12 that he graded and let,
X = { the number of projects from the second class among the first 12 graded }
(a) What type of random variable is X and what are the associated parameters?
The random variable X is a hypergeometric random variable with parameters:
N
= 50 = the total number of objects in the population, (i.e. the students)
M
= 27 = the total number of “successes” in the population, in this case the number
of students from the second class
n = 12 = the number of objects sampled in this case the number of graded exams
(b) What is the probability that among the first 12 graded, exactly 5 are from the second
class? (Give an exact numerical answer time permitting)
Using the formula for the pmf for a hypergeometric random variable we have that:
23 27
p(x) = P (X = x) =
x
12−x
50
12
27
5
P (X = 5) = p(5) =
23
7
50
12
,
=
x = 0, 1, 2, 3, ...12
27! 23!
5!22! 7!16!
50!
12!38!
≈ 0.163
(c) What is the probability that at least 2 of the first 12 graded are from the second class?
(Give an exact numerical answer time permitting)
We want P (X ≥ 2) here but it is easier to compute the probability of the compliment:
P (X ≥ 2) = 1 − P (X ≤ 1) = 1 − p(0) − p(1) = 1 −
27! 23!
0!27! 12!11!
50!
12!38!
−
27! 23!
1!26! 11!12!
50!
12!38!
≈ .999688
(d) What are the mean and standard deviation of the number of projects from the second
class among the first 12 graded?
Finally we use the formulas for computing the expected value and variance for a hypergeometric random variable:
M
27
E(X) = n
= 12 = 6.48
N
50
M
M
N −n
27
23
38
V (X) = n
1−
= 12
≈ 2.3116
N
N
N −1
50
50
49
p
√
σX = V (X) = 2.3116 ≈ 1.5204
```