Review of Part IV
279
Review of Part IV
1. Quality Control.
Construct a Venn diagram of the disjoint outcomes.
a)
P
(defect) =
P
(cosm.) +
P
(func.) –
P
(cosm. and func.)
= 0.29 + 0.07 – 0.02 = 0.34
Or, from the Venn: 0.27 + 0.02 + 0.05 = 0.34
b)
P
(cosm. and no func.)
=
P
(cosm.) –
P
(cosm. and func.)
= 0.29 – 0.02 = 0.27
Or, from the Venn: 0.27 (the region inside Cosmetic circle, yet outside Functional circle)
c)
P
P
P
(func.
cosm.)
func.
cosm.)
(cosm.)
=
∩
=≈
(.
.
.
002
029
0 069
From the Venn, consider only the region inside the Cosmetic circle.
The probability that
the car has a functional defect is 0.02 out of a total of 0.29 (the entire Cosmetic circle).
d)
The two kinds of defects are not disjoint events, since 2% of cars have both kinds.
e)
Approximately 6.9% of cars with cosmetic defects also have functional defects.
Overall, the
probability that a car has a cosmetic defect is 7%.
The probabilities are estimates, so these
are probably close enough to say that they two types of defects are independent.
2. Workers.
Organize the counts in a twoway
table.
a) i)
P
female)
==
90
150
06
ii)
PP
P
P
((
(
(
.
female
production)
female)
production) 
female and production)
∪=
+
=+−=
90
150
117
150
72
150
09
iii)
Consider only the production row of the table.
There are 72 women out of 117
production workers.
72/117 ≈ 0.615.
Or, use the formula:
P
P
P
(female
production)
(female
production)
(production)
=
∩
72
150
117
150
0 615
.
Male
Female
Total
Management
7
6
13
Supervision
8
12
20
Job
Type
Production
45
72
117
Total
60
90
150
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Part IV
Randomness and Probability
iv)
Consider only the female column.
There are 72 production workers out of a total of 90
women.
72/90 = 0.8.
Or, use the formula:
P
P
P
(production
female)
(production
female)
(female)
=
∩
==
72
150
90
150
08
.
b)
These data suggest that holding a production position may be associated with gender.
60% of the plant employees are women, but 61.5% of the production workers are women.
However, this is a small difference, and may be due to sampling error.
3. Airfares.
a)
Let
C
= the price of a ticket to China
Let
F
= the price of a ticket to France.
Total price of airfare = 3
C +
5
F
b)
µ
=+
=
+
=
+
=
EC F
EC
EF
()
(
)
(
)
(
)
(
)
$
3
5
3
5
3 1000
5 500
5500
σ
=
+
=
+
≈
SD
C
F
Var C
Var F
(
)
((
)
)
)
)
(
)
(
)
$.
3
5
3
5
3 150
5 100
672 68
2
2
2222
c)
=−
=
−
=
−=
(
)
(
)
$
1000
500
500
=
+
≈
SD C
F
Var C
Var F
(
)
(
)
150
100
180 28
22
d) No assumptions are necessary when calculating means.
When calculating standard
deviations, we must assume that ticket prices are independent of each other for different
countries but all tickets to the same country are at the same price.
4. Bipolar.
Let
X
=
the number of people with bipolar disorder in a city of
n
= 10,000 residents.
These may be considered Bernoulli trials.
There are only two possible outcomes, having
bipolar disorder or not having bipolar disorder.
Psychiatrists estimate that the probability
that a person has bipolar is about 1 in 100, so
p
= 0.01.
We will assume that the cases of
bipolar disorder are distributed randomly throughout the populations.
The trials are not
independent, since the population is finite, but 10,000 people represent fewer than 10% of
all people.
Therefore, the number of people with bipolar disorder in a city of 10,000 may be
modeled by
Binom
(10000, 0.01).
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 Spring '08
 WEINSTEIN
 Probability, Standard Deviation, Variance, Probability theory, µ

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