Interpret Probability Random Variables and Probability Distributions

DATA FILE 3
ASSIGNMENT 3
1 Of six cars produced at a particular factory between 8 am, and 10 am last Monday morning, test runs revealed three of them to be “lemons.” Nevertheless, three of the six cars were shipped to Dealer A and the other three to Dealer B. Dealer A received all three lemons. What is the probability of this event occurring if the three cars shipped to Dealer A were selected at random from the six produced?
The sample points are lemons and not lemons
P(l lemons only)=3/6=0.5
2 Carbon monoxide (CO) is an odorless, colorless, highly toxic gas which is produced by fires as well as by motor vehicles and appliances that use carbon-based fuels. A study on unintentional CO poisoning of Colorado residents for six years identified 981 cases. Each case was classified as fatal or nonfatal and by source of exposure. The number of cases occurring in each of the categories is shown in the accompanying table. Assume that one of the 981 cases of unintentional CO poisoning is randomly selected.
Source of Exposure

Fatal

Nonfatal

Total

Fire

63

53

116

Auto exhaust

60

178

238

Furnace

18

345

363

Kerosene or space heater

9

18

27

Appliance

9

63

72

Another gas-powered motor

3

73

76

Fireplace

0

16

16

Other

3

19

22

Unknown

9

42

51

Total

174

807

981

1 List all sample points for this experiment.
Fire-Fatal and Nonfatal
Auto exhaust- Fatal and Nonfatal
Furnace - Fatal and Nonfatal
Kerosene or space heater - Fatal and Nonfatal
Appliance- Fatal and Nonfatal
Another gas-powered motor- Fatal and Nonfatal
Fireplace- Fatal and Nonfatal
Other- Fatal and Nonfatal
Unknown- Fatal and Nonfatal
2 What is the set of all sample points called?
The sample space is the set or collection of all the sample points, where the possibilities of all the outcomes that could occur are identified (Kinney, 2015).
3 Let A be the event that CO poisoning is caused by fire. Find P(A).
P (A) = 116/981= 0.11825
4 Let B be the event that CO poisoning is fatal. Find P(B).
P (B) =174/981 = 0.17737
5 Let C be the event that CO poisoning is caused by auto exhaust. Find P(C).
P(C) =238/981= 0.2426
6 Let D be the event that the CO poisoning is caused by auto exhaust and is fatal. Find P(D).
P (D) =60/238=0.2521
7 Let E be the event that the CO poisoning is caused by fire but is nonfatal. Find P(E).
P (E) =53/116= 0.4569
3 Security analysts are professionals who devote full-time efforts to evaluating the investment worth of a narrow list of stocks. The following variables are of interest to security analysts. Which are discrete and which are continuous random variables?
“Random variables that can assume a countable number of values are called discrete” and “random variables that can assume values corresponding to any of the points contained in an interval are called continuous” (McClave & Sincich, 2012)
8 The closing price of a particular stock on the New York Stock Exchange.
Discrete can assume a countable number of values.
9 The number of shares of a particular stock that are traded each business day.
Discrete can assume a countable number of values.
10 The quarterly earnings of a particular firm.
A continuous variable can take any value in the interval.
The percentage change in yearly earning between 2015 and 2016 for a particular firm.
A continuous variable can take any value in the interval, and an infinite number of outcomes are possible in these variables (Hoerl &Snee, 2012).
The number of new products introduced per year by a firm.
Discrete can assume a countable number of values.
The time until a pharmaceutical company gains approval from the US Food and Drug Administration to market a new drug.
A continuous variable can take any value in the interval.
4 Explain why each of the following is or is not a valid probability distribution for a discrete random variable x:
a.
x

0

1

2

3

p(x)

0.1

0.3

0.3

0.2

This is not a valid probability distribution because the sum of the probabilities is less than 1 at 0.9
b.
x

-2

-1

0

p(x)

0.25

0.50

0.25

This is a valid probability distribution because the probabilities are all non-negative and their sum is 1
c.
x

4

9

20

p(x)

-0.3

0.4

0.3

This is not a valid probability distribution because the sum of the probabilities is less than 1 at 0.4 and there is a negative probability.
d.
x

2

3

5

6

p(x)

0.15

0.15

0.45

0.35

This is not a valid probability distribution because the sum of the probabilities is more than 1 at 1.1
SOURCE: MBS, Section 4.2, Problem 4.12, p. 167.
5 The random variable x has the following discrete probability distribution:
x

1

3

5

7

9

p(x)

0.1

0.2

0.4

0.2

0.1

Find P(x ≤ 3).
0.1+0.2=0.3
Find P (x < 3).
0.1
Find P(x = 7).
0.2
Find P(x ≥ 5).
0.4+0.2+0.1=0.7
Find P(x > 2).
0.7
Find P(3 ≤x ≤9).
0.9
6 Consider the probability distribution shown here.
x

-4

-3

-2

-1

0

1

2

3

4

p(x)

0.02

0.07

0.10

0.15

0.30

0.18

0.10

0.06

0.02

Calcul...