Textbook Notes
(363,135)

Canada
(158,215)

Ryerson University
(11,185)

Quantitative Methods
(72)

QMS 230
(8)

Chapter 5

# Chapter 5 Worksheet

Unlock Document

Ryerson University

Quantitative Methods

QMS 230

Douglas Mc Kessock

Fall

Description

StatisticsForManagement-CQMS204 Chapter5
CHAPTER 5 - Probability
The chance that an event will occur
1. Discuss the meaning of “10% chance of snowing this late afternoon”.
- Low chance.
2. For a true or false question, what is the chance of getting the correct answer if
I completely guess the answer? Why?
- 50%, because it’s true or false.
3. What is the purpose of buying insurance?
- To reduce your risk.
4. For a multiple-choice question with four choices, which are A, B, C and D,
what is the chance of getting the correct answer if I completely guess the
answer? Why?
- 25%.
5. Give 2-3 examples of situations where you use probability concept in your
daily life.
PROBABILITY DISTRIBUTION
Outcome:
Understand the concepts of discrete and continuous random
variables and their associated distributions
Definitions
1. random variable – a variable whose value is determined by the
outcomes of a random experiment (Random = equal chance)
2. discrete random variable – a random variable that can take on any
value from a list of distinct possible values (Whole number)
3. continuous random variable – a random variable that can take on
any value from a continuous range. (Decimal number)
4. probability distribution for a discrete random variable – a list of all
the possible values of the random variable and their associated
probabilities
Example1
Which of the following random variables are discrete, and which are
continuous?
a. The number of passengers on a flight from Toronto to Hong Kong. Discrete
b. The number of students in Business Statistics class. Discrete
c. The time required driving from home to Ryerson University in the morning.
Continuous
Winter2011 Page # 1 StatisticsForManagement-CQMS204 Chapter5
d. The number of traffic fatalities per year in Toronto. Discrete
Probability distributions
1.can be represented by a.table b. graphical chart
2.can be summarized with
a. expected values(or means), which are measures of centre tendency
(location) of the probability distribution
b. standard deviations, which are measures of variability of the probability
distribution
Expected value = mean = E ( X ) ∑ xP ( x(Use probability for this
equation) 2 2
Standard deviation = ∑ x P ( x ) − μ
Example2
The probability distribution for a random variable X is as shown in the following
table.
x 1 2 3 4 5
P(X=x) 0.20 0.25 0.40 0.10 0.05
Determine
a. the expected value b. the variance c. the standard deviation
(Probability is always list 2)
Standard deviation = n-1 (TEST QUESTION: Why is Sx = 0? N is 1, and 1-1 = 0.)
a. E(X) = 2.55
b. 1.14 (stand. Dev.)^2
c. 1.07
Example3
The probability distribution for a random variable X is as shown in the following
table.
x 1 2 3 4 5
P(X=x) 0.1 0.3 y 0.2 0.1
Determine
a. the value of y b. the expected value c. the standard deviation
a. Y = 0.3
b. 2.9
c. 1.14
Winter2011 Page # 2 StatisticsForManagement-CQMS204 Chapter5
Example#4
Amanda, the marketing manager for Third Cups Coffee Store is implementing a
new marketing strategy that offers free refills on all coffee order. To review her
strategy, she gathered the following information on coffee refills.
Refills 0 1 2 3 4
Percent 0.21 0.37 0.32 0.10 0
Compute
a. the probability that a customer refill his/her coffee one or more times
b. the expected value, variance and standard deviation for the distribution of
number of refills.
a. 0.79
b. 1.31, 0.83, 0.91
Example5
A fast food company plans to install a new ice-cream dispensing unit in one of
the two store locations. The company figures that the probability of a unit being
successful in a location is 5/8 and the annual profit in this case is $185,000. If it is
not successful there will be losses of $36,800. At the location B the probability of
succeeding is ½ but the potential profit and loss are $250,000 and $58,000
respectively.
a. Where should the company locate to maximize expected profit?
b. Which location is less risky, i.e has the lowest relative variability?
(create a table)
X 185,000 -36,800
P 0.625 0.375 (1 – 0.625)
E(x) = 101825
Standard Deviation = 107378.463
X 250,000 -58,000
P 0.5 0.5
E(x) = 96,000
Standard Deviation = 154,000
a. Answer is Location A because of higher expected profit.
b. Use coefficient of variation formula (Sx divided by E(x))
i. Location A: 1.0545
ii. Location B: 1.604 Therefore; Location A!
Diff. = 160.41% - 105.45% = 55%
10% of smaller = 0.1(105.45) = 10.5%
Diff > 10% of smaller, therefore, location B has a higher variation. Therefore,
Location A is less risky.
Winter2011 Page # 3 StatisticsForManagement-CQMS204 Chapter5
Example6
The normal weekly demand of a certain perishable product sold by Ryerson Inc.
is given by the following distribution
Demand 21 22 23 24 25
Probability 0.4 0.2 0.2 0.1 0.1
The product costs Ryerson $9 each. The product sells for $16 each. If not sold
by the end of the week, the leftover units must be scrapped.
The supplier only has 23, 24 or 25 units for Ryerson to purchase. How many
would you recommend Ryerson to purchase based on expected profit?
23 units
Demand Profit (Revenue - COGS) (list 1) P(X) (List 2)
21 * Remaining 2 units you have to add to4 (you get these values
the cost (since it has to be scrapped;om the above table)
demand = 21, supplier only has 23
units)
21(16) – 23(9) = 129
22 22(16) – 23(9) = 145 0.2
23 23(16) – 23(9) = 161 0.2
23(16) – 23(9) = 161 -> Supplier has 0.1
24 only 23
25 23(16) – 23(9) = 161 0.1
E(x) = $145
24 units
Demand Profit P(X)
21 21(16)-24(9) = 120 change to 24 0.4
units
22(16)-24(9) = 136 0.2
22
23 23(16)-24(9) = 152 0.2
24 24(16)-24(9) = 168 0.1
25 24(16)-24(9) = 168 Supplier only has1
24
E(x) = $139.20
25 units
Demand Profit P(X)
21 21(16)-25(9) = 111 change to 25 0.4
units
22(16)-25(9) = 127 0.2
22
23 23(16)-25(9) = 143 0.2
24 24(16)-25(9) = 159 0.1
Winter2011 Page # 4 StatisticsForManagement-CQMS204 Chapter5
25 25(16)-25(9) = 175 0.1
E(x) = $131.80
Answer: 23 Units is the best because it has the highest expected
profit.
THE BINOMIAL PROBABILITY DISTRIBUTION (discrete)
Outcome:
Recognize situations when the binomial probability distribution
applies, and use formula to calculate binomial probabilities
Binomial probability distribution
1. Often applies when we are interested in the number of times a particular
characteristic turns up. Example: We might want to know how many of the
next 50 customers who come into the store will purchase some
merchandise.
2. The count (how many out of a particular number), which can also be
expressed as a percentage or proportion.
3. This count is a random variable, because its outcome is determined by
chance.
4. A binomial random variable counts the number of times one of only two
possible outcomes takes place (thus the bi-part of the binomial’s name).
5. A binomial experiment must satisfy the following four conditions.
a. there are n identical trials
b. each trial has only two possible outcomes
c. the probabilities of the two outcomes remain constant
d. the trials are independent (don’t effect each other)
6. mean and standard deviation of a binomial distribution
mean = μ = np , standard deviation = σ = npq
Binomial Formula
x n − x
P (x)= n Pxq where
C = combination

More
Less
Related notes for QMS 230