Chapter VI
Introduction to Probability and
Probability Distribution
... ...
…...
Objetive
Chapter
The student will be able to:
Define experiment, outcome, event,
probability and equally likely. Restate
the formula for finding the probability of
an event. Determine the outcomes and
probabilities for experiments.
Probability distributions may be
summarized in ways that enable
researchers to easily make objective
decisions based on samples drawn from
the populations that the distributions
represent.
6.1 Introduction
The theory of probability provides the foundation for statistical inference. We define probability as the likelihood that an event will occur, the probability that an event will occur based an analysis in which each measure is based on a recorded observation, rather than a subjective estimate. Probability makes possible the prediction of particular outcomes, in a large general area of interest.
In this chapter we develop probability models that can be used to study Mathematical Science for which future outcomes are unknown.
Consider chance experiments with a finite number of possible outcomes ω1, ω2, . . . , ωn. For example,
we roll a die and the possible outcomes are 1, 2, 3, 4, 5, 6 corresponding to the side that turns up. We toss a coin with possible outcomes H (head) and T (tail). It is frequently useful to be able to refer to an outcome of an experiment.
The question always is “how likely is it that a particular event will occur?”
For example: Which country will be the winner between RDC Congo, Egypt, Senegal and Tunisia based on FIFA ranking?
6.2 What is probability?
The concept of probability is the chance that something happens or will not happen. In statistics it is denoted by the capital letter P and is measured on an inclusive numerical scale of 0 to 1. If we are using percentages, then the scale is from 0% to 100%. If the probability is 0, then there is absolutely no chance that an outcome will occur. For example I live in Rwanda, but I born in Peru, the probability of I becoming president in Rwanda is 0. At the top end the probability scale is 1 which means that it is certain the outcome will occur.
Characteristics of a probability distribution
a. The probability of a particular outcome is between 0 and 1 inclusive. b. The outcomes are mutually exclusive events.
The opposite of random experiment is deterministic where the outcome is certain on the assumption that the input data is reliable. For example if revenues are 20000 Rwf and cost are 18000 Rwf then it is sure that the gross profit is 2000 Rwf.
The probability of any outcome of a random phenomenon is the proportion of times the outcome would occur in a very long series of repetitions.
Value of Probability
Probability always has a value between 0, impossibility, and 1, certainty. Probability of 0.0 corresponds to ‘impossible’
0.1 corresponds to ‘extremely unlikely’ 0.5 corresponds to ‘evens chance’ 0.8 corresponds to ‘very likely’ 1.0 corresponds to ‘certainty’
6.3 Terminology
Experiment is a process that leads to the occurrence of one and only one of the several possible results or observations.
Random experiment. A random experiment is a process leading to two or more possible outcomes, with uncertainly as to which outcome will occur. Examples of random experiments are:
roll a die,
measuring the amount of rainfall in Kigali in April,
a customer enters a store and either purchases a shirt or does not. the daily change in an index of stock market prices is observed.
a bag of cereal is selected from a packing line and weighed to determine if the weight is above or below the stated package weight.
some number of persons will be admitted to a hospital emergency room during any hour. betting on the result between the two best football teams in Rwanda.
Outcome isa particular result of an experiment.
Sample Space. The sample space Ω, which is the set of all possible outcomes of an experiment. The symbol W will be used to denote the sample space.
Example:
What is the sample space for the roll of a single six-sided die?
Solution: The basic outcomes are the six possible face numbers, and the sample space is All 6 faces of a die: W = {1,2,3,4,5,6} Þ n(W) = 6
The sample space contains six basic outcomes. No two outcomes can occur together, and one of the six must occur.
Note: In many cases we are interested in some subset of the basic outcomes and no the individual outcomes. For example, for the roll of a die we might be interested in whether the outcome is even – that is, 2, 4, or 6.
Event
It is a collection of one or more outcomes of an experiment. An Event, E, is any subset of basic outcomes from the sample space. An event occurs if the random experiments results in one of its constituent basic outcomes. The null event represents the absence of a basic outcome and is denoted by Ø.
Events are often represented by uppercase letters, such as A, B, or C.
Notation: The probability that event E will occur is written P(E) and is read “the probability of event E.” Example: A die is rolled, W = {1, 2, 3, 4, 5, 6}, let A be the event “Number resulting is even” and B the event “Number resulting is at least 4”. Then:
A= {2, 4, 6}; and B= {4, 5, 6}
a. Find the complement of each event
The complements of these events are, respectively, = {1,3,5}; and = {1,2,3}
b. The intersection of A and B is the event “Number resulting is both even and at least 4” and so
c. The union of A and B is the event “Number resulting is either even or at least 4, or both” and so d. Intersection of and B
Since the only outcome that is both “not even” and “at least 4” is 5, it follows that:
Example
Two dice are thrown. What is the probability that the sum of their faces is four? 1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6
A: When throwing two dice there are 6×6 equiprobable events. (Equiprobability is a property for a collection of events that each have the same probability of occurring ).From these, only the events (1,3), (3,1), (2,2) are favorable. Therefore:
What is the probability that the sum of their faces is four? P(A) = 3/36 = 0.083
6.4 Elementary properties of probability
1. Nonnegativity P(A) ≥ 0, for every event A. Given some process (or experiment) with n mutually exclusive or disjoint events, A1; A2;...; An, the probability of any event Ai is assigned a nonnegative
number. That is, P(Ai) ³ 0.
In other words, all events must have a probability greater than or equal to zero, a reasonable requirement in view of the difficulty of conceiving of negative probability. A key concept in the statement of this property is the concept of mutually exclusive outcomes. Two events are said to be mutually exclusive if they cannot occur simultaneously.
2. Additivity If A and B are two disjoint events, then the probability of their union satisfies.
P(A∪B) = P(A) + P(B). Furthermore, if the sample space has an infinite number of elements and A1, A2,...is a sequence of disjoint events, then the probability of their union satisfies.
P(A1 U A2∪···) = P(A1) + P(A2)+···
3. Normalization The sum of the probabilities of the mutually exclusive outcomes is equal to 1. P(A1) + P(A2) + … P(An) = 1 , that is, P(W) = 1
This is the property of exhaustiveness and refers to the fact that the observer of a probabilistic process must allow for all possible events, and when all are taken together, their total probability is 1.
The requirement that the events be mutually exclusive is specifying that the events A1; A2;...; An do
not overlap; that is, they cannot be occur at the same time.
6.5 Approaches to Assigning Probabilities
Probability measures the extent to which an event is likely to occur and can be calculated from the ratio of favorable outcomes to the whole number of all possible outcomes.
There are two main ways of assessing the probability of a single outcome occurring: The objective approach and the subjective approach. See the following figure:
6.5.1 Objective probability
The objective probability is a probability that an event will occur based an analysis in which each measure is based on a recorded observation, it is subdivided into:
Classical probability Empirical probability
Classical or A Priory Probability
Examples
1. The record of weather stations report shows that out of the past 95 consecutive days, its weather forecast was correct 65 times. Find the probability that on a given day:
a. it was correct b. it was not correct. Solution:
Total number of days = 95
Number of correct weather forecast = 65
Number of not correct weather forecast = 95 - 65 = 30 a. Probability of ‘it was correct forecast’
Number of favorable outcomes
P(X) =Total number of possible outcome= 65/95 = 13/19 b. Probability of ‘it was not correct forecast’
Number of favorable outcomes P(Y) =Total number of possible outcome = 30/95 = 6/19
2. Consider an experiment of rolling a six-sided die. What is the probability of the event “an even number of spots appear face up”?
There are three “favourable” outcomes (2, 4, 6) in the collection of six equally likely possible outcomes. Therefore:
Probability of an even number = 3. Rolling a pair of dice
(all possibilities shown below - known as total sample space) 1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6
Find the probability of:
P(getting a sum of 8 points by throwing a die) = 5/36 P(double six) =
P(any double) = P(sum less than 4) =
P(the sum more than 8) = P ( sum > 8 ) = P(at least one six) =
P(the sum of the rolls is odd) =
Example:
When 3 unbiased coins are tossed once. What is the probability of: a. getting all heads
b. getting two heads c. getting one head d. getting at least 1 head e. getting at least 2 heads f. getting at most 2 heads Solution:
In tossing three coins, the sample space is given by Ω = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} And, therefore, n(Ω) = 8.
a. getting all heads
Let E1= event of getting all heads. Then, E1= {HHH} and, therefore, n(E1) = 1.
Therefore, P(getting all heads) = P(E1) = n(E1)/n(Ω) = 1/8. b. getting two heads
Let E2= event of getting 2 heads. Then,
E2= {HHT, HTH, THH} and, therefore, n(E2) = 3.
Therefore, P(getting 2 heads) = P(E2) = n(E2)/n(Ω) = 3/8. c. getting one head
Let E3= event of getting 1 head. Then,
E3= {HTT, THT, TTH} and, therefore, n(E3) = 3.
Therefore, P(getting 1 head) = P(E3) = n(E3)/n(Ω) = 3/8. d. getting at least 1 head
Let E4 = event of getting at least 1 head. Then, E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH} and, therefore, n(E4) = 7.
Therefore, P(getting at least 1 head) = P(E4) = n(E4)/n(Ω) = 7/8. e. getting at least 2 heads
Let E5 = event of getting at least 2 heads. Then, E5 = {HHT, HTH, THH, HHH}
and, therefore, n(E5) = 4.
Therefore, P(getting at least 2 heads) = P(E5) = n(E5)/n(Ω) = 4/8 = 1/2. f. getting at most 2 heads
Let E6= event of getting at most 2 heads. Then,
E6= {HHT, HTH, HTT, THH, THT, TTH, TTT} and, therefore, n(E6) = 7. Therefore, P(getting at most 2 heads) = P(E6) = n(E6)/n(Ω) = 7/8
Mutually Exclusive the occurrence of one event means that none of the other events can occur at the same time
Collectively Exhaustive At least one of the events must occur when an experiment is conducted
Empirical Probability (Probability from Relative Frequency)
Empirical probability or relative frequency probability is based on the number of times an event occurs as a proportion of a known number of trials.
The relative frequency of an event is simply the proportion of the possible times it has occurred.
The relative frequency probability is the limit of the proportion of times that event A occurs in a large number of trials, n:
Where nA is the number of A outcomes and n is the total number of trials or outcomes.
The probability is the limit as n becomes large (or approaches infinity).
The empirical approach to probability is based on what is called the law of large numbers. The key to establishing probabilities empirically is that more observations will provide a more accurate estimate of the probability.
Law of large numbers as an experiment is repeated over and over, the experimental probability of an event approaches the theoretical probability of the event. The greater the number of trials the more likely the experimental probability of an event will equal its theoretical probability.
Example:
Niyomwungeri Lydiane is considering an opportunity to establish a new car dealership in Kigali City, which has a population of 150,000 people. Experience from many other dealerships indicates than in similar areas dealerships will be successful if at least 40% of the households have annual incomes over $500. She has asked Shingiro Eric, a marketing consultant, to estimate the proportion of family incomes above $500, or the probability of such incomes.
Solution: After considering the problem, Eric decides that the probability should be based on the relative frequency. He first examines the most recent census data and finds that there were 54,345 households in Kigaly City and that 31,496 had incomes above $500. Eric computed the probability for event A, “Family income greater than $500” as
Since Eric knows that there are various errors in census data, he also consulted similar data published by Sales Management magazine. From this source he found 55,100 households, with 32,047 having incomes above $500. Eric computed the probability of event A from this source as
6.5.2. Subjective probability
Subjective probability is a probability measure resulting from intuition, educated guesses, and estimates.
Therefore, there is no formula to calculate it. Usually found by consulting an expert. Illustration of subjective probability
Estimating the likelihood you will be involved in a motorcycle accident in the next year
Estimating the likelihood the Rwandans budget deficit will be reduce by half in the next 10 years What is the probability that you will be married by age 30?
6.6 Rules for computing probabilities 6.6.1 Rules of Addition
Two events are mutually exclusive if they cannot occur at the same time. Another word that means mutually exclusive is disjoint.
If two events are disjoint, then the probability of them both occurring at the same time is 0. P(A U B) = P(A or B)
If A and B are mutually exclusive: P(A U B) = P(A)+P(B), in general: P(A1 U A2 U.... U An) = P(A1)+P(A2)+....+P(An)
Example.
A day of the week is chosen at random. What is the probability of choosing a Monday or Tuesday? P(Monday or Tuesday)= P(Monday) +P(Tuesday)=1/7+1/7=2/7
Non-Mutually Exclusive Events
In events which aren't mutually exclusive, there is some overlap. When P(A) and P(B) are added, the probability of the intersection (and) is added twice. To compensate for that double addition, the intersection needs to be subtracted.
a) If A, and B are not mutually exclusive P(A U B) = P(A)+P(B) - P(A ∩ B) b) If A, B and C are not mutually exclusive P(A U B U C) =
P(A U B U C) = P(A)+P(B)+P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B∩ C)
Example
Routine physical examinations are conducted annually as part of a health service program to General Concrete Inc. employees. It was discovered that 8% of the employees need corrective shoes, 15% need major dental work, and 3% need both corrective shoes and major dental work.
a. What is the probability that an employee selected at random will need either corrective shoes or major dental work?
b. Show this situation in the form of a Venn diagram. Solution
a. Need for corrective shoes is event A. Need for major dental work is event B. P(A U B)= P(A)+P(B) - P(A ∩ B)
6.6.2 Rules of Multiplication
Multiplication Rule of Probability means to find the probability of the intersection of events, the rules multiplication refers to the product of events.
Independent events If the occurrence of one event A does not affect, nor is affected by, the occurrence of another event B then we say that A and B are independent events. In such cases we say that A and B are independent events. The multiplication rule for two independent events, then, may be written as
P(A|B) = P(A) P(A)≠0 P(B|A) = P(B) P(B)≠0
Then, using the last key point formula we have, for independent events: If A and B are independent events then P(A ∩ B) = P(A)P(B)
In words ‘The probability of independent events A and B occurring is the product of the probabilities of the events occurring separately.’
Example: Suppose we roll one die followed by another and want to find the probability of rolling a 4 on the first die and rolling an even number on the second die.
Solution: P(A∩B) = P(A) * P(B)=1/6 * 3/6 =1/12
Example: From experience, about Goodyear Wrangler Radial Tire knows the probability is .94 that a particular XB-70 tire will last 70,000 miles before it becomes bald or fails. And adjustment is made on any tired that does not last 70,000 miles. You purchase four XB-70s. What is the probability all four tires will last at least 70,000 miles?
Solution: Each tired is independent event, therefore P(for one tired is)= .94 Then (.94)(.94) (.94)(.94) = 0.7807
The events that are not Independent or Dependent events by the definition of conditional probability we have P(A1∩A2)=P(A1)P(A2/ A1).
When you want to know the probability of two events occurring, that is called the intersection of the two events. The Multiplication Rule of Probability is used to find the intersection of two different sets of events, called independent and dependent events. Independent events are when the probability of an event is not affected by a previous event. A dependent event is when one event influences the outcome of another event in a probability scenario. To find the intersection of two events, whether they are independent or dependent, multiply the two probabilities together.
We can generalize this to n intersections A1∩A2∩···∩An, this gives the product rule of probability
P(A1 ∩ A2 ∩… ∩ An)=P(A1) P(A2/A1) P(A3/A1∩A2)···P(An|A1∩···∩An−1).
Assuming that all of the conditioning events have positive probability, we have
Note that any intermediate node along the path also corresponds to some intersection event and its probability is obtained by multiplying the corresponding conditional probabilities up to that node. For example, the event A1 ∩A2 ∩A3 its probability is
For the case of just two events, A1 and A2, the multiplication rule is simply the definition of conditional
probability. Example
A golfer has 12 golf shirts in his closet. Suppose nine of these shirts are white and the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it on. He plays golf two days in a row and does not launder and return the used shirts to the closet. What is the probability both shirts selected are white?
The event that the first shirt selected is white is A1, then the probability is P(A1)=9/12
The event that the second shirt selected is also white is identified as A2. The conditional probability that
the second shirt selected is white, given that the first shirt selected is also white, is P(A2 / A1) = 8/11
Why is this so? Because after the first shirt is selected there only 11 shirts remaining in the closet and 8 of these are white.
To determine the probability of two white shirts being selected: P(A1∩A2)=P(A1)P(A2/ A1).= (9/12) (8/11)= .55
So the likelihood of selecting two shirts and finding them both to be white is .55
6.7 Contingency tables
Another alternative is that probabilities can be estimated from contingency tables.
Contingency table is a table used to classify sample observations according to two or more identifiable categories or classes. In the others words is a table display all possible combined outcomes and the frequency with which each of them has happened in the past. These frequencies are used to calculate future probabilities.
Example. A new supermarket is to be built in City Town. In order to estimate the requirements of the local community, a survey was carried out with a similar community in a neighboring area. Part of the results of this is summarized in the table below:
Expenditure on groceries
Mode of travel None Under 20 Rwf At least 20 Rwf Total
On foot 40 20 10 70
By bus 30 35 15 80
By car 25 33 42 100
Total 95 88 67 250
Suppose we put all the till invoices into a drum and thoroughly mix them up. If we close our eyes and take out one invoice, we have selected one customer at random. Calculate:
a. P(customer spends over Rwf 20) = 67/250 = 0.268 b. P(customer will travel by car) =100/250=0.40
From the ‘cells’ we can calculate, for example:
c. P(customer spends over Rwf 20 and travels by car) = 42/250 = 0.168 (or using the rule of multiplication P( A and B) = P (A) P (B/A) = (67/250) (42/67) = 0.168
d. P(customer arrives on foot or spends no money on groceries) = (40 + 20 + 10 + 30 + 25)/250 =0.50 or (70+95-40)/250= 0.50
e. P(customer spends over Rwf 20 or travels by car)
Sometimes we need to select more than one row or column:
In the examples above all the customers have been under consideration without any condition being applied which might exclude any of them from the overall ratio. Often not all are included as some condition applies.
6.8 Conditional Probability
If the probability of the outcome of a second event depends upon the outcome of a previous event then the second is conditional on the result of the first. This does not imply a time sequence but simply that we are asked to find the probability of an event given additional information - an extra condition.
Example
We toss a fair coin three successive times. We wish to find the conditional probability P(A|B) when A and B are the events
A = {more heads than tails come up}, B = {1st toss is a head}. The sample space consists of eight sequences,
Ω={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT},
What we assume to be equally likely. The event B consists of the four elements HHH, HHT, HTH, HTT so its probability is
The event A∩B consists of the three elements outcomes HHH, HHT, HTH, so its probability is
Thus, the conditional probability P(A|B) is
Because all possible outcomes are equally likely here, we can also compute P(A|B) using a shortcut. We can bypass the calculation of P(B) and P(A∩B), and simply divide the number of elements shared by A and B (which is 3) with the number of elements of B (which is 4), to obtain the same result 3/4. Another example. Using the previous example about a new supermarket is to be built in City Town: If we need P(a customer spends at least Rwf 20, if it is known that he/she travelled by car)
We eliminate from the choice all those who did not arrive in a car, i.e. we are only interested in the third row of the table.
a. P(a customer spends at least Rwf 20 / he/she travelled by car) = 42/100 = 0.420
b. P(a customer came by car / he/she spent at least Rwf 20 on groceries) = 42/67 = 0.627 Note that
P(spending ³ Rwf 20, | he/she travelled by car) ¹ P(coming by car | he spent ³ Rwf 20 on groceries) c. P(a customer spends at least Rwf 20, | he/she travelled by bus) =
d. P(a customer came on foot, | he/she spent at least Rwf 20 on groceries) =
Example
Employment
Men 40 460 500
Women 260 140 400
Total 300 600 900
Adult is chosen randomly, what is the probability that: a. Be unemployed
b. Be unemployed as a woman? c. Be unemployed as a men? Solution:
Interpretation: When choosing an adult at random, the probability that the unemployed as a woman is 0.65.
Example
In a certain high school class, consisting of 60 girls and 40 boys, it is observed that 24 girls and 16 boys wear eyeglasses. If a student is picked at random from this class, the probability that the student wears eyeglasses, P(E). What is the probability that a student picked at random wears eyeglasses, given that the student is a boy?
Solution
By using the formula for computing a conditional probability, we find this to be
6.8 Tree Diagrams
The tree diagram is a graph that is helpful in organizing calculations that involve several stages. Each segment in the tree is one stage of the problem. The branches of a tree diagram are weighted by probabilities.
Steps:
1. Draw the appropriate tree diagram 2. Assign probabilities to each branch
3. Multiply the probabilities along branches to find the probability of the outcome at the end of each branch
4. Add the probabilities of the relevant outcomes, depending on the event. Example
Tree diagram for the toss of a coin: There are two "branches" (Heads and Tails)
How do we calculate the overall probabilities? We multiply probabilities along the branches We add probabilities down columns
The probability of "Head, Head" is 0.5×0.5 = 0.25
All probabilities add to 1.0 (which is always a good check)
The probability of getting at least one Head from two tosses is 0.25+0.25+0.25 = 0.75
6.9 Random Variable
Is a variable whose value is a number determined by the outcome of a random experiment is denoted by capital letters: X, Y, Z. and their individual values by the corresponding lowercase letter, if X (is a random variable) is, then their values are: X1, X2,... Xn
If every value of the random variable its probability we associate obtain the probability distribution of that variable.
Example: Let an experiment of tossing 2 coins and let us note the results of the head.
We define the random variable “Y” as the number of heads obtained in the 2 releases. Thus, taking the variable values may be 0, 1, 2 depending on the results of the experiment.
Sample space Random Variable
TT 0
HT 1
TH 1
HH 2
Then the probability distribution of this variable will be
Random Variable Probability
0 ¼
1 2/4
2 ¼
Total 1
6.10 Expected values
The expected value of each decision is defined by:
E(x) = åpx
Example: Two independent operations A and B are started simultaneously. The times for the operations are uncertain, with the probabilities given below:
Operation A Operation B
Duration (days) (x)
Probability (p)
Duration (days) (x)
Probability (p)
1 0 1 0.1
2 0.5 2 0.2
3 0.3 3 0.5
4 0.2 4 0.2
Determine whether A or B has the shorter expected completion time. Operation A
E(x) = åpx = (1 x 0.0) + (2 x 0.5) + (3 x 0.3) + (4 x 0.2) = 2.7 days Operation B
E(x) = åpx = (1 x 0.1) + (2 x 0.2) + (3 x 0.5) + (4 x 0.2) = 2.8 days Hence Operation A has the shorter completion time.
6.11 Principles of Counting
If the number of possible outcomes in an experiment is small, it is relatively easy to count them. If, however, there are a large number of possible outcomes, it would be tedious to count all the possibilities. To facilitate counting, we discuss two formulas:
1. Permutation formula 2. Combination formula
The Permutation formula
A permutation is an arrangement in which the order of the objects selected from a specific pool of objects is important.
Suppose that we have a collection of n objects, C = { C1, C2,... , Cn}. We want to make r selections from
C. How many possible ordered selections can we make?
If we are sampling with replacement, then we have r experiments, and each has n possible (equally likely) outcomes, and so by the multiplication principle, there are
n × n ×···× n = n
r ways of doing this. If we are sampling without replacement, then we have r experiments. The first experiment has n possible outcomes. The second experiment only has n−1 possible outcomes, as one object has already been selected. The third experiment has n−2 outcomes and so on until the rth experiment, which has n−r+1 possible outcomes. By the multiplication principle, the number of possible selections is:This is a commonly encountered expression in combinatorics, and has its own notation. The number of ordered ways of selecting r objects from n is denoted , where:
Permutations without repetition
r is total number of objects selected We refer to as the number of permutations of r out of n objects. Permutations with repetition:
Note. Factorial it is written n! and means the product of n(n-1)(n-2)(n-3)…(1). For instance, 4!=4 x 3 x 2 x 1 = 24
By definition, zero factorial, written 0! = 1
1! = 1 Example
We have a, b, and c three objects. Calculate the number of different permutations of three objects taken 2 times with and without repetition.
Permutations with repetition: n = 3 and r = 2
(a,b), (b,a), (b,c), (c,b), (a,c), (c,a), (a,a), (b,b), (c,c) Permutations without repetition: n = 3 y r = 2
(a,b), (a,c), (b,a), (b,c), (c,a), (c,b)
Example
Let us count the number of words that consist of four distinct letters. This is the problem of counting the number of 4-permutations of the 26 letters in the alphabet. The desired number is
The Combination Formula
A combination is an arrangement where the order of the objects selected from a specific pool of objects is not important.
There are n people and we are interested in forming a committee of k. How many different committees are there?
More abstractly, this is the same as the problem of counting the number of k-element subsets of a given n-element set. Notice that forming a combination is di erent than forming a k-permutation,ff
because in a combination there is no ordering of the selected elements. Thus for example, whereas the
2-permutations of the letters A, B, C, and D are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC, the combinations of two out four of these letters are AB, AC, AD, BC, BD, CD.
There is a close connection between the number of combinations and the binomial coe cient that wasffi
introduced in the next chapter.
The number of possible combinations, is given by
Example
The seven employees taken three at time would create 35 different teams.
When the number of permutation or combinations is large, the calculations are tedious. Computer software and handheld calculators have functions to compute these numbers. The Excel output is shown below:
PROBABILITY DISTRIBUTIONS
6.12 Introduction to probability distributions
Probability distributions are a mathematical model which can be used to describe a set of data.
You should by now be able to recognize whether the data is discrete or continuous (discrete: If the set of all possible values, when pictured on the number line, consists only integer values; continuous: if the set of all values, when pictured on the number line, consists of intervals) and a brief look at a histogram should give you some idea about its shape. The choice of an appropriate distribution depends upon the 'shape' of the data and whether it is discrete or continuous. (See in the following graph)
Probability distribution for Discrete Random Variable:
Binomial Distribution: yes/no experiments (two possible outcomes) Poisson Distribution
Hypergeometric Distribution
Probability distribution for Continuous Random Variable:
Normal Distribution T Student Distribution F Snedecor Distribution Exponential Distribution Chi-Square Distribution
Probability distribution for Discrete Random Variable: 6.13 Binomial Distribution
only one of two mutually exclusive outcomes, such as dead or alive, sick or well, full-term or premature, the trial is called a Bernoulli trial.
Conditions
1. The experiment consists on repeated trials.
2. Each trial results in one of two possible, mutually exclusive, outcomes. One of the possible outcomes is denoted (arbitrarily) as a success,and the other is denoted a failure (hence the name,
binomial).
3. The probability of a success, denoted by p, remains constant from trial to trial. The probability of a failure, 1- p, is denoted by q.
4. The trials are independent; that is, the outcome of any particular trial is not affected by the outcome of any other trial.
Some example questions that are modeled with a Binomial distribution are:
• From the children born in a given hospital on a given day, how many of them will be girls? • How many students in a given classroom will have computer?
We conduct n repeated experiments where the probability of success is given by the parameter p and add up the number of successes. This number of successes is represented by the random variable X. The value of X is then between 0 and n. When a random variable X has a Binomial Distribution with parameters p and n we write it as: X ~ Bin (n, p) {reads: “X is distributed binomially with parameters n and p} and the probability mass function is given by the equation:
n = the number of trials
x = 0, 1, 2, ..., n (# successes out of n trials) p = the probability of success in a single trial
q = the probability of failure in a single trial (i.e. q = 1 − p)
The Shape of the Binomial Distribution
Shape is determined by values of n and p Only truly symmetric if p = 0.5
Approaches normal distribution if n is large, unless p is very small.
Mean (expected value) and Variance of Binomial Distribution
If p is the probability of success and q is the probability of failure in a binomial trial, then the expected number of successes in n trials (i.e. the mean value of the binomial distribution) is:
Mean = µ → E(x) = np (number of “successes”)
The variance of the binomial distribution is Variance of distribution is
sx2 = V(x) = npq (q=1-p)
sx = Sd(x)=
Note: In a binomial distribution, only 2 parameters, namely n and p, are needed to determine the probability.
Example
Solution
The number of people asked, n=5. The probability of any given person answering "yes", p=0.3. (Remember, I said that 30% of people believe in life on other planets!)
We’re asking for the probability that exactly 2 people answer "yes" so x=2. This gives the equation:
Using Excel
Only 31% probability that 2 people of 5 confess believe that there is life on other planets. Similarly, the probability that any one person does not believe in extraterrestrial life is 69%, or 0.69, so the probability that any three people do not believe in extraterrestrial life is 0.69.
Example
In King Faisal Hospital records show that of patients suffering from a certain disease, 75% die of it. a. What is the probability that of6randomly selected patients, 4 will recover?
b. What is the expected value and variance of the number of patients will recover? Solution
a. This is a binomial distribution because there are only 2 outcomes (the patient dies, or recover). Let X = number who recover.
n=6 and x=4.
p=0.25 (success, that is, they live), q=0.75 (failure, i.e. they die). The probability that 4 will recover:
It means that out of the 6 patients chosen, the probability that four will recover is 0.032959 (3% of patients will recover)
b. E(X) = 6 (.25) = 1.5; Var(X) = 6 (.25) (.75) =1.125; Sd(X) = 1.06
Cumulative Binomial
Suppose that 80% of the employees in a large corporation participate in the voluntary group health insurance program. We take a representative sample of 16 employees from the corporation. Use the cumulative binomial distribution table to find the probability that the number participating is:
a. at most 7, answer =0.001 b. at least 7, answer =1 c. exactly 7, answer =0.001 d. fewer than 8, answer =0.001 e. more than 11, answer =0.798
f. in the interval from 7 to 13 (including both endpoints)
g. in the interval from 7 to 13 (not including either endpoint), answer =0.400
h. in the interval from 5 to 9 (including the first endpoint, but not the second one), answer =0.007 i. in the interval from 8 to 11 (not including the first endpoint, but including the second one), answer
=0.195
Solution (because the question are cumulative, so the answers is better with Excel or Binomial table)
6.14 Poisson Distribution
The name comes from the mathematician Siméon-Denis Poisson (1781-1840). The Poisson Distribution is very similar to the Binomial Distribution. We are examining the number of times an event happens (when we want to find probability of a certain number of events happening in a period of time), in other hands we used to calculate the probability of an event occurring over a certain interval. The interval can be time, area, volume or distance. Whereas the Binomial Distribution looks at how many times we register a success over a fixed total number of trials. Let’s examine a couple of experiments or questions that might have an underlying Poisson nature.
• Number of file server virus infections at a data center during a 24-hour period. • In the last week, how many e-mails viruses did your firewall deflect?
•The number of people entering a supermarket per hour.
• The number of customers who call to complain about a service problem per month • Number of customers arriving at a bank ATM in a given minute
• How many defects will there be per 100 meters of rope sold?
What’s a little different about this distribution is that the random variable X which counts the number of events can take on any non-negative integer value.
Instead of having a parameter p that represents a component probability like in the Bernoulli and Binomial distributions, this time we have the parameter "lambda" or λ which represents the "average or expected" number of events to happen within our experiment. The probability mass function of the Poisson is given by
Where
P(X=x) = is the probability for a specified value of x
e = Euler’s constant ≈2.718282 (based on the Napierian logarithm system) λ=Mean or expected value of the variable
x=Number of success in the event !=factorial
Conditions:
1. Each successful event must be independent.
2. The probability of success over a short interval must equal the probability of success over a longer interval.
Mean and Variance of Poisson Distribution (Expectation and Variance) E[X] =λ = np
σ2 = Var(X ) = λ = np
Example
At a call centre the average number of calls received per minute is 7. Calculate the probability that, in the next minute, there are:
a. four calls b. eight calls c. zero calls
d. at most three calls
e. What is the expected number of calls in a minute? Solution
Let X denote the number of calls received by the call centre in a minute
The number of calls received by the call centre per minute follows an average rate of 7. Therefore X∼Po(7) and so:
a.
b. c.
d.
Example
Twenty sheets of aluminum alloy were examined for surface flaws. The frequency of the number of sheets with a given number of flaws per sheet was as follows:
Number
of flaws Frequency
0 4
1 3
2 5
3 2
4 4
5 1
6 1
Total 20
What is the probability of finding a sheet chosen at random which contains 3 or more surface flaws?
Solution
The total number of flaws is given by:
(0x4)+ (1x3)+ (2x5)+ (3x4)+ (4x4)+ (5x1)+ (6x1)=46
So the average number of flaws for the 20 sheets is given by:
The required probability is:
Interpretation. The probability that we find a sheet that chosen at random which contains 3 or more surface flaws is 0.40396
P(X≥3) = 1- P(X≤2) =1 - 0.596 = 0.404
When the value of n is very large it can be very tedious and time-consuming to calculate probabilities be used as an approximation to the Binomial distribution, where larger values of n and smaller values of p result in a better approximation.
The Poisson approximation to the Binomial distribution has mean λ=np (where n is the number of trials and p is the probability of success for a Binomial distribution) and the same variance λ=np.
Example
Suppose in a big town there are 5000 business and a fire occurs in any given business with probability 0.001 over a decade. What’s the probability that there is at most one fire in the next decade?
Solution
Let X denotes the number of fires in the city in the next decade. We wish to find P(x≤1) = P(x=0) +P(x=1)
λ= E(x) = np = 5000x0.001 = 5 so X~Po(5) P(x≤1) = P(x=0) +P(x=1) = 0.040
An insurance company could use this approximation to calculate how likely it is a business will need to claim for fire damage or the expected number of claims likely to be made.
6.15 Hypergeometric Distribution
The hypergeometric distribution is a probability distribution that’s very similar to the binomial distribution. In fact, the binomial distribution is a very good approximation of the hypergeometric distribution as long as you are sampling 5% or less of the population.
Supose a population consists of N items, K of which are successes. And a random sample drawn from that population consists of n items, x of which are successes. Then the hypergeometric probability is:
Where:
K is the number of successes in the population. x is the number of successes in the sample. N is the number of size in the population
n is the size of the sample or the number of trials.
A sample of n elements are selected at random without replacement from a population of N items. Conditions
1. An outcome on each trial of an experiment is classified into one or two mutually exclusive categories (a success or failure).
2. The random variable is the number of successes in a fixed number of trials. 3. The trials are no independent.
4. We assume that we sample from a finite population without replacement and n/N > 0.05. So, the probability of success changes for each trial.
Mean (expected value) and Variance of Hypergeometrics Distribution
A wallet contains 3 $100 bills and 5 $ 1 bills. You randomly choose 4 bills. What is the probability that you will choose exactly 2 $ 100 bills?
Using Excel
Example
Boxes contain 200 items of which 10% are defective. Find the probability that no more than 2 defectives will be obtained in a sample of size 10. (Without Replacement Sampling)
Solution
Probability distribution for Continuous Random Variable:
6.16 Normal Probability Distributions
A symmetric distribution defined on the range -¥ to + ¥ whose shape is defined by two parameters, the mean, denoted µ, that centers the distribution, and the standard deviation, σ, that determines the spread of the distribution.
Where
X is the value of any particular observation or measurement. m is the mean of the distribution
s is the standard deviation of the distribution
are physical constants,
The Normal distribution is found to be a suitable model for many naturally occurring variables which tend to be symmetrically distributed about a central modal value - the mean.
Also called a “Gaussian” distribution
Centered around the mean (
m
)
with a width determined by the standard deviation (s
)
Total area under the curve = 1.0If the variable described is continuous, its probability function is described as a probability density function which always has a smooth curve under which the area enclosed is unity.
The Characteristics of any Normal Distribution
The normal distribution approximately fits the actual observed frequency distributions of many naturally occurring phenomena e.g. human characteristics such as height, weight, IQ etc. and also the output from many processes, e.g. weights, volumes, etc.
There is no single normal curve, but a family of curves, each one defined by its mean, µ, and standard deviation, s; µ and s are called the parameters of the distribution.
As we can see the curves may have different centers and/or different spreads but they all have certain characteristics in common:
The curve is bell-shaped,
It is symmetrical about the mean (µ), The mean, mode and median coincide.
The Area beneath the Normal Distribution Curve
No matter what the values of µ and s are for a normal probability distribution, the total area under the curve is equal to one. We can therefore consider partial areas under the curve as representing probabilities.
Note that the curve neither finishes nor meets the horizontal axis at m ± 3s, it only approaches it and actually goes on indefinitely.
Even though all normal distributions have much in common, they will have different numbers on the horizontal axis depending on the values of the mean and standard deviation and the units of measurement involved. We cannot therefore make use of the standard normal tables at this stage.
The Standard Normal Probability Distribution
• It is tedious to ‘integrate’ a new normal distribution for every population, so use a ‘standard normal distribution’ with standard tabulated areas.
• Convert your measurement x to a standard normal value (Transformation of each x-value to a z-value)
If X is normally distributed , the standardized variable Z has a Standard Normal Distribution. Its mean is 0 and standard deviation is 1, denoted N(0,1). (Areas tabulated in any statistics text book, even in Excel)
No matter what units are used to measure the original data, the first step in any calculation is to transform the data so that it becomes standardized normal following the standard distribution which has a mean of zero and a standard deviation of one. The effect of this transformation is to state how many standard deviations to particular given value is away from the mean.
Use of Standard Normal Distribution
The formula for calculating the exact number of standard deviations (z) away from the mean (m) is: ;
The process of calculating z is known as standardizing, producing the standardizedvalue which is usually denoted by the letter z, though some tables may differ.
Knowing the value of z enables us to find, from the Normal tables, the area under the curve between the given value (x) and the mean (m) therefore providing the probability of a value being found between these two values.
Finding Probabilities under a Standard Normal Curve
The steps in the procedure are: draw a sketch of the situation,
standardize the value of interest, x, to give z,
use the standard tables or excel to find the area under the curve associated with z, if necessary, combine the area found with another to give the required area,
convert this to: a probability, using the area as found since total area = 1; or a percentage, multiplying the area by 100 since total area = 100%; or a frequency, multiplying the area by the total frequency; as required by the question.
Notation
Suppose X has a normal distribution with mean m and standard deviation s, denoted X ~ N(m, s). Then a new random variable defined as Z=(X- m)/ s, has the standard normal distribution, denoted Z ~ N(0,1).
How do you use the Standard Normal Distribution?
Use the area UNDER the normal distribution
For example, the area under the curve between x=a and x=b is the probability that your next measurement of x will fall between a and b
Example: Between which values is 95% of the data, if you have a normal distribution with an average of 75 and a standard deviation of 10.
95% of the area is within 1.96 standard deviations of the mean.
The shaded area contains 95% of the area and extends from 55.4 to 94.6.A normal distribution with a mean of 75 and a standard deviation of 10. For all normal distributions for two tails, 95% of the area is within 1.96 standard deviations of the mean.
The shaded area contains 95% of the area and extends from 55.4 to 94.6.
How do you get m and s?
• To draw a normal distribution you must know m and s
In practice, you have a finite number of measurements with mean x and standard deviation s
• For now, m and s will be given; later we’ll use x and s to estimate m and s Example
In order to estimate likely expenditure by customers at a new supermarket, a sample of invoices from a similar supermarket describing the weekly amounts spent by 500 randomly selected customers was analyzed. These data were found to be approximately normally distributed with a mean of Rwf 50 and a standard deviation of Rwf 15. Using this knowledge we can find the following information for shoppers at the new supermarket.
The probability that any shopper selected at random: a. spends more than Rwf 80 per week,
b. spends less than Rwf 50 per week.
The percentage of shoppers who are expected to: c. spend between Rwf 30 and Rwf 80 per week,
d. spend between Rwf 55 and Rwf 70 per week. The expected number of shoppers who will:
e. spend less than Rwf 70 per week, and calculate expected number of shoppers who will spend less than Rwf 70 per week
f. spend between Rwf 37.50 and Rwf 57.50 per week. The value below which:
g. 70% of the customers are expected to spend, Solution
We assume that expenditure by customers is normally distributed. In this way, we use the Normal Distribution as a model for measurement.
First, we transform expenditure by customers into a Z-score, using the z-score transformation equation:
a. The probability that a shopper selected at random spends more than Rwf 80 per week
Find P(Z > a). The probability that a standard normal random variable (z) is greater than a given value (a) is easy to find. The table shows the P(Z < a). The P(Z > a) = 1 - P(Z < a).
Suppose, for example, that we want to know the probability that a z-score will be greater than 3.00: From the Normal probabilistic table, we find that P(Z < 3.00) = 0.9987
Therefore, P(Z > 3.00) = 1 - P(Z < 3.00) = 1 - 0.9987 = 0.0013. In our example:
We need P(x > Rwf 80), the probability that a customer spends over Rwf 80
Descriptive Statistics – Dr. Rosa Padilla de Casamayor m = £50, s
= £15,
x = £80.
5 20 35 50 65 80 95
Q
1z
1Q
2z
2z
m = Rwf 50,First standardize:
From tables: (z in the margin, Q in the body of the tables) From Excel:
Therefore: P(x > Rwf 80) = P(z > 2) = 1-P(z ≤ 2) = 1-0.9773 = 0.0237
b. The probability that a shopper selected at random spends less than Rwf 50 per week
No need to do any calculations for this question. The mean is Rwf 50 and, because the distribution is normal, so is the median. Half the shoppers, 250, are therefore expected.
Therefore: P(x < Rwf 50) = P(z <0) = 0.500
c. The percentage of shoppers who are expected to spend between Rwf 30 and Rwf 80 per week
Find P(a < Z < b). The probability that a standard normal random variable is between two values is also easy to find. The P(a < Z < b) = P(Z < b) - P(Z < a).
In our example, we first need P(30 < x < 80) = P(x≤80) - P(x≤30)
Therefore
The whole area is equivalent to100% so 0.8860 of it = 88.6% The expected number of shoppers who will:
e. spend less than Rwf 70 per week, First:
Second: Expected number of shoppers who will spend less than Rwf 70 per week: 500*0.9088 = 454.39
Finding values from given proportions
In this example in parts (a) to (f), we were given the value of x and had to find the area under the normal curve associated with it. Another type of problem gives the area, even if indirectly, and asks for the associated value of x, in this case the value of the shopping basket. Carrying on with the same example we shall find the following information:
g. 70% of the customers are expected to spend,
The value below which 70% of the shoppers are expected to spend 70% of the total area, 1.00, is below x Þ
From Excel we ask for NORMSINV in Function:
Approximately to 70% in our normal standard distribution, correspond the value Z=0.524 We now know the value of z and need to find x.
Using standardizing formula:
The value below which 70% of customers spend is 57.867 Rwf
The value below which:
h. 350 of the shoppers are expected to spend,
Review problems of chapter
1. What is probability? Discuss three applications of probability in everyday life.
2. The center for Child Care reports on 539 children and the marital status of their parents. There are 333 married, 182 divorced, and 24 widowed parents. What is the probability a particular child chosen at random will have a parent who is divorced? Which approach did you use? Answer: P(d)=.338, approach empirical probability
3. What is the probability that you will save one million dollars by the time you retire? Which approach to probability did you use to answer this question? Answer: Subjective probability
4. A survey of 34 students at the Wall College of Business showed the following majors:
Accounting 10
Finance 5
Economics 3
Management 6
Marketing 10
Suppose you select a student and observe his or her major.
a. What is the probability he or she is a management major? Answer: .176
b. Which concept of probability did you use to make this estimate? Answer: Empirical approach 5. A sample of 40 oil industry executives was selected to test a questionnaire. One question about
a. What is the experiment? Answer: The survey of 40 people about environmental issues b. List one possible event Answer: 20 or more respond yes, for example
c. Ten of the 40 executives responded yes. Based on these sample responses, what is the probability that an oil industry executive will respond yes? Answer: 10/40=.25
d. What concept of probability does this illustrate? Answer: empirical
6. We throw two dice. Given that the sum of the eyes is 10, what is the probability that one 6 is cast? Answer = 2/3.
7. A machine fill plastic bags whit a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight. A check of 4000 packages filled in the past month revealed:
Weight Event Number of package
Underweight A 100
Satisfactory B 3600
Overweight C 300
Total 4000
What is the probability that a particular package will be either underweight or overweight? Answer: .10 (the events are mutually exclusive)
8. In a pet store, there are 6 puppies, 9 kittens, 4 gerbils and 7 parakeets. If a pet is chosen at random, what is the probability of choosing a puppy or a parakeet? Answer: 13/26=.50
9. A number from 1 to 10 is chosen at random. What is the probability of choosing a 5 or an even number? Answer: 6/10
10. The Event A and B are mutually exclusive. Suppose P(A) = .30 and P(B) = .20. a. What is the probability of either A or B occurring? Answer: .50
b. What is the probability that neither A nor B will happen? Answer: .50 11.A study of 200 advertising firms reveled their income after taxes:
Income after Taxes Number of Firms
Under $1 million 102
$ 1 million to $20 million 61
$20 million or more 37
a. What is the probability an advertising firm selected at random has under $1 million in income after taxes? Answer: .51
b. What is the probability an advertising firm selected at random has either an income between $1 million and $20 million, or an income of $20 million or more? What rule of probability was applied? Answer: 0.49. Rule of addition
12. The probabilities of the events A and B are .20 and .30, respectively. The probability that both A and B occur is .15. What is the probability of either A or B occurring? Answer: .35
13.The table bellow summarizes the community types and degree of pollution for 268 locations in Uganda
Location/pollution Low Moderate High Total
Rural 33 23 9 65
Suburban 8 23 20 51
Urban 7 10 73 90
Commercial 3 11 48 62
Total 51 67 150 268
a. Suburban? (Answer: p=0.19) b. Not commercial? (Answer: p=0.77)
c. Rural, given that pollution is low? (Answer: p=0.65)
d. Highly polluted, given that the location is urban? (Answer: p=0.81) e. Urban or highly polluted? (Answer: p=0.62)
f. Commercial and moderately polluted? (Answer: p=0.04) [P(C and Moderate pollution)=P(C)P(M/C)] multiplication rule
g. Rural and not highly polluted? (Answer: p=0.21)
14.An automobile dealership in Nairobi has compiled the following sales data over the past year:
Number of cars sold per day
Relative
frequency E(x)
0 0.2 0
1 0.3 0.3
2 0.3 0.6
3 0.15 0.45
4 0.05 0.2
Y 1 1.55
a. What’s the probability there will be more than 2 cars sold in a day? (Answer: p=0.20) b. What is the probability that at least 1 car will be sold in a day? (Answer: p=0.80) c. What is the expected value of the number of cars sold in a day? (Answer: E(x)=1.55)
d. Based on the expected value, how many cars would the dealer expect to sell in a 31 day month? (Answer: 31*1.55=48.05. The dealer should expect to sell about 48 cars a month)
15. A marketing manager is considering whether it would be more profitable to distribute his company's product on a national or on a more regional basis. Given the following data, what decision should be made?
National distribution Regional distribution
Level of Demand
Net profit £m (x)
Prob. that demand is
met (p)
E(x)=px DemandLevel of Net profit£m (x)
Prob. that demand is met (p)
E(x)=px
High 4 0.5 High 2.5 0.5
Medium 2 0.25 Medium 2 0.25
Low 0.5 0.25 Low 1.2 0.25
Expected profits for National distribution =2.625 Expected profits for Regional distribution =2.050
Review problems of Probability Distribution
16. For each of the following, is the random variable discrete or continuous? a. The time required to complete a phone call to a prospective client. b. The number of articles purchased by the client.
c. The shipping weight of the articles. d. The distance the articles are shipped. e. The purchase price of house
17. If I toss a coin 20 times, what’s the probability of getting exactly 10 heads? Answer: 0.176197 18. If the probability of being a smoker among a group of cases with lung cancer is .6,
b. More than 5? Answer: 0.3153
c. What are the expected value and variance of the number of smokers? E(x)= 4.8, V(x)=1.92, Sd=1.38
19. There are five flights daily from East Africa via RwandAir, Ethiopian Airways, Kenya Airways to the International Airport Kigali. Suppose the probability that any flight arrives late is .20.
a. What is the probability that none of the flights are late today? Answer: 0.3277 b. What is the probability that exactly one of the flights is late today? Answer: 0.4096
c. Determine the mean and standard deviation of the number that will arrive late. Answer: E(x)=1, v(x)=.80
20. The ratio of boys to girls at birth in Singapore is quite high at 1.09:1.
What proportion of Singapore families with exactly 6 children will have at least 3 boys? (Ignore the probability of multiple births.). The probability of getting a boy is:
[Interesting and disturbing trivia: In most countries the ratio of boys to girls is about 1.04:1, but in China it is 1.15:1 (this means that for every 1.15 boys, there is 1 girl in China).] Answer: 0.69565 21. A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because
they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain. a. no more than 2 rejects? Answer:
b. at least 2 rejects? Answer:
22. PlayTime Toys, Inc. employs 50 people in the assembly department. Forty of the employees belong to a union and ten do not. Five employees are selected at random to form a committee to meet with management regarding shift starting times. What is the probability that four of the five selected for the committee belong to union? (The sampling is done without replacement).
Answer: 0.4313. Interpretation. The probability of selecting 5 assembly workers at random from the 50 workers and finding 4 of the 5 are union members is 0.4313.
23. The human resources department of Kigali Bank plans to hire 5 new financial analysts this year. There is a pool of 12 approved applicants, and the department of Human Resources, decides to randomly select those who will be hired. There are 8 men and 4 women among the approved applicants. What is the probability that 3 of 5 hired are men? Answer: 0.4242
24. A life insurance salesman sells on the average 3 life insurance policies per week. Use Poisson's law to calculate the probability that in a given week he will sell
a. Some policies. Answer: 0.95021
"Some policies" means "1 or more policies". We can work this out by finding 1 minus the "zero policies" probability:
b. 2 or more policies but less than 5 policies. Answer:0.61611
c. Assuming that there are 5 working days per week, what is the probability that in a given day he will sell one policy? Answer: 0.329287
25. If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week. Answer: 0.989814
"Not more than one failure, means we need to include the probabilities for “0 failure" plus “1failure". 26. Vehicles pass through a junction on a busy road at an average rate of 300 per hour.
a. Find the probability that none passes in a given minute. Answer: 0.006738 b. What is the expected number passing in two minutes? Answer: 10
27. In order to estimate likely expenditure by customers at a new supermarket, a sample of invoice from a similar supermarket describing the weekly amounts spent by 500 randomly selected customers was analyzed. These data were found to be approximately normally distributed with a mean of Rwf 50 and a standard deviation of Rwf 15. Using this knowledge we can find the following information for shoppers at the new supermarket.
The probability that any shopper selected at random: a. spends more than Rwf 60 per week. Answer: 0.2514 b. spends less than Rwf 40 per week. Answer: 0.2514)
The percentage of shoppers who are expected to:
c. spend between Rwf 30 and Rwf 60 per week. Answer: 65.68% d. spend between Rwf 55 and Rwf 80 per week. Answer: 34.67%
The expected number of shoppers who will:
e. spend less than Rwf 75 per week. Answer: 476
f. spend between Rwf 40.5 and Rwf 60.50 per week. Answer: 247
The value below which:
g. 400 of the shoppers are expected to spend Answer=62.75 Rwf h. 60% of the customers are expected to spend. Answer: 53.80 i. 45% of the customers are expected to spend. Answer: 48.115
28. If scores are normally distributed with a mean of 30 and a standard deviation of 5, what percent of the scores is:
a. greater than 30?. Answer: 50% b. greater than 37?. Answer: 8.08% c. between 28 and 34?. Answer: 44.36%
29. A Company services copiers. A review of its records shows that the time taken for a service call can be represented by a normal random variable with mean 75 minutes and standard deviation 20 minutes.
a. What proportion of service calls taken less than 1 hour. Answer: 22.66% b. What proportion of service call takes more than 90 minutes? Answer: 22.66%
30. Molly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (Assume that test scores are normally distributed.). Answer: we estimate that 18.41 percent of the students tested had a higher score than Molly.
31. You have just received your score on a test of intelligence. If your score was 78 and you know that the mean score on the test was 67 with standard deviation of 5, how does your score compare with the distribution of all test scores? Answer: The score is higher than 98.61%, of the all test scores, in other hands you did pretty very well.
