Unit5 Statistics MATH2(D) Ikbn

7/29/2019 Unit5 Statistics MATH2(D) Ikbn

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UNIT 5

STATISTICS

5.1 Introduction

The concept of statistics is introduced in this topic. It covers the definition and the

application involving thepermutations and combinations, probability of any event,

Binomial distribution and Normal distribution.

Objectives

At the end of the topic, you will be able to: Differentiate between permutations and combinations.

Calculate the problem involving permutations and combinations.

Estimate the probability of any event.

Calculate the probability of events for Binomial and Normal distribution.

5.2 Permutations and Combinations

5.2.1 Method of Counting

The Multiplication Rule

Consider an experiment with two stages, one stage following the other. Suppose there

are m possible outcomes in the first stage and that for each of the possible outcomes in

the first stage, there are n possible outcomes in the second stage.

The total number of possible outcomes for the experiment is

m x n.

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In general, the multiplication rule is used to find the number of possible outcomes for an

experiment that consists of at least two stages.

Example 5.1

A traveller wants to go to Town D from Town A via Town B and town C. Suppose

there are 3 different routes from Town A to Town B, 2 different routes from Town B to

Town C and only 1 route from Town C to Town D. How many different routes can the

traveller takes?

Solution:

For this experiment, there are 3 stages;

First stage (From Town A to Town B): 3 possible outcomes (routes).

Second stage (From Town B to Town C): 2 possible outcomes (routes).

Third Stage (From Town C to Town D): 1 possible outcome (route).

Therefore, under the multiplication rule, the number of different routes the traveller can

take = 3 x 2 x1 = 6 or 3!, which is called three factorial.

Hence; 3! = 3 x 2 x 15! = 5 x 4 x 3 x 2 x 1

Generally, for positive integer n,

n! = n x (n-1) x (n-2) x x 3 x 2 x 1 .

5.2.2 Permutation

Permutation is an ordered arrangement of a number of distinct

objects

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Our objective is to find the number of different ways of ordering or arranging a number

of objects selected from a complete set or an original set. Selection from the original set

is either selection with repetition (sampling with replacement) or selection without

repetition. (sampling without replacement).

Consider the following experiments:

Case 1 : A box contains 10 balls numbered 1, ....10.

First, one ball is selected at random from the box and its number is noted. The ball is

then put back in the box and a ball is selected and its number is asked. The process is

continues until a desired number of balls are selected in this ways. This process is called

selection with repetition. It is assumed that each of the 10 balls is equally likely to be

selected at each selection and that all selections are made independently of each other.

Case 2 : A box contain 10 balls numbered 1, .. 10.

First, one ball is selected at random from the box and its number is noted. The ball is

removed from the box. A ball is then selected from the remaining balls and the number

is noted. Then, process continues until a desired number of balls are selected in this

way, a maximum of 10 balls. This process is called selection without repetition.

Definition 1

The number of permutations of n different object selected with

repetition from a set of n different objects is nn .

Definition 2

The number of permutations of n different objects selected

without repetition from a set of n different objects is n!.

Symbol used : .nPn = n!

Definition 3

The number of permutations of r different objects selected with

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repetition from a set of n different objects is nr.

Definition 4

The number of permutations of r different objects selected with

repetition from a set of n different objects is n!_(n-r) !

Symbol used: nPr = n!(n-r)!

Definition 5

The number of permutations of n different objects selected

without repetition of which p objects are of the first kind, q

objects are of the second kind and r objects are of the third kind

where p + q + r = n is given by

Symbol used: = __n!__p! q! r!

Example 5.2

Suppose an original set A is given as A = {2, 4, 6}. How many 3digit numbers can

possibly be formed from set A if :a. Each digit can be used more then once?

b. Each digit can be used only once?

Solution:

a. Since each digit can be used more than once, the selections from set A are with

repetition:

1st digit 2nd digit 3rd digit

_________________________________________________________

3 ways to determine 3 ways to determine 3 ways to determine

(record and returned) (record and returned) (record and returned)

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Therefore, the number of possible ways to form 3-digit numbers from a set of 3

different digits with repetition

= 3 x 3 x 3

= 33 = 27 .

b. Since each digit can be used only once, the selection from set A is without repetition


________________________________________________________

3 ways to determine 2 ways to determine 1 way to determine

(record and removed) (record and removed) (record and removed)

Therefore, the number of possible ways to from 3-digit numbers from a set 3 different

digits without repetition

= 3 x 2 x 1

= 3! = 6 .

Example 5.3

How many ways can the alphabets in the word STATISTICS be arranged?

Solution:

The number of ways

= 10! = 50,400,0003! 3! 2! 1! 1!

Note: 10! Number of ways to arranging the 10 alphabets STATISTICS.

3! Number of ways of arranging the 3 alphabets S and 3 alphabets T.

2! Number of ways of arranging the 2 alphabets I.

1! Number of way of arranging 1 alphabet A and alphabets C.

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Practice 5.1

Suppose an original set A is given as A = {1, 2, 3, 4, 5}. How many 3digit numbers

can possibly be formed from set A if each digit can be used only once?

Solution

Since each digit cannot be used more than once, the selections from set A are without

repetition:


_________________________________________________________

_ ways to determine _ ways to determine _ ways to determine

(record and returned) (record and returned) (record and returned)

Therefore, the number of possible ways to form 3-digit numbers from a set of 3

different digits without repetition

= _ x _ x _

= __

Practice 5.2

Suppose an original set A is given as A = {2, 4, 6}.How many 2-digit numbers can

possibly be formed from set A if:

a. Repetition is allowed?

b. Repetition is not allowed?

Solution

a. Selection with repetition.

1st digit 2nd digit

_____________________________________

_ ways to determine _ ways to determine

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(Record and returned) (Record and returned)

Therefore, the number of possible ways to from 2-digit numbers from a set of 3

different digits with repetition = __ x __

= 32 = __ .

b. Selection without repetition.

1st digit 2nd digit

_____________________________________

_ ways to determine _ ways to determine

(Record and removed) (Record and removed)

Therefore, the number of possible ways to from 2- digit numbers from a set of 3

different digits without repetition = _ x _ = _

or =!1

!3=

)!23(

!3

(from the definition) .

Practice 5.3

Suppose that 18 red beads, 12 yellow beads, 8 blue beads and 12 black beads are to be

strung in a row. How many different arrangements of the beads can be formed?

Solution

The number of different arrangements of 50 beads of which 18 are red, 12 yellow, 8

blue and 12 black

= ___________ = _______

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5.2.3 Combination

Our objective is to find the number of different subset containing a certain number of

elements from an original set. It is important to note that no two combinations will

consist of exactly the same elements since the arrangement of the element in each

subset is irrelevant. For example, the subjects {a, b} and {b, a}are considered identical

and are treated as one combination. However, the subset are considered as two

permutations because {a, b} and {b, a} are two different arrangements.

Definition

The number of combinations of r different objects selectedwithout repetition from a set of n different objects is

)!(!

!

rnr

n

Symbol used: nCr =)!(!

!

rnr

n

Example 5.4

Suppose that committee composed of 8 people is to be selected from a group of 20

peoples. Find the number of different groups can be formed.

Solution:

Therefore, the number of different groups is,

C20, 8 = __20!_____8! (20-8)!

Combination is a subset of elements (objects) selected

from a original set.

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= __20!__8! 12!

= 125,970Example 5.5

A team of 4 students is to be selected from 4 male and 6 female students. Find thenumber of different groups could be formed consisting 2 male and 2 female students.

Solution:

For the team to consist 2 male and 2 female students, we have to select 2 male from 4

male students and 2 female from 6 female students.

Therefore:

The number of ways of choosing 2 male students = 4C2

The number of ways of choosing 2 female student = 6C2

The number of ways the team consisting of 2 male and 2 female students can be formed

= 4C2 x 6C2

= 6 x 15

= 90 .

Practice 5.4

A team of 4 students is to be selected from 4 male and 6 female students. Find the

number of different groups could be formed consisting at least 2 male students.

Solution

The team can be formed in the following ways:

Male Female No. of ways

i 2 and __ _____________ = __

or ii 3 and 1 4C3 x 6C1 = 24

or iii 4 and 0 _____________ = __

Total number of different groups could be formed consisting at least 2 male students

= __+ 24 +__

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= ___

Practice 5.5

A team of 3 students is to be selected from 5 Malay and 3 Chinese students. Find the

number of different groups could be formed consisting at most 2 Malay students.

Solution

The team can be formed in the following ways:

Malay Chinese No. of ways

i 0 and __ 5C0 x 3C3 = 1

or ii 1 and 2 _____________ = __

or iii 2 and __ _____________ = __

Total number of different groups could be formed consisting at least 2 male students =

1+ __ +__ = ___

5.3 Probability

5.3.1 Introduction

A probability is the numeric value representing the chance, likelihood or possibility a

particular event will occur such as a rainy day, the price of a stock increasing, a

nonconforming unit of production or the outcome head in one toss of a coin. In these

entire events, the probability attached is a proportion or fraction whose value ranges

between 0 and 1 inclusively. There are 3 approaches to the subject of probability;

(i) Classical probability

(ii) Empirical probability

(iii) Subjective probability

For the classical probability, the probability of success is based on prior knowledge of

the process involved where each outcome is equally likely. In the empirical probability,

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the outcomes are based on the observed data or experiment, not on prior knowledge of a

process. The subjective probability differs from the other two approaches because

subjective probability differs from person to person. To deal with problem of

probability, it is helpful to know the concepts of experiment, samplespace and event.

Experiment: Any process whose outcome is not known in advance with certainty

but the entire possible outcome can be specified.

Sample space: A collection or set of all possible outcomes of an experiment and

each outcome is an element of the sample space.

Event: A collection or set of specific outcomes of an experiment.

Example:

Experiment: Roll a balanced die.

[The outcomes is not known in advance]

Sample space: The possible outcomes of rolling a die are the numbers 1, 2, 3, 4, 5 and

6. [If we denote S be the sample space of the experiment, therefore

S = {1, 2, 3, 4, 5, 6}

Event: The event that an even number is defined byA= {2, 4, 6}.

The event B that a number greater than 3 is defined by B={4, 5, 6}.

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5.3.2 Operations of Set Theory

Suppose thatA and B are any two events of an experiment with a sample space, S.

a. Unions : The unions of A and B, denoted by A U B, is defined to be the event

containing all outcomes that belongs to A alone, to B alone, or to both A

and B.

The event A U B

Properties: A U B = B UA

A U =A A U B = B ifAB

b. Intersections: The intersections ofA and B, denoted byAB, is defined to be the

event containing all outcomes belonging both toA and to B.

The eventAB

Properties: AB = BA A = AB =A ifAB

c. Complements: The complement of A, denoted by A, is defined to be the event

containing all outcomes in the sample space, S, which do not belong

toA.

The eventA

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A B

A B

A

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Properties: (A) =A

AA = S

AA =

d. Disjoint Events : A and B are disjoint, or mutually exclusive if A and B have no

outcomes in common. If follows thatAA =

The disjoint

events ofA and B

5.3.3 Probability Calculation Techniques

Single Event

The probability of an eventA, is given by:

P(A) = n(A)n(S)

where n(A) is the number of elements in setA,

n(S) is the number of elements is sample space

Under this technique, the sample space must be a finite simple sample space. The

number of elements in the finite simple space is countable. It is convenient to have a

method of determining the total number of outcomes in the sample space and in various

events without compiling a list of all these outcomes.

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A B


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Probability Theorems

Marginal, Joint and Conditional Probabilities

Marginal probability is simply the probability that an event will occur regardless of

whether or not another event occurs. Therefore, P(A) and P(B) are marginal

probabilities.

SupposeA and B are any two events of on experiment.P(A) andP(B) are such that;

P(A) = Probability that event A occurs regardless of whether or not event B

occurs.

P(B) = Probability that event B occurs regardless of whether or not event A

occurs.

Joint probability is the probability that two or more events will occur at the same time.

Therefore,P(AB) is a joint probability.

P(AB) or P(AB) : Probability that both events A and B occur at the same

time.

Conditional probability is the probability of an event occurring given that another event

has occurred. P(A|B)and P(B|A)are conditional probabilities. P(A) and P(B) are also

known as unconditional probabilities.

1. P(S) = 1 and P() = 0

2. For any eventA, 0 P(A) 1

Probability has value between 0 and 1.

3. For any eventA, P(A) = 1 -P(A)

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It called the Law of Addition.

Special Cases for the Law of Addition

IfA and B are independent events, then:

)(). BPPPPB)P(A ()(+)(=

IfA and B are mutually exclusive or disjoint events, then:

)(+)(= PPB)P(A

Example 5.6

Given thatA and B are two events with probabilitiesP(A) = 0.4,P(A|B) = 0.2 andP(B)

= 0.15. Find;

a. P(B|A)

b. P(AB)

Solution:

a. Given;P(A|B)=)(

)(

BP

BAP = 0.2

)( BAP = )(BP x 0.2 = 0.15 x 0.2 = 0.03

Then,P(B|A)= )(

)(

AP

BAP =

4.003.0 = 0.075

b. B)PPPB)P(A ()(+)(=

= 0.4 + 0.15 0.03 = 0.52

B)PPPB)P(A ()(+)(=

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Example 5.7

Rand Tare defined in a sample space with probabilities P(R) = 0.35, P(T) = 0.28 and

P(RT) = 0.06. Find;

a. P(RT) c.P(RT)

b. P(RT)

Solution :

a. P(RT) =P(R) +P(T) - P(RT) = 0.35 + 0.28 0.06

= 0.57

b. P(RT) =P(RT) -P(T)

= 0.57 0.28 = 0.29

c. P(RT) = 1 -P(RT)

= 1 - 0.57

= 0.43

Practice 5.6

If the probability that students A will fail a certain statistics examination is 0.5, the

probability that student B will fail the examination is 0.2, and the probability that the

both student A and student B will fail the examination is 0.1, what is the probability

that:

a. At least one of these two students will fail the examination?

b. Neither studentA nor student B will fail the examination?

c. Exactly one of these two students will fail the examination?

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R T

TR


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Solution

Let :A be the event that studentA fails,A be the event that studentA passes

P(A) = 0.5, P(A) = 1-P(A) = 1- 0.5 = 0.5

Let : B be the event that student B fails, B be the event that student B passes

P(B) = 0.2,P(B) = 0.8

Give also,P(AB) = 0.1

a. P(at least one will fail) = P(AB)

= P(A) +P(B) -P(AB)

= __ + __ __ = 0.6

b . P(neither students will fail) = P(both will pass )

= P(AB)

= 1-P(AB)

= 1- __ = __

c. P(exactly one fails) = P(A fail and B passed) +P(A passed and B fail)= P(AB) + P(AB)

= __ + __ = ___

Practice 5.7

156

AB

0.4 0.1 0.1

0.4


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Two studentsA and B are both registered for a certain course. If studentA attends class

70% of the time and student B attends class 80% of the time. If the absences of the two

students are independent, what is the probability that at least one person attends the

class?

Solution

LetA be the studentA attends class and B be the student B attends class.

GivenP(A) = 0.7,P(B) = 0.8

SinceA and B are independent,

)( BAP =P(A).P(B) = __ x __ = 0.56

Then,P(at least one person attends the class) or,

B)PPPB)P(A ()(+)(=

= __ + __ - __

= ___

5.4 Binomial Distribution

Binomial probability distribution is one of the most useful mathematical models for

discrete random variable. This distribution needs repetition of a Bernoulli experiment.

Bernoulli distribution has only two possible outcomes. Consider the experiment of

tossing a coin and we record only two outcomes i.e. head (H) or tail (T). This

experiment is called Bernoulli experiment. If we tossing two times, the sample space for

the new experiment is

S = {HH, HT, TH, TT}.

The experiment consists of repeating two or more times where each trial had two

possible outcomes is called a Binomial experiment.

Properties of Binomial Experiment:

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a. The experiment consists of n trials where n at least two trials.

b. There are only two possible outcomes on each trial. Example: Head or Tail, Alive

or Dead, Good or Defective Success or Failure etc.

c. The probability of success on a single trial (p) and failure (q) remain the same for

each trial.

d. The outcome for each trial is independent of the outcome of any other trial.

The binomial random variable, X, can be written as ),(~ pnBX . The value of

)( rXP = could be obtained from formula and Cumulative Binomial Probabilities

Table (J. Murdoch and J.A.Barnes).

(a) Formula

rnrpp

r

nrXP

== )1()( or rnrrn qpC

where=p

probability of success on a single trial,q = probability of failure on a single trial or pq =1 ,

=n number of experiments or trials,=r number of successes in n trials varies from 0 to n

(b) Cumulative Binomial Probabilities Table

When n is getting large, computation using formula will be become tedious. To avoid

computational drudgery, you can find many binomial probabilities directly from the

Cumulative Binomial Probabilities Table written by J. Murdoch and J.A.Barnes. The

table gives the probability of obtaining r or more successes in n independent trials i.e.

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Interpretation from English Language to Statistics Symbol

Let X be the number of heads.

English Language Statistics Symbol

Probability of getting three heads )3( =XP

Probability of getting more than three heads )3( >XPProbability of getting less than three heads )3(


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Since n = 3,X= 0, 1, 2, 3

The sample space is S = {HHH,HHT,HTH,HTT, THH, THT, TTH, TTT}

The probability of success in getting head for each trial is p = and q = 1 - =

Xcan be written as ),3(~ 21BX .

By using formula,

)3()2()2( =+== XPXPXP

33

213

21

3

323

212

21

2

3 )()()()( += CC

= 0.375 + 0.125

= 0.5

Example 5.10

Given X is binomial random variable with n = 5 and p = 0.20. By using Cumulative

Binomial Probabilities Table, find:

(i) )4( XP

(ii) )4( >XP

(iii) )4( XP

(iv) )4(


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n = 2

n = 5

.

0=r12

345

. . .

1.0000.6723.2627

.0579.0067

.0003

(i) The value of )4( XP is directly obtained from the table i.e. )4( XP =0.0067

(ii) )4( >XP = )5( XP = 0.0003

(iii) )4( XP = 1 - )5( XP = 1 0.0003 = 0.9997

(iv) )4(


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(b) )2(


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=____

____ )()(C +

______

__ )()(C }+____

____ )()(C +

______

__ )()(C

= ______ (formula)

@P(at most one head or one tail)

= )5()2(1)5()1( +=+ XPXPXPXP

= 1 - _____+_____

= ______ (table)

Practice 5.9

Thirty percent of the voters in a small city are opposed to a purposed development. If

fifty voters are selected at random, find

(a) the probability that at least twenty voters are opposed to a purposed development.

(b) the mean and variance of the voters who are opposed to a purposed development.

Solution

LetXbe the number of voters who are opposed to a purposed development.

n = 50,p = 0.30

(a) __)( XP = _____ (from table)

The computation using formula will be become tedious and are not suggested.

(b) The mean, np= = __x__ = ___

The variance, 2 = npq =__x__x__ = ___

Practice 5.10

The probability that a patient recovers from a delicate heart operation is 0.7. What is the

probability that exactly four out of the five patients having this operation will survive?

Solution

LetXbe the number of patients having heart operation will survive.

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n = 5,p = 0.7

)4( =XP =____

____ )3.0()7.0(C = ______ (formula)

If we want to use table, a certain thing should be change.

LetXbe the number of patients having heart operation will die.

n = 5,p = 0.3

If 4 patients will survive, its mean 1 patients will die.

)1( =XP = __)( XP - __)( XP

= ____ - ____

= ____(table)

5.5 Normal Distribution

The normal distribution(sometimes referred to as the Gaussian distribution) is the most

common continuous distribution used in statistics. The normal distribution is vitally

important in statistics for three main reasons:

1. The normal distribution can be used to approximate various

discrete probability distributions such as the Binomial and Poisson distribution.

2. The normal distribution provides the basis for classical

statistical inference because of its relationship to the central limit theorem.

3. Numerous continuous variables in the business world have

distributions that closely resemble the normal distribution.

The normal distribution has several properties:

1. x can take on any real number.

2. It is bell-shaped in its appearances, and thus symmetrical.

3. Its measures of central tendency (mean, mode and median) are all identical.

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After using transformation formula, the probability can be found by using the standard

normal probability table. If we want to find ( )kZP > , Z score is a distance along the

horizontal scale of the standard normal distribution (refer to the leftmost column and

top row of standard normal probability table). While area or probability is a region

under the curve (refer to the values in the body of standard normal probability table).

z0

Since the standard normal probability table is a cumulative distribution, any symbol of

mathematics should be transformed into ( )kZP > .

Note:

1. )()( aZPaZP => (Direct from the table)

2. )()( aZPaZP >=< since its symmetry.

- a a

3. )(1)( aZPaZP >==>

- a a

5. )()()( bZPaZPbZaP >>=>=


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- a b a b

For )()()||( aZPaZPaZP =>

)()||( aZaPaZP


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(b) The area to the right of a is 33.067.01 =

a

Thus a = 0.44.

Example 5.13

Given thatXhas the normal distribution with mean 30 and variance 36, find;

(a). )40( >XP (b). )4020( XP

Solution:

Xis the normal distribution,X~N(30, 36). Then 30= and 6= .

(a).

>=>

6

3040)40( ZPXP

0475.0

)67.1(=

>= ZP

(b)

=

6

3040

6

3020)4020( ZPXP

905.0)2(0475.01

)67.0()67.1(1

)67.167.1(

==

=

=

ZPZP

ZP

X 0.00 0.04

0.0

4.0 0.33

170

0.67 0.33


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X~N(2.5, 1). So 5.2= and 1=

(i) ( )___)2(


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From normal probability table, a = 1.645 =

k

Then, 1.645 =__

__k

1.645 (___) = __k

=k ____ + ____ = ______

The probability 5% to the right indicates that a student taking the test scores must

be higher than _____.

EXERCISE

1. Suppose an original set B is given as B = {1, 2, 3, 4, 5, 6, 7}. How many 4digit

numbers can possibly be formed from set B if:

(a) Each digit can be used more then once?

(b) Each digit can be used only once?

2. How many ways can the alphabets in the word REGRESSION be arranged?

3. A team of 3 students is to be selected from 5 male and 7 female students. Find the

number of different groups could be formed consisting at most 2 male students.

4. Given thatA and B are two events with probabilities P(A) = 0.5, P(A|B) = 0.2 and

P(B) = 0.4. Find

(a) P(AB) . (b) P(B|A) .

5. Aand B are defined in a sample space with probabilities P(A) = 0.5,P(B) = 0.4 and

P(AB) = 0.2. Find

(a) P(AB) . (b)P(AB) . (c)P(AB) .

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6. If the probability that studentsA will fail mathematic subject is 0.3, the probability

that student B will fail mathematic subject is 0.2, and the probability that the both

studentA and student B will fail the examination is 0.1, what is the probability that:

(a) At least one of these two students will fail the examination?

(b) Both students will pass the examination?

7. GivenXis binomial random variable with n = 20 andp = 0.20. By using Cumulative

Binomial Probabilities Table, find

(i) )4( XP .

(ii) )41( < XP . (iv) )127(


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(a). What are her chances of having her claim granted if she is in fact only guessing?

(b). What are the chances of having her claim rejected if she in fact does have the

claimed ability?

12. Find the following probabilities by using calculator or table;

(a) )25.3( ZP

(c) ).55.2||( >ZP

(d) ).53.0||(


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(c) Assume that the number of staff on the shift is 60 and that the deadline for

completion of each task is 90 minutes. How many staff do you expect will be

unable to complete the task in the allotted time?

Answers to exercise:

1. (a) 2,401 (b) 840

2. 453,600

3. 210

4. (a) 0.08 (b) 0.16

5. (a) 0.7 (b) 0.3 (c) 0.3

6. (a) 0.4 (b) 0.6

7. (a) 0.4114 (b) 0.5604 (c) 0.0006 (d) 0.0866

8. 0.2460 and 0.6230

9. 0.3020

10. (a) 0.2503 (b) 0.0563 (c) 0.9996

11. (a) 0.1094 (b) 0.4661

12. (a) 0.0006 (b) 0.9595 (c) 0.0108 (d) 0.4039 (e) 0.9144

13. (a) 0.8536 (b) 0.1333 (c) 0.0377

14. (a) 0548.0 (b) 47725.0

15. (a) 0.0228 (b) 0.2858 (c) 9

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Activity

You are instructed to do a comparison between the classical probability and empirical

probability. The experiment chosen is you need to toss a coin. In classical probability,

the probability of getting a head or tail in a single experiment is equal. We can denote as

2

1)()( ==== TXPHXP . In empirical probability, you have to do an experiment to

find the )( HXP = or )( TXP = . In this case, tossing a coin for 30 times to find

)( HXP = or )( TXP = . Answer the following questions;

(a) Which type of probability shows the higher )( HXP = ?

(b) Let say the number of experiment is 50. Without conducting an experiment, find the

probability of getting at most 20 heads and at most 10 heads by using Binomial

distribution (Compare between classical probability and empirical probability).

(c) Then, compare your results among 4 students around you.

(d) What are your conclusions?

Unit5 Statistics MATH2(D) Ikbn

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