CHAPTER 4 PROBABILITY  확률(確率)

4.1 Getting Started with Probability

Five basic concepts form the basis for understanding probability. You need to learn these concepts before you can learn about the role that probability and probability distributions play in supporting inferential statistics.

Event 사건(事件)

Concept: An outcome of an experiment or survey.  어느 실험이나 조사결과

Example: Rolling a die and turning up six dots, an individual who votes for the incumbent candidate in an election.

Interpretation: When discussing probability, many statisticians use the word experiment broadly to include surveys, so you can use the shorter definition "an outcome of an experiment" if you understand this broader usage of experiment.

Elementary Event 본원적 사건

Concept: An outcome that satisfies only one criterion.

Example: A red card from a regular deck of cards, etc.

Interpretation: Elementary events are distinguished from joint events, which meet two or more criteria such as a card that is a red ace.

Random Variable 확률변수(確率變數)

Concept: A variable whose numerical values represent the events of an experiment.

Example: Throwing a die, asking voters for their preferred candidate.

Interpretation: You use the phrase random variable to discuss a variable that has no data values until an experimental trial is performed or a survey question is asked and answered.

Random variables are either discrete, in which the possible numerical values are a set of integers; or continuous, in which the possible values are any number within a specific range.

Probability 확률

Concept: A number that represents the likelihood that a particular event will occur for a random variable.

Example: Odds of winning a random drawing, chance of rolling a seven when rolling two dice, likelihood of an incumbent winning reelection, etc.

Interpretation: Probability determines the likelihood that a random variable will be assign a specific value. Probability considers things that may occur in the future, and its forward-looking nature provides a bridge to inferential statistics.

Probabilities are formally stated as decimal numbers in the range of 0 to 1. A probability of 0 indicates an event that never occurs (such an event is known as a null event). A probability of 1 indicates a certain event, an event that must occur. Probabilities can also be stated informally as the "percentage chance of (something)" or as quoted odds, such as "50-50 chance."

Collectively Exhaustive Events

Concept: A set of events that includes all the possible events

Example: Head and Tails in the toss of a coin, male and female, all six faces of a die.

Interpretation: When you have a set of collectively exhaustive events, one of them must occur. The coin must land on either heads or tails; the person must be male or female; etc. The sum of the individual probabilities associated with a set of collectively exhaustive events is always 1.

4.2 Some Rules of Probability  확률의 법칙

A number of rules govern the calculation of the probabilities of elementary and joint events.

RULE 1 The probability of an event must be between 0 and 1. The smallest possible probability value is 0. You cannot have a negative probability.  The largest possible probability value is 1.0. You cannot have a probability greater than 1.0.   [eqn: 0 <= P <= 1]

RULE 2 The event that [eqn: A`] does not occur is called [eqn: A`] complement or simply not [eqn: A`], and is given by the symbol [eqn: A`']. If [eqn: P(A)`] represents the probability of event A occurring, then [eqn: 1-P(A)`] represents the probability of event [eqn: A`] not occurring.  여(餘)사건

RULE 3 If two events [eqn: A`] and [eqn: B] are mutually exclusive, the probability of both events [eqn: A`] and [eqn: B] occurring is 0.  상호배타적(相互排他的)

RULE 4 If two events [eqn: A`] and [eqn: B] are mutually exclusive, the probability of either event [eqn: A`] or event [eqn: B] occurring is the sum of their separate probabilities. 두 확률의 합

RULE 5 If events in a set are mutually exclusive and collectively exhaustive, the sum of their probabilities must add up to 1.0.  확률의 합=1

RULE 6 If two events [eqn: A`] and [eqn: B] are not mutually exclusive, the probability of either event [eqn: A`] or event [eqn: B] occurring is the sum of their separate probabilities minus the probability of their simultaneous occurrence (called joint probability). 결합(結合)확률

Example: For rolling a single die, turning up a face with an even number of dots is not mutually exclusive to turning up a face with fewer than five dots, because both events include these (two) elementary events: turning up the face with two dots and turning up the face with four dots. To determine the probability of these events, you can express this as follows:

P(even face or face with fewer than five dots)

= P(even face) + P(face with fewer than five dots) - P(even face and face with fewer than five dots) =[eqn: 3 over 6 + 4 over 6 - 2 over 6 = 5 over 6 cong 0.833]

RULE 7 If two events [eqn: A] and [eqn: B] are independent, the probability of both events occurring is equal to the product of their respective probabilities. Two events are independent if the occurrence of one event in no way affects the probability of the second event.

Example: To determine the probability that two consecutive rolls both tuen up the face with five dots, you would multiply the probability of turning up that face on roll one (1/6) by the probability of turning up that face on roll two (also 1/6). You can express this as follows:  두 사건이 독립인 경우: 두 확률의 곱

P (face with five dots on roll one and face with five dots on roll two)

= P(face with five dots on roll one ) x P(face with five dots on roll two) [eqn: = 1 over 6 times 1 over 6 = 1 over 36 cong 0.028]

RULE 8 If two events [eqn: A] and [eqn: B] are not independent, the probability of both events [eqn: A] and [eqn: B] occurring is the product of the probability of event [eqn: A] times the conditional probability of event [eqn: B] occurring, given that event [eqn: A] has occurred. 독립이 아닌 경우: 조건부(條件附)확률

Example: During the taping of a television game show, contestants are randomly selected from the audience (30 women and 20 men) of the show. After a particular person has been selected, he or she does not return to the audience and cannot be chosen later, therefore making this a case in which a conditional probability occur. What would be the probability that the first two contestants chosen are male? You can express this as follows:

P(male selection first and male selection second)

= P(male selection first ) x P(male selection second) = [eqn: 20 over 50 times 19 over 49 = 380 over 2,450 cong 0.155]

4.3 Assigning Probabilities

There are three distinct approaches for assigning probabilities to the events of a random variable: the classical approach, the empirical approach, and the subjective approach.

Classical Approach 고전적 확률개념 선험적 지식

Concept: Assigning probabilities based on prior knowledge of the process involved.

Example: Assigning the probability of rolling a die and turning up the face with three dots.

Interpretation: Classical probability often assumes that all elementary events are equally likely to occur 동등발생 . When this is true, the probability that a particular event will occur is defined by the number of ways the event can occur divided by the total number of elementary events.

Empirical Approach  장기적 상대 도수 개념

Concept: Assigning probabilities based on frequencies obtained from empirically observed data.

Example: Probabilities determined by polling or market surveys.

Interpretation: The empirical approach does not use theoretical reasoning or assumed knowledge of a process to assign probabilities. Similar to the classical approach when all elementary events are equally likely, the empirical probability can be calculated by dividing the number of ways [eqn: A] can occur by the total number of elementary events.

Subjective Approach  주관적 확률 개념

Concept: Assign probabilities based on expert opinions or other subjective means such as "gut" feelings or hunches.

Example: Commentators stating odds about a political candidate's chance of winning.

Interpretation: In this approach, you use yoyr own intuition to judge the likeliest outcomes. You use the subjective probabilities when either the number of elementary events or actual data are not available for the calculation of relative frequencies. Because of the subjectivity, different individuals might assign different probabilities to the same event.

CHAPTER 5 PROBABILITY DISTRIBUTIONS 확률분포(確率分布)

5.1 Probability Distribution for Discrete Variables

Probability distributions for a random variable summarize or model the probabilities associated with the events for that random variable. Such distributions take different forms depending on whether the random variable is discrete or continuous.

Discrete Probability Distribution 이산(離散) 확률분포

Concept: A listing of all possible distinct (elementary) events and their probabilities of occurring for a random variable.

Interpretation: In a probability distribution for a discrete variable, the sum of the probability of all the events always equals 1, an indirect way of saying that the (elementary) events listed are always collectively exhaustive (that is, that they have used up all possible occurrences).

Problem 1. You seek to determine the probability of getting 0, 1, 2, or 3 heads when you toss a fair coin three times in a row. You can form a table of all possible outcomes of tossing a fair coin three times as follows: 동전 3회 던지는 실험. 앞면이 0,1,2,3회 나타나는 확률

Outcome	1st Toss	2nd Toss	3rd Toss

1	Head	Head	Head

2	Head	Head	Tail

3	Head	Tail	Head

4	Head	Tail	Tail

5	Tail	Head	Head

6	Tail	Head	Tail

7	Tail	Tail	Head

8	Tail	Tail	Tail

From this table, you can make summary table as follows:

<Table 5.1> Probability Distribution for Tossing a Fair Coin Three Times

No. of Head	Number of Outcomes with That Number of Heads	Probability
0 1 2 3	1 3 3 1	1/8=0.125 3/8=0.375 3/8=0.375 1/8=0.125

From this table, you determine that the chance of rolling 3 heads in a row is 0.125 and that the sum of the probabilities is 1.0, as it should be for a distribution for a discrete variable.

Another way to obtain these probabilities is to extend RULE 8 to three events; you have:

[eqn: P(H _{1```} `and````H _{2````} and````H _{3}) = P(H sub 1 ) times P(H sub 2 ) times P(H sub 3 )= (0.5)(0.5)(0.5)=0.125]

The Expected Value of a Random Variable 기대치 (期待値)

Concept: The sum of the products formed by multiplying each possible event in a discrete probability distribution by its corresponding probability.

Interpretation: The expected value tells you the value of the random variable that you could expect in the "long run," after many experimental trials.  장기적으로 취할 것으로 기대하는 값

Problem 1. From <Table 5.1> you can calculate the expected value of the number of heads as is done in <Table 5.2>

<Table 5.2> Computing the Expected Value of a Probability Distribution

Number of Head	Probability	(Number of Heads) x (Probability)
0 1 2 3	1/8=0.125 3/8=0.375 3/8=0.375 1/8=0.125	0(0.125)=0 1(0.375)=0.375 2(0.375)=0.750 3(0.125)=0.375

Average value = Sum of [each value x the probability of each value]

기대치=[(확률변수값) x (확률)]의 합  [eqn: mu = 0(0.125) + 1(0.375) + 2(0.375) + 3(0.125) = 1.50]

Notice that in this example, the average or expected value of the number of heads is 1.5, a value for the number of heads that is impossible.  This average tells you that, in the long run, if you toss three fair coins many times, the average number of heads you can expect is 1.5.  평균 값 해석에 주의

Standard Deviation of a Random Variable [eqn: (sigma)] 확률변수의 표준편차

Concept: The measure of variation around the expected value of a random variable, calculated by first summing the products formed by multiplying the squared difference between each value and the expected value by its corresponding probability and then taking the square root of that sum.

Example: From <Table 5.1>, you can calculate the standard deviation of the number of heads as is done in <table 5.3>

<Table 5.3> Computing the Standard Deviation of a Probability Distribution

No. of Head	Probability	(Number of Heads - Average number of Heads)[eqn: `sup 2] x (Probability)
0 1 2 3	0.125 0.375 0.375 0.125	(0-1.5)[eqn: `sup 2](0.125)=2.25(0.125)=0.28125 (1-1.5)[eqn: `sup 2](0.375)=0.25(0.375)=0.09375 (2-1.5)[eqn: `sup 2](0.375)=0.25(0.750)=0.09375 (3-1.5)[eqn: `sup 2](0.125)=2.25(0.125)=0.28125
Total=0.75

and [eqn: sigma = sqrt {0.75} = 0.866]

Interpretation: In financial analysis, you can use the standard deviation of a random variable about investment evaluations to assess the degree of risk of an investment,

Problem 1 Suppose that you are deciding between two alternative invest- ments, A and B.  You estimate the returns (per $1,000 investment) for each investment alternative under three economic condition events, and also provide your subjective probability of the occurrence of each economic condition as follows:

Probability	Economic Event	Investment A	Investment B
0.2 0.5 0.3	Recession Stable Economy Expanding Economy	-$100 +$100 +$250	-$200 + $50 +$350

The mean return (expected value) for the two investments is as follows:

Mean for A = -100(0.2)+100(0.5)+250(0.3)=105

Mean for B = -20090.2)=50(0.5)+350(0.3)=90

You can calculate the standard deviation for the two investments as follows:

[eqn: sigma sub A = sqrt {(-100-105)sup 2 (0.2)+(100-105) sup 2 (0.5)+(250-105)sup 2 (0.3)} =121.35]

[eqn: sigma sub B = sqrt {(-200-90) sup 2 (0.2)+(50-90) sup 2 (0.5)+(350-90) sup 2 (0.3)} =194.68]

The investment A has a higher mean return than the investment B and also has a lower standard deviation, indicating less variation in the return under the different economic conditions. Having a higher mean return with less variation makes the investment A a more desirable investment than the investment B.

The mean of a probability distribution: [eqn: mu = sum _{i=1} ^{N} `X sub i `P(X sub i )]

The Standard deviation of probability distribution: [eqn: sigma = sqrt {sum _{i=1} ^{N}(X sub i - mu ) sup 2 `P(X sub i )}]

5.2 The Binomial and Poisson Probability Distribution  이항(二項)분포와 포아송 분포

Some probability distributions for certain types of discrete random variables can be modeled using a mathematical formula. This section looks at two widely used distributions that can be used to estimate probabilities. The first probability distribution, the binomial, is used for random variables that have only two mutually exclusive events. The second probability distribution, the Poisson, is used when you are counting the number of outcomes that occur in a unit.

The Binomial Distribution 이항분포의 조건:

Concept: The probability distribution for a discrete random variable that meets these criteria:

- The random variable is for a sample that consists of a fixed number of experimental trials.  Bernouilli 시행

- The random variable has only two mutually exclusive and collectively exhaustive events, typically labeled as success and failure. 두 가지 사건(성공/실패)

- The probability of an event being classified as a success, [eqn: p], and the probability of an event being classified as a failure, [eqn: 1-p], are both constant in all experimental trials.

성공/실패확률

- The event (success or failure) of any single experimental trials is independent of (not influenced by) the event of any other trial. 각 시행은 서로 독립

Example: The coin tossing experiment described in Problem 1 on page 23.  p.23, 문제1의 예:

Interpretation: Binomial distributions can be symmetrical or skewed. Whenever [eqn: p=0.5], the binomial distribution will be symmetrical regardless of how large or small the value of n. However, when [eqn: p!=0.5], the distribution will be skewed. If [eqn: p<0.5], the distribution will be negative or left-skewed. The closer [eqn: p] is to 0.5 and the larger the sample size, [eqn: n], the more symmetrical the distribution will be.

[eqn: p=0.5]이면 표본크기에 관계없이 이항분포는 대칭분포를 따른다.

Binomial Distribution Characteristics 이항분포의 특성:  평균=[eqn: np],  분산=[eqn: npq],  표준편차=[eqn: sqrt {npq}]

Mean : The sample size times the probability of success or [eqn: n times p], remembering that the sample size is the number of experimental trials.

Variance: The product of these three: sample size, probability of success, and	probability of failure, or [eqn: n times p times (1-p)]

Standard deviation: The square root of the variance, or[eqn: sqrt { np(1-p)}]

For the equation for the binomial distribution, you take the symbols [eqn: X] (random variable), [eqn: n] (sample size), and [eqn: p] (probability of success) and add these symbols:

- [eqn: x], which represents the number of successes in the sample.

- [eqn: P(X=x` | `n, p)], which represents the probability of the value [eqn: x], given sample size [eqn: n] and

probability of success [eqn: p].

You use these symbols to form two separate expressions. One expression represents the number

of ways you can get a certain number of successes in a certain number of trials:

[eqn: `sub n C sub x = {n!} over {x!(n-x)!}]

The second expression represents the probability of getting a certain number of successes in a

certain number of trials in a specific order:  [eqn: p sup x times (1-p) sup n-x]

Using these expressions forms the following equation:

[eqn: P(X=x`|`n,p)= {n!} over {x!(n-x)!}`` p sup x` (1-p) sup n-x]

Using symbols previously introduced, you can write the equation for the mean and standard

deviation as follows:  [eqn: mu= np]  and [eqn: sigma = sqrt {np(1-p)} = sqrt {npq}]

Problem 1. Whether a would-be shoppers stays and views a retail Web site for more than one minute is one of the measures of success of such sites. Suppose that the probability that the shopper does stay for more than one minute is 0.16. What is the probability that at least four (either 4 or five) of the next five shoppers will stay for more than one minute as well?  문제1. 인터넷 쇼핑몰에 1분 이상 머무는 경우=성공. 1분 이상 머물 확률=0.16.

5명중 최소 4명이 1분 이상 머물 확률?

answer: [eqn: P(X>4`|`n=5,`p=0.16)=0.0029] (or 0.29%)

The Poisson Distribution  포아송 분포의 조건.  1837. S. D. Poisson

Concept: The probability distribution for a discrete random variable that meets these criteria.

- You are counting the number of times a particular event occurs in a unit.

단위 시간(구간, 면적)

- The probability that an event occurs in a particular unit is the same for all other units.

각 단위구간 독립적

- The number of events that occur in a unit is independent of the number of events that occur in other units.

- As the unit gets smaller, the probability that two or more events will occur in that unit approaches zero.  소 구간 내에서 2개 이상의 사건 발생할 확률은 거의 0

예:

Example: Number of computer network failures per day, number of surface defects per square meter of floor coverings, etc.

Interpretation: To use the Poisson distribution, you define an area of opportu- nity, a continuous unit of area, time, or volume in which more than one occurrence of an event can occur.

You determine Poisson probabilities by applying the formula, by using a table of Poisson values, or by using software functions.

Poisson Distribution Characteristics  Poisson 분포의 특성:  평균=분산=[eqn: lambda]

Mean: The population mean, [eqn: lambda]

Variance: The population mean, [eqn: lambda], that in the Poisson distribution is equal to the variance.

Standard deviation: The square root of the variance, or [eqn: sqrt {lambda}]

For the equation for the Poisson distribution, you take the symbols [eqn: X] (random variable), [eqn: n] (sample size), and [eqn: p] (probability of success) and add these symbols:

- [eqn: e], which represents the mathematical constant approximated by the value 2.71828.

- [eqn: lambda], which represents the average number of times that the event occurs per area of

opportunity.

- [eqn: x], which represents the number of times the event occurs per area of opportunity.

- [eqn: P(X=x` | `lambda)], which represents the probability of [eqn: x], given [eqn: lambda]

Using these symbols forms the following equation:[eqn: P(X=x`|`lambda) = {e sup {- lambda} lambda sup x } over {x!}]

As an example, the calculations for determining the Poisson probability of exactly 2 arrivals

in the next minute given an average of three arrivals per minute is as follows:

[eqn: P(X=2 ` | ` lambda=3)= {e sup -3 (3) sup 2 } over {2!} = {(2.71828) sup -3 (9) } over {2}=0.224042]

5.3 Continuous Probability Distributions and the Normal Distribution

연속확률분포와 정규분포(正規分布)

Probability distributions for a continuous random variable differ from discrete distributions in several important ways: 이산 확률변수의 확률분포와 다른 점:

- An event can take on any value within the range of the random variable and not just integers.

- The probability of any specific value is zero

- Probabilities are expressed in terms of an area under a curve that represents the continuous distribution.

One continuous distribution, the normal distribution, dominates statistics, because it can model many different types of continuous random variables. Probabilities associated with such diverse things as physical characteristics such as height and weight, scores on standardized exams, and the dimension of industrial parts tend to follow a normal distribution.  Under certain circumstances, the normal distribution also approximates various discrete probability distributions such as the binomial and Poisson and provides the basis for classical statistical inference.

경우에 따라 이항분포와 포아송 분포의 근사치로사용

For these reasons, the normal distribution is the focus of this section.  정규분포의 중요성

Normal Distribution  정규분포,  Gaussian distribution  밀도함수:  [eqn: f(x)=` {1} over {sigma sqrt {2 pi }} e ^{- {1} over {2} ( {X- mu } over {sigma } ) ^{2}}]

Concept: The probability distribution for a continuous random variable that meets these criteria:

- The graphed curve of the distribution is bell-shaped and symmetrical. 종 모양,  좌우 대칭

- The mean, median, and mode are all the same value.  평균=중앙값=최빈값

- The population mean, [eqn: mu], and the population standard deviation, [eqn: sigma], determine probabilities.

모수는 [eqn: mu]와 [eqn: sigma]

- The distribution extends from negative to positive infinity.

(The distribution has an infinite range.) 범위: 무한대

Probabilities are always cumulative and expressed as inequalities, such as [eqn: P<X] or[eqn: P>=X], where [eqn: X] is a value for the variable.

You determine normal probabilities by using a table of normal probabilities or by using software functions. Normal probability tables and some software functions use a standardized normal distribution that require you to convert an [eqn: X] value of a variable to its corresponding [eqn: Z] score.  You perform this conversion by subtracting the population mean [eqn: mu] from the [eqn: X] value and dividing the resulting difference by the population standard deviation [eqn: sigma], expressed as follows:  평균과 분산에 따라 곡선의 형태와 위치 변화 -> 무수한 정규분포

[eqn: Z`=`{X-mu} over sigma]              [eqn: ->] 표준정규분포 이용  [eqn: f(z)=` {1} over {sqrt {2 pi }} e ^{- {1} over {2} z ^{2}}]

Note that when the mean is 0 and the standard deviation is 1, the [eqn: X] value and [eqn: Z] score will be the same and no conversion is necessary.  [eqn: X` SIM `N( mu , ``sigma ^{2} `)]이면  [eqn: Z` SIM N`(0,1)]

Problem 1. A certain machine uses ball bearings that must be between 0.49cm (lower) and 0.51cm (upper) in diameter. Past experience has indicated that the actual diameter of ball bearings used is approximately normally distributed with a mean [eqn: mu=0.503]cm and a standard deviation [eqn: sigma=0.004]cm. Suppose that you want to determine the probability that a single ball bearing used will have a diameter between 0.503 and 0.507cm using the table of the probabilities of the cumulative standardized normal distribution.

You must first convert the diameters to their [eqn: Z] scores by using the formula, as shown here:

[eqn: Z ```(lower)`=`{0.503-0.503} over 0.004 = 0]     [eqn: Z ```(upper)`=`{0.507-0.503} over 0.004 = 1.0]

Therefore, you need to determine the probability that corresponds to the area between 0 and +1 [eqn: Z] units (standard deviations). To do this, you take the cumulative probability associated with +1[eqn: Z] units. Using Appendix Table, you determine that these probabilities are 0.8413 and 0.5000, respectively. Therefore, the probability of obtaining a single ball bearing that is between 0.503 and 0.507cm is 0.3413(=0.8413-0.5000).

0.5000     0.3413

0.503       0.507              Diameter

0          +1               Standard Deviation Units

0.8413

Using Standard Deviation Units  이항분포와 정규분포의 관계:

Because of the equivalence between [eqn: Z] scores and standard deviation units, probabilities of the normal distribution are often expressed as ranges of plus-or-minus standard deviation units. Such probabilities can be determined directly from the table of the probabilities of the cumulative standardized normal distribution.

이산확률을 연속확률로 근사시킬 때 연속성수정필요.  [eqn: P(X=x)=>P(x-0.5 <= X <= x+0.5)]

For example, to determine the normal probability associated with the range plus-or-minus 3 standard deviations, you would use Appendix Table to look up the probabilities associated with [eqn: Z=-3.00] and [eqn: Z=+3.00]:

Area between [eqn: -3 sigma] and [eqn: +3` sigma]

is 0.9973

표준정규곡선 하의 면적

0.00135                                                0.00135

[eqn: -3 sigma]                     0                  [eqn: +3` sigma]

0.99865

Therefore, the normal probability associated with the range plus-or-minus 3 standard deviations is 0.9973 (=0.99865-0.00135). Stated another way, there is the probability of 0.0027 that a value will not be within the range of plus- or-minus 3 standard deviations. The following table summarizes probabilities for several different ranges of standard deviation units.

Normal Probabilities for Selected Number of Standard Deviation Units

-----------------------------------------------------

Standard Deviation           Probability or Area          Probability or Area

Unit Rages                  Outside These Units         Within These Units

-----------------------------------------------------

[eqn: -1 sigma] to [eqn: +1 sigma]                  0.3174	0.6826

[eqn: -2 sigma] to [eqn: +2 sigma]                  0.0455	0.9545

[eqn: -3 sigma] to [eqn: +3 sigma]                  0.0027	0.9973

[eqn: -6 sigma] to [eqn: +6 sigma]                  0.000000002	0.999999998

-----------------------------------------------------

Finding the [eqn: Z] Value from the Area Under the Normal Curve

Each of the previous examples involved using the normal tables to find an area under the normal curve that corresponded to a specific [eqn: Z] value. There are many circumstances when you want to do the opposite of this and find the [eqn: Z] value that corresponds to a specific area.  For example, you might want to find the [eqn: Z] value that correspond to a cumulative area of 1%, 5%, 95%, or 99% of the area under the curve is contained.

Problem 1. You want to find the [eqn: Z] values such that 95% of the normal curve is contained between a lower [eqn: Z] value and an upper [eqn: Z] value with 2.5% below the lower [eqn: Z] value, and 2.5% above the upper [eqn: Z] value.

[eqn: P(Z`<`-1.96)=0.025],   [eqn: P(Z`>`+1.96)=0.025],

[eqn: P(-1.96`<`Z``<` +1.96)`=`0.95]

5.4 The Normal Probability Plot

You need to establish that a set of data values follows a normal distribution in order to use many inferential statistical methods. One technique for showing that the data follow the normal distribution is the normal probability plot.

Concept: A graph that plots the relationship between ranked data values and the Z scores to which those values would correspond if the set of data values follows a normal distribution. If the data values follows a normal distribution, the graph will be linear (a straight line).

Example:

Left-Skewed                Normal                Right-Skewed

Interpretation: Normal probability plots are based on the idea that each ranked value will have a Z score greater than its immediate predecessor and that Z scores increase at a predictable rate in data that follow a normal distribution. The exact details to produce a normal probability plot can vary, but one common approach is called the quantile-quantile plot. In this method, each ranked value is transformed to a Z score and plotted along with the ranked values of the variable.  If the data are normally distributed, a plot of the data in order from lowest to highest will follow a straight line. If the data are left-skewed, the curve will rise more rapidly at first, and then level off. If the data are right-skewed, the data will rise more slowly at first, and then rise at a faster rate for higher values of the variable being plotted.

Problem 1. You seek to determine whether the viscosity measurements taken from 120 manufacturing batches (chemical), first presented on page 9, follows a normal distribution.

Consistent with the results of the histogram on page 10, the approximate straight line that the data follow in this normal probability plot appears to indicate that the viscosity data are approximately normally distributed.