Chapter 5Probability and the

Normal Curve

Introduction to Part II

• In Part I, we learned to categorize data to see basic patterns and trends.

• Measures of central tendency and variability allowed for more descriptions.

• In Part II, we’ll move towards using statistics to assist us in decision making.

• The concept of probability

Probability• (P) varies from 0 to 1.0• Able to use percentages rather than decimals to

express the level of probability– A 0.50 probability (or 50 chances out of 100) is

sometimes called a 50% chance– For statistical use, decimal form is more appropriate

• A probability of 0 implies that something is impossible

• A probability of 1 constitutes certainty

Rules of Probability• Probability: the relative likelihood of occurrence

of any given outcome or event

• Probability

• For example:– Choosing at random a woman from a room of three men and

seven women– Drawing a single card (e.g ., the queen of spades)

Types of Probability

• Theoretical– Chance or randomness• For example: Flipping a coin

• Empirical– Depends on observation to determine or estimate

the values• For example: Number of baby boys born at Lowell

General in 2011

Converse Rule• The probability that something will not occur• Subtract the probability that something will

occur from 1 to find the probability that the event will not occur.

• P = 1 – p• For example:– The probability of not drawing a queen

Addition Rule• The probability of obtaining any one of

several different and distinct outcomes equals the sum of their separate probabilities

• Mutually exclusive• For example: – The probability of drawing the queen of spades or

queen of clubs

Addition Rule - Example• Suppose a high school consists of 25%

freshmen, 35% sophomores, 25% juniors, and 15% seniors. The relative frequency of students who are either juniors or seniors is 40%. We can add the relative frequencies of juniors and seniors because no student can be both a junior and a senior.

• P (J or S) = 0.25 + 0.15 = 0.40 = 40%

Addition Rule – Example 2• A student goes to the library. The probability that she

checks out (a) a biography is 0.40, (b) a science fiction is 0.10, and (c) a mystery is 0.50. What is the probability that the student checks out a work of mystery or science fiction?

Solution: Let S = the event that the student checks out science fiction, and let M = the event that the student checks out mystery. Then, based on the rule of addition:

P(S) + (P)M = 0.10 + 0.50 = 0.60

Addition Rule - Practice• A card is drawn randomly from a deck of ordinary playing cards.

You win $10 if the card is a spade or an ace. What is the probability that you will win the game?

Solution: Let S = the event that the card is a spade; and let A = the event that the card is an ace. We know the following:

There are 52 cards in the deckThere are 13 spades so P(S) = 13/52There are 4 aces so P(A) = 4/52There is 1 ace that is also a spade so P(A&S) = P(1/13 * 1/4) = 1/52Therefore, based on the rule of addition

P (S or A) = P(S) + P(A) – P(Both) =13/52 + 4/52 – 1/52 = 16/52 = 0.31

Multiplication Rule• Sometimes we want to know the probability

of successive outcomes. • Multiplication rule: combination of

independent outcomes equals the product of their separate probabilities– For example:• What is the probability of getting heads on both coin

flips? That is, heads on the first and heads on the second flip?

Multiplication Rule - Example• A jar contains 6 red marbles and 4 black marbles. Two

marbles are drawn without replacement from the jar? What is the probability that both of the marbles are black?

• Solution: Let A = the event that the first marble is black and let B = the event that the second marble is black. We know the following: – In the beginning, there are 10 marbles in the jar; 4 of which are

black. Therefore, P = 4/10– After the first selection, there are 9 marbles in the jar, 3 of which

are black. Therefore, P(B/A) = 3/9

Based on the rule of multiplication:(4/10) x (3/9) = 12/20 = 2/15 = 0.13

Multiplication Rule - Practice• Suppose we repeat the experiment but we select marbles

with replacement. That is, we select one marble, note its color, and then replace it in the jar before making the second selection. When we select the replacement, what is the probability that both of the marbles are black?

Therefore, based on the rule of multiplication:

(4/10) x (4/10) = 16/100 = 4/25 = 0.16

• Solution: Let A = the event that the first marble is black; and let B = the event that the second marble is black. We know the following:i. In the beginning, there are 10 marbles in the jar; 4 of which

are blacks. Therefore, P = 4/10ii. After the first selection, we replace the selected marble; so

there are still 10 marbles, 4 of which are black. Therefore, P(B/A) = 4/10.

Independent Outcomes• What is the probability that two tails occurs when two coins are

tossed? Let A represent the occurrence of a tail on the first coin and B represent the occurrence of a tail on the second coin. In this example, the occurrence of A is not dependent upon the occurrence of B and vice versa. Events A and B are said to be independent. That is, the outcome of the first toss has no effect on the outcome of the second toss. The probability of the simultaneous occurrences of two independent events is the product of the probabilities of each event:

P(A and B) = P(A) x P(B)P(A) = ½P(B) = ½P (A and B) = ½ x ½ = ¼ = 0.25

Independent Variables - Practice• Suppose we have two dice. A is the event that

4 shows on the first die, and B is the event that 4 shows on the second die. If both dice are rolled at once, what is the probability that two 4s occur?

P(A) = 1/6 P(B) = 1/6 P(A and B) = P(A) x P(B) P(A and B) = 1/6 x 1/6 = 1/36 = 0.03

Copyright © Pearson Education, Inc., Allyn & Bacon 2009

Probability Distributions• Probability distributions analogous to

frequency distributions• Probability distributions based on

probability theory

For each problem, identify the rule used (if applicable) and compute the answer for the following situation: there are 25 marbles in a jar: 13 red, 8 blue, 3 green, and 1 yellow.

a. What is the probability that a green marble is chosen?

b. What is the probability that two green marbles are chosen?

c. What is the probability that a red marble is not chosen?

d. What is the probability a blue marble and a green marble are chosen?

Probability vs. Frequency Distributions

Table 2: Probability Distribution for a Single Coin Flip

Event Probability (P)

Heads 0.50

Tails 0.50

Total 1.00

Table 3: Probability Distribution for the Number of Heads in Two Flips

X Probability (P)0 0.25

1 0.50

2 0.25

Total 1.00

Figure 1: Probability Distribution for Number of Heads in Two Flips

0 1 20

0.1

0.2

0.3

0.4

0.5

0.6

# of Heads

Frequency Distributions

Table 4: Frequency Distribution of 10 Flips of Two Coins

# of Heads f %

0 3 30

1 6 60

2 1 10

Total 10 100

Table 5: Frequency Distribution of 1,000 Flips of Two Coins

# of Heads f %

0 253 25.3

1 499 49.9

2 248 24.8

Total 1000 100

Mean of a Probability Distribution• Probability distribution has a mean – known as an

expected value

• Greek letter mu (μ) used to indicate the mean in a probability distribution

• Don’t confuse μ with in a frequency distribution.

• A probability distribution also has a standard deviation

• Symbolized by the Greek letter sigma, σ• Variance in probability distributions is shown

by σ2

• s2 for observed data; σ2 for theoretical distribution

Standard Deviation of a Probability Distribution

Mean and Standard Deviation• Computing the mean of the frequency

distribution in Table 4 for 10 flips of two coins:• (0 + 0 + 0 + 1 + 1 + 1 + 1 + 1 + 1 + 2)/10 = 0.8

• Computing the mean of the frequency distribution in Table 5 for 1,000 flips of two coins:• (253)(0) + (499)(1) + (248)(2)/1,000 = 0.995

Probability Distribution

• We are interested in finding the theoretical probability distribution of the number of clearances for three aggravated assaults. According to the UCR in 2010, aggravated assaults have a clearance rate of roughly 56%.

25

Probability and the Normal Curve:Review

• Probability varies from __ to __ ?• Name the three basic rules of probability.• What are the two types of probabilities? – Give an example of each.

End Day 1

The Normal Curve as a Probability Distribution• Theoretical or ideal model • The normal curve is a theoretical ideal because it is a probability

distribution.Uses:– Describing distributions of scores – Interpreting the standard deviation – Making statements about probability

• Sometimes referred to as a bell-shaped curve – perfectly symmetrical.• Normal curve is unimodal.• Mean, median, mode coincide.• Both continue infinitely ever closer but without touching the baseline.

The Model and the Reality of the Normal Curve

– The normal curve is a theoretical ideal.– Some variables do not conform to the normal

curve.– Many distributions are skewed, multi-modal, and

symmetrical but not bell-shaped.– Consider wealth – more “haves” than “have nots”– Often, radical departures from normality

• 100% of the cases in a normal distribution fall under normal curve

• A constant proportion of the total are under the normal curve will lie between the mean and any given distance from the mean as measured in sigma units.

Any sigma distance above the mean contains the identical proportion of cases as the same sigma below the

mean

• What must we do to determine the percent of cases for distances lying between any two score values? – For instance: How do we determine the percent of

total frequency that falls between the mean and a raw score located 1.40 standard deviations above the mean?

Figure 7: The Position of a Raw Score that Lies 1.40 Standard Deviation Above the Mean

|+1.40

The Following is a Portion of Table A(a)z

(b)Area between Mean and z

(c) Area beyond z

0.00 0.00 50.000.01 0.40 49.600.02 0.80 49.200.03 1.20 48.800.04 1.60 48.400.05 1.99 48.010.06 2.39 47.610.07 2079 47.210.08 3019 46.81

Copyright © Pearson Education, Inc., Allyn & Bacon 2009

Using Table A• Method exists to determine distance from the

mean for standard deviations that are not whole numbers.

• Table A in Appendix C (p. 377)• Sigma distances labeled • Values for one side of the normal curve given

because of symmetry

Standard Scores and the Normal Curve• It is possible to determine

area under the curve for any sigma difference from the mean.

• This distance is called a z score or standard score

• z score – indicates direction and degree that any raw score deviates from the mean in sigma units

• z scores by obtaining the deviation

X

z

Where:•µ = mean of a distribution•σ = standard deviation of a distribution•z = standard score•X = raw score

Standard Scores• How do we determine the sigma

distance of any given raw score? That is, how do we translate raw scores into units of standard deviation?

For example: Imagine a raw score of 6 from the distribution in which the mean is 3 and the standard deviation is 2.

– Difference of the raw score and the mean – Divide this distance by the standard

deviation

X

z

6 – 32Z = = 2

Figure 8: The Negative Position of z = -0.57 for the Raw Score 50

Level of satisfaction of hospital patients with nursing services. Scores have a mean of 58 and a standard deviation of 14.

| 50 58 72 86 100

Z = (50-58) / 14Z = -0.57

An Illustration: Probability Under the Normal Curve

Step 1: Translate the raw score into a z-scoreStep 2: Use Table A

What is the probability that a hospital with a raw score of less than 50 will be chosen?

What is the probability that a hospital with a score between 50 and the mean will be chosen?

Above 50?

Figure 9: The Portion of Area Under the Normal Curve for Which We Seek to Determine the Probability of Occurrence

| 50 58 72 86 100

Z = -0.57

Finding Scores from Probability Based on the Normal Curve• Using the formula below, it is possible to calculate score values

• Step 1: Locate in Table 1 the z-score associated with the closest percent.• Step 2: Convert the z value to its raw score equivalent.• Step 3: Interpretation

zX

Suppose a particular police department has data showing that the 911 response time from receiving the call to police arrival is normally distributed with a mean of 5.6 minutes and a standard deviation of 1.8 minutes. The chief wants to know how much time is required for..

A) 75% of all calls to be handledB) 90% of all calls to be handled

Summary• Introduction to probability• Foundation for decision making in statistics• Normal curve is a theoretical ideal• Use of the normal curve assists in understanding

standard deviation• z scores used to determine area between and

beyond a given sigma distance from the mean• Also able to calculate score values for a given z

• Find the probability that an individual would not pull a diamond out of a deck of cards.

• Find the probability that an individual would pull either a king or queen out of a deck of cards.

• Find the probability that an individual would pull a five and six or seven without replacement out of a deck of cards? With replacement?

Practice: Probability

Practice: Z ScoresFind the (a) z score for each quiz, the (b) probability of getting a higher score, and the (c) probability of getting a score between the raw score and the mean?

Score Z Score b c9

5.54

μ = 7.2 σ = 1.5

X

z

Score Z Score b c

9 1.20 11.51% 38.49%

5.5 -1.13 87.08% 37.08%

4 -2.13 98.34% 48.38%

PracticeThe SAT is standardized to be normally distributed with a mean μ = 500 and a standard deviation σ = 100. What percentage of SAT scores falls

a. 500 and 600?b. between 400 and 600?c. between 500 and 700?d. between 300 and 700?e. above 600?f. below 300?g. between 350 and 700?