AP Statistics – Probability

AP Statistics – Probability

Mrs. LernerCharlotte Catholic High School

With some updates byDr. Davidson

Mallard Creek High School

Random Phenomena

1. Are uncertain in the short run2. Exhibit a consistent pattern in

the long run Note the dual aspect

Note: This is the Condition for the Probability we study

Definitions: An event is an outcome or a set of outcomes of random phenomenon (RP).

Outcome(s) of interest are “Success(es)” [even if they are bad news!!!!]

Probability of an Event : (note dual aspect) Probability of an event =1.proportion of times a success

occurs in the long run (measure by experiment or simulation)

2.[number of ways for a success to occur]/[all possible events (specified by a model of the RP)

If there are n ways to do a first event & m ways to do a second event Then the number of all possible outcomes=nm

This is called the multiplication rule

All Possible Outcomes

Probability – Example # 1

Consider the sum of two dice that are rolled

Let’s say we are interested in the Probability the sum is a 4, so SUCCESS = ?

Find # of ways to get a sum = 4 [a success]is 3: 1+3, 2+2, 3+1 [3 ways]

Probability The probability of a specific outcome is [# of ways to get a success]/ [all

possible outcomes e.g. Probability of getting a sum =4 is: 3/36 = 1/12 ~ .0833 = 8.33 %

1. This is the chance the next outcome is a success

2. This is the proportion of times that successes occur in a large repetitive # of identical trials

The probability P(A) of any event A satisfies 0≤P(A)≤1

Thus the probability of any event is between 0 & 1

Probability Rules - 1

The Sample Space is the set of all possible outcomes

If S is the sample space in a probability model, then P(S)=1

The probabilities of all possible outcomes must

add up to 1


The complement of any event A is

1. the event that A does not occur

2. written as Ac

The complement rule states that

P(Ac)=1-P(A)The probability that an

event does not occur is 1 minus the probability that it does occur


Two events A and B are disjoint 1. if they have no outcomes in

common 2. cannot occur at the same time.

If A and B are disjoint, then

P(A or B) = P(A) + P(B).If two events have no outcomes in common, then the probability of

either one occurring is the sum of their individual probabilities


For any two events A and B:

P(A or B) = P(A) + P(B) – P(A and B)

The probability of either A or B occurring is the sum of their individual probabilities minus the probability that they occur at the same time


Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent, then P(A and B) = P(A)P(B).

If two events are independent, then the probability that they both occur is the product of their individual probabilities

Note – Disjoint events are always NOT INDEPENDENT, for if one occurs, then we know the other can not occur!


And = Joint = Intersection Or = Both (1, or other, or 2) = Union

Remember…

and B

or

P A P A B

P A B P A B

When P(A)>0, the conditional probability of B given A is:

P(B|A)=P(A and B)/P(A)

The probability of B occurring, given that A occurs, is the probability of A & B jointly occurring divided by the probability of the stated condition (A) occurring


For any two events A and B,

P(A and B) = P(B|A) P(A)P(A and B) = P(A|B) P(B)

The probability of A and B occurring jointly is the probability that B occurs given that A has already occurred multiplied by the probability that A occurs


Example - 2:Suppose that for a certain Caribbean island

in any 3-year period:1. the probability of a major hurricane

is .252. the probability of water damage is .44

3. and the probability of both a hurricane and water damage is .22.

What is the probability of water damage given that there is a hurricane?

Suppose that for a certain Caribbean island in any 3-year period the probability of a major hurricane is .25, the probability of water damage is .44 and the probability of both a hurricane and water damage is .22. What is the probability of water damage given that there is a hurricane?

water damage and hurricanewater damage|hurricane

hurricane

.22

.25.88

PP

P

If three people, Joe, Betsy, and Sue, play a game in which

Joe has a 25% chance of winning Betsy has a 40% chance of

winning What is the probability that Sue

will win?

2. If three people, Joe, Betsy, and Sue, play a game in which Joe has a 25%

chance of winning and Betsy has a 40% chance of winning, what is the probability that Sue will win?

Sue 1 .25 .4

1 .65

.35

P

4. A summer resort rents rowboats to customers but does not allow more than four people to a boat. Each boat is designed to hold no more than 800 pounds.

Suppose the distribution of the weight of adult males who rent boats, including their clothes and gear, is normal with a mean of 190 pounds and standard deviation of 10 pounds.

If the weights of individual passengers are independent, what is the probability that a group of four adult male

passengers will exceed the acceptable weight limit of 800 pounds?

190

10X

X

4

4 190

760

T X

190

10X

X

4

4 190

760

T X

2 2 2 2 2

2 2 2 210 10 10 10

400

20

T X X X X

T

190

10X

X

4

4 190

760

T X

2 2 2 2 2

2 2 2 210 10 10 10

400

20

T X X X X

T

exceed 800 lbs 0.023P

800,1 99,760,20normalcdf E

The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city.

Dem. Repub.Indep.18-30 25 18 1231-40 32 21 1041-50 17 25 17Over 50 14 32 15

_____________________________________________________________

5. What proportion of the Republicans are over 50?

a. 61/238b. 32/96c. 96/238d. 32/61e. Cannot be determined.



_____________________________________________________________

5. What proportion of the Republicans are over 50?

a. b. 32/96c. d. e.



_____________________________________________________________

6. If one adult citizen is chosen at random, what is the probability that this person is a Democrat between the ages of 41 and 50?

a. 17/238b. 17/88c. 61/238d. 17/61e. 88/238



_____________________________________________________________

6. If one adult citizen is chosen at random, what is the probability that this person is a Democrat between the ages of 41 and 50?

a. 17/238b. c. d. e.



_____________________________________________________________

7. Given that a person chosen at random is between 31 and 40, what is the probability that this person is an Independent?

a. 10/238b. 10/63c. 10/54d. 54/238e. 63/238



_____________________________________________________________

7. Given that a person chosen at random is between 31 and 40, what is the probability that this person is an Independent?

a. b. 10/63c. d. e.



_____________________________________________________________

8. What proportion of the citizens sampled are over 50 or Independent?

a. 54/238b. 61/238c. 100/238d. 115/238e. Cannot be determined



_____________________________________________________________

8. What proportion of the citizens sampled are over 50 or Independent?

a. b. c. 100/238d. e.

Example - Assume an multiple-choice examination

consists of questions, each having five possible answers.

Linda estimates that she has probability 0.75 of knowing the answer to any question that may be asked.

If she does not know the answer, she will guess, with conditional probability 1/5 of being correct. What is the probability that Linda gives the correct answer to a question? (Draw a tree diagram to guide the calculations.)

An examination consists of multiple-choice questions, each having five possible answers. Linda estimates that she has probability 0.75 of knowing the answer to any question that may be asked. If she does not know the answer, she will guess, with conditional probability 1/5 of being correct. What is the probability that Linda gives the correct answer to a question? (Draw a tree diagram to guide the calculations.)

P(correct) = .75 +.25*.2 = .8

A random variable assumes any of several different values as a result of some random phenomenon

Random Variables

A Discrete RV – has a countable number of possible values

◦ The probabilities must satisfy two requirements Every probability is between 0 and 1 The sum of all probabilities is 1

◦ We can use a probability histogram to look at the probability distribution.

Discrete Random Variables [RV]

Value of X

x1 x2 x3 … xk

Probability p1 p2 p3 … pk

Mean of a Discrete R. V. – (also called expected value) –

Discrete Random Variables

ii

kkX

px

pxpxpxpx

332211

Variance of a Discrete R. V. –

Discrete Random Variables

iXi

kXkXXXX

px

pxpxpxpx2

23

232

221

21

2

Continuous R. V. – takes all values in an interval of numbers

◦We look at its distribution using a density curve

◦The probability of any event is the area under the density curve in that interval.

Continuous Random Variables

1. If X is an R. V. and a & b are fixed numbers, then themean μa+bX = a +bμX

2. If X and Y are R. V.‘s, then μX±Y = μX ± μY

Rules for Means

If X is an R. V. and a and b are fixed numbers, then σ2

a+bX = b2σX2

◦ Note that multiplying by a constant changes the variance but adding a constant does not.

If X and Y are independent R. V.’s, then σ2

X±Y =σX2+σY

2

“Pythagorean Theorem of Statistics”

For STANDARD DEVIATION:square ‘em, add ‘em, take the square

root

Rules for Variances

11. Suppose X and Y are random variables with μX = 10, σX = 3, μY = 15, and

σY = 4. Given that X and Y are independent, what are the mean and standard deviation of the random variable X+Y?

11. Suppose X and Y are random variables with μX = 10, σX = 3, μY = 15, and

σY = 4. Given that X and Y are independent, what are the mean and standard deviation of the random variable X+Y?

μX+Y = μX + μY σX+Y = √σ2X + σ2

Y

=10 + 15 = √9+16= 25 = √25

= 5

12. You roll a die. If it comes up a 6, you win $100. If not, you get to roll again. If you get a 6 the second time, you win $50. If not, you lose.

a. Create a probability model for the amount you

will win at this game.


a. Create a probability model for the amount you

will win at this game.Winnings $100 $50 $0

Probability 1/6 (5/6)(1/6)5/36

(5/6)(5/6)25/36


b. Find the expected amount you’ll win.


b. Find the expected amount you’ll win.

1 5 25

100 50 06 36 26

$23.61

X E X

Law of Large Numbers – The long run relative frequency of repeated independent trials gets closer and closer to the true relative frequency as the number of trials increases.

Law of Large Numbers

Binomial Distribution – the distribution of the count X successes in the binomial setting.

B(n,p) where n is the number of

observations and p is the probability of success

Binomial Distributions

Use binompdf(n,p,X) to find the probability of a single value of X, such as P(X = 3).

Use binomcdf(n,p,X)to find the probability of at most X successes

for example P(X ≤ 3).



pnp

np

1

13. Pepsi is running a sales promotion in which 12% of all bottles have a “FREE”

logo under the cap. What is the probability that you find two free cans in a 6-

pack?



pack? 2 46

2 .12 .882

P X



pack?

2 462 .12 .88

2P X

6,.12,2binompdf

2 .13P X

14. The National Association of Retailers reports that 62% of all purchases are now made by credit card; you think

this is true at your store as well. On a typical day you make 20 sales.

a. Let X represent the number of customers who use a credit card on a typical day. What is the probability model for X?

14. The National Association of Retailers reports that 62% of all purchases are now made by credit card; you think

this is true at your store as well. On a typical day you make 20 sales.

a. Let X represent the number of customers who use a credit card on a typical day. What is the probability model for X?

The model is = B(20, .62) [i.e. B(n,p)] Please Explain Why

14. The National Association of Retailers reports that 62% of all purchases are now made by credit card; you think this is true at your store as well. On a typical day you make 20 sales.

b. Find the mean and standard deviation.


b. Find the mean and standard deviation.

20 .62

12.4

X np

1

12 .62 .38

2.17

X np p


c. What is the probability that on a typical day at least half of your customers

use a credit card?


c. What is the probability that on a typical day at least half of your customers

use a credit card? 10 1 9

.9077

P X P X

1 (20,.62,9)binomcdf

19. The volumes of soda in quart soda bottles can be described by a Normal model with a mean of 32.3 oz and a standard deviation of 1.2 oz. What is the probability that a randomly selected bottle has a volume less than 32 oz?

There are typos in the next slide: The z score calculation should read:

P(x< 32) = P(z < ([32-32.3]/1.2) = -.25

AndNormalcdf(-E99, 32, 32.3, 1.2)

19. The volumes of soda in quart soda bottles can be described by a Normal model

N(32.3, 1.2) a mean of 32.3 oz and a standard deviation of 1.2 oz.

What is the probability that the volume of a randomly selected bottle has a less than 32 oz?

32.3 3232

1.2

.1429

.4013

P x P z

P z

( 1 99,32,32.1,1.2)normalcdf E

20. A bank's loan officer rates applicants for credit. The ratings can be described by a Normal model with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, what is the probability that the rating is between 200 and 275?

20. A bank's loan officer rates applicants for credit. The ratings can be described by a Normal model with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, what is the probability that the rating is between 200 and 275?

200 275 .4332P x

Sampling distribution – the distribution of values taken by a statistic in all possible samples of the same size from the same population

Sampling Distributions

Provided that the sampled values are independent and the sample size is large enough, the sampling distribution of is modeled by a Normal model with meanand standard deviation .

Sampling Distributions of Sample Proportions

p̂

p̂ p

1ˆ

p pSD p

n

Assume that 12% of students at a university wear contact lenses. We randomly pick 200 students. ◦ What is the mean of the proportion of students in

this group who may wear contact lenses? ◦ What is the standard deviation of the proportion

of students in this group who may wear contact lenses?

Assume that 12% of students at a university wear contact lenses. We randomly pick 200 students. ◦ What is the mean of the proportion of students in

this group who may wear contact lenses? ◦ What is the standard deviation of the proportion

of students in this group who may wear contact lenses?

.12

.12 .88

200.023

Suppose that x-bar is the mean of an SRS of size n drawn from a large population with mean μ and standard deviation σ.

Then the◦ mean of the sampling distribution of xbar is μ

(hence xbar is an unbiased indicator of μ)◦ standard deviation of the sampling distribution of

xbar is

σ /√n.

Sampling Distribution of a Sample Mean

The scores of individual students on the ACT have a normal distribution with mean 18.6 and standard deviation 5.9. At Northside High, 76 seniors take the test. If the scores at this school have the same distribution as national scores, what are the mean and standard deviation of the distribution of sample means for these 76 students?

The scores of individual students on the ACT have a normal distribution with mean 18.6 and standard deviation 5.9. At Northside High, 76 seniors take the test. If the scores at this school have the same distribution as national scores, what are the mean and standard deviation of the average (sample mean) for the 76 students?

18.6

5.9

76.6768

AP Statistics – Probability

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