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Mat 211 Dr. Firoz

7-8: Probability and Statistics

Chapter 7 Probability

Definition: Probability is a real valued set function P that assigns to each event A in the

sample space S a number P(A), called the probability of A, such that the following

properties hold:

a) 1 ( ) 0P A

b) ( ) 1P S

c) ( ) ( ) ( ) ( ) 1 ( )P S P A P A P A P A , where A is the complement of A.

d) For events ,andA Bwe have ( ) ( ) ( ) ( )P A B P A P B P A B

e) If A and B are mutually exclusive (means A B , ( ) ( ) 0P A B P , then

( ) ( ) ( )P A B P A P B

f) For three events A, B, and C verify that

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )P A B C P A P B P C P A B P B C P C A P A B C

Example 1. Roll a die once. The total outcomes are namely 1, 2, 3, 4, 5, and 6. The

sample space is the set {1, 2, 3, 4, 5, 6}S . The event E rolling a 3 is the singleton set

{3}E . Now the probability of the event E for rolling a 3 is 1

( ) (rolling a 3)6

p E p ,

because there are 6 total outcomes and 1 desired outcome.

Example 2. Roll a die once. Write the sample space. Find the following probabilities:

a) ( ) (rolling a 3 or a 5)P E P

b) ( ) (rolling a 3 or more)P E P

c) ( ) (rolling a number greater than 10)P E P

d) number)even an rolling()( PEP

e) o(E) = odds for the event that the roll is a 3 or a 5

Answer: a) 1/3 b) 2/3 c) 0 d) ½ e) 2 : 4

Example 3. Roll a die twice. Write the sample space. Find the following probabilities:

a) ( ) (rolling a sum 3 or a sum 5)P E P

b) ( ) (rolling a sum 3 or more)P E P

c) ( ) (rolling a sum greater than 10)P E P

d) ( ) (rolling a sum greater than 12)P E P

The sample space of the event of rolling a die twice:

2

11 12 13 14 15 16

21 22 23 24 25 26

31 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

Answer: a) 6/36 b) 35/36 c) 3/36 d) 0

Example 4. Roll a die twice. Write the sample space. Find the following probabilities:

a) 3) aor 1 aeither are rolls twoall()( PEP

b) 2)not are rolls twoall()( PEP

c) 3) above are rolls twoall()( PEP

d) 5) a are allor 1 a are all()( PEP

The sample space of the event of rolling a die twice:

11 12 13 14 15 16

21 22 23 24 25 26

31 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

Answer: a) Look at the sample space, we have all three rolls either a 1 or a 3 are 11, 13,

31, and 33. Thus 3) aor 1 aeither are rolls twoall()( PEP = 4/36.

We could find this probability using independent event and multiplication principle as

the probability of getting first outcome either 1 or 3 is 2/6, then the probability of getting

second outcome either 1 or 3 is again 2/6. By multiplication principle the probability of

getting all two outcomes either a 1 or a 3 is 36/46/26/2

b) 2)not are rolls twoall()( PEP = 25/36, you may verify looking at the sample space,

that we have 12, 21, 22, 23, 24, 25, 26, 32, 42, 52, 62 with outcome 2’s. There are 25

outcomes with no 2’s.

Using independent event and multiplication principle we have

2)not are rolls twoall()( PEP = 36/256/56/5

c) 3) above are rolls twoall()( PEP = 9/36

Using independent event and multiplication principle we have

3) above are rolls twoall()( PEP = 36/96/36/3

d) 5) a are allor 1 a are all()( PEP = 36/26/16/16/16/1 . You look at the

sample space we have only two outcomes 11 and 55.

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Example 5. Roll a die three times. How many outcomes do we have in the sample space?

Find the following probabilities:

a) ( ) (all three rolls are either a 1 or a 3)P E P

b) ( ) (all three rolls are not 2)P E P

c) ( ) (all three rolls are above 3)P E P

d) 5) a are allor 1 a are all()( PEP

Answer: a) 216/86/26/26/2 b) 216/1256/56/56/5

c) 216/276/36/36/3 d) 216/26/16/16/16/16/16/1

Example 6. Roll a die five times. How many outcomes do we have in the sample space.

Find the following probabilities:

a) ( ) (all five rolls are either a 1 or a 3)P E P

b) ( ) (all five rolls are not 2)P E P

c) ( ) (all five rolls are above 3)P E P

d) 5) a are allor 1 a are all()( PEP

Answer: a) 55 6/2 b) 55 6/5 c) 55 6/3 d) 56/2

Methods of Enumeration

1. A true/false test contains 10 questions. In how many ways can a student answer

the questions? If a student makes random guesses, what is the probability that the

student will make exactly 5 questions correct? Answer: 102 1024, (10,5) /1024 0.2461c .

2. How many three letter words (without meaning) are possible when repetition of

letters is not allowed? What is the probability that those words will not start with

a vowel? Answer: (26,3), 0.8077P

3. How many three letter words (without meaning) are possible when repetition of

letters is allowed? What is the probability that those words will not start with a

vowel? Answer: 326 , 0.8077

4. A coin is tossed 9 times. What is the probability of getting at least 2 heads?

Answer: 0.9805

5. A company has 9 women and 8 men. What is the probability that a 7 person

committee will have 4 men and 3 women? Answer: 0.3023

6. If the letters in the word POKER are rearranged, what is the probability that the

word will begin with a K and ends with an O? Answer: 0.05

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7. A computer retail store has 12 personal computers in stock. A customer wants to

purchase three of the computers. Assume that of the 12 computers, 4 are

defective. If the computers are selected at random what is the probability that

exactly one of the purchased computers is defective? Answer: 0.509

8. A computer retail store has 12 personal computers in stock. A customer wants to

purchase three of the computers. Assume that of the 12 computers, 4 are

defective. If the computers are selected at random what is the probability that at

least one of the purchased computers is defective? Answer: 0.7455

Independent Probabilities:

Example 1. A card is chosen at random from a deck of 52 cards. It is then replaced, the

deck reshuffled and a second card is chosen. What is the probability of getting a jack and

an eight?

Solution. The event is independent. The probability of drawing first card a jack is 4/52

and second card an eight is 4/52. Also drawing a first card an eight is 4/52 and second

card a jack is 4/52. The probability of drawing a jack and an eight is

169/252/452/452/452/4

Exercise: A card is chosen at random from a deck of 52 cards. It is then replaced, the

deck reshuffled and a second card is chosen.

a) What is the probability of getting a jack and then an eight? Ans: 1/169

b) What is the probability of getting a diamond and then a heart? Ans: 1/16

Example 2. A family has two children. Using b to stand for boy and g for girl in ordered

pairs, give each of the following.

a) the sample space b) the event E that the family has exactly one daughter.

c) the event F the family has at least one daughter

d) the event G that the family has two daughters

e) p(E) f) p(F) g) p(G)

Example 3. A group of three people is selected at random. 1)What is the probability that

all three people have different birthdays. 2) What is the probability that at least two of

them have the same birthday?

1) The probability that all three people have different birthdays is

992.0365

363

365

364

365

365

365

33653

P

2) The probability that at least two people have same birthday is

008.0992.01365

363

365

364

365

3651

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Conditional probability. A conditional probability is a probability whose sample space

has been limited to only those outcomes that fulfill a certain condition. A conditional

probability of an event A, given event B is )(

)()|(

Bn

BAnBAp

Example 1. In a newspaper poll concerning violence on television, 600 people were

asked, “what is your opinion of the amount of violence on prime time television – is there

too much violence on television?” Their responses are indicated in the table below.

Yes (Y) No (N) Don’t know Total

Men (M) 162 95 23 280

Women (W) 256 45 19 320

Total 418 140 42 600

Suppose we label the events in the following manner: W is the event that a response is

from a woman, M is the event that a response is from a man, Y is the event that a

response is yes, and N is the event that a response is no, then the event that a woman

responded yes would be written as Y | W and p(Y | W) = 256/320 = 0.8.

Use the given table to answer following questions.

a) p(N) b) p(W) c) p(N | W) d) p(W | N) e) )( WNp

f) )( NWp g) p(Y) h) p(M) i) p(Y | M) j) p(M | Y)

k) )( MYp l) )( YMp m) )( YWp n) )|( YWp

Answer: a) 0.23 b) 0.53 c) 0.14 d) 0.32 e) 0.08

f) 0.08 g) 0.70 h) 0.47 i) 0.58 j) 0.39 k) 0.27

l) 0.27 m) 0.43 n) 0.61

Independent Events

Independent Events: Two events A and B are called independent if and only if

)()()( BPAPBAP , otherwise A and B are dependent.

Example 1. In two tosses of a single fair coin show that the events “A head on the first

toss” and “A head on the second toss” are independent.

Solution: The sample space S },,,{ TTTHHTHHS , the event with a head on the first

toss },{ HTHHA and an event with a head on the second toss },{ THHHB .

Now show that )()()( BPAPBAP .

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Bayes’ Formula: Let A, B, C are mutually exclusive events whose union is the sample

space S. Let E be the arbitrary event in S such that 0)(EP , then

)(

)()|(

EP

EAPEAP ,

)(

)()|(

EP

EBPEBP ,

)(

)()|(

EP

ECPECP

where )()|()( APAEPEAP and so on.

General form of Bayes’ Theorem (Page # 348)

m

i

ii

kk

k

BAPBP

BAPBPABP

1

)|()(

)|()()|( , where mk ......,3,2,1 . The conditional probability

)|( ABP k is often called the posterior probability of kB .

Example 1. A company produces 1,000 refrigerators a week at three plants. Plant A

produces 350 refrigerators a week, plant B produces 250 refrigerators a week, and plant C

produces 400 refrigerators a week. Production records indicate that 5% of the

refrigerators at plant A will be defective, 3% of those produced at plant B will be

defective, and 7% of those produced at plant C will be defective. All refrigerators are

shipped to a central warehouse. If a refrigerator at the warehouse is found to be defective,

what is the probability that it was produced a) at plant A? b) at plant B? c) at plant C?

We consider D as defective and D’ as non defective.

0.05 D

A

0.95 D’

0.35

0.03 D

Start 0.25 B

0.97 D’

0.40

0.07 D

C

0.93 D’

We now answer all questions from the tree diagram.

a) 3

1

)07.0(40.0)03.0(25.0)05.0(35.0

)05.0(35.0

)(

)()|(

DP

DAPDAP

You now try to find b) ?)07.0(40.0)03.0(25.0)05.0(35.0

)03.0(25.0

)(

)()|(

DP

DBPDBP

c) ?)07.0(40.0)03.0(25.0)05.0(35.0

)07.0(40.0

)(

)()|(

DP

DCPDCP

More examples on Bayes’ Theorem:

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1. The Belgian 20-frank coin (B20), the Italian 500-lire coin (I500), and the Hong

Kong 5-dollar (HK5) are approximately the same size. Coin purse one (C1)

contains six of each of these coins. Coin purse two (C2) contains nine B20s, six

I500s, and three HK5s. A fair four-sided die is rolled. If the outcome is {1}, a

coin is selected randomly from C1. If the outcomes belong to {2, 3, 4}, a coin is

selected randomly from C2. Find

a) )500(IP , the probability of selecting an Italian coin

b) )500|1( ICP , the conditional probability that the coin selected from C1,

given that it was an Italian coin.

Solution: We have 4/1)1(CP and 4/3)2(CP . We know that (you may draw a

tree diagram) 3/118/6)500|1( ICP , 3/118/6)20|1( BCP and

3/118/6)5|1( HKCP , and you can find for C2 in the same way.

a)

3/112/44/3*3/14/1*3/1)2()2|500()1()1|500()500( CPCIPCPCIPIP

b) 4/13/1

4/1*3/1

)500(

)1|500()1()500|1(

IP

CIPCPICP

2. The Belgian 20-frank coin (B20), the Italian 500-lire coin (I500), and the Hong

Kong 5-dollar (HK5) are approximately the same size. Coin purse one (C1)

contains six of each of these coins. Coin purse two (C2) contains nine B20s, six

I500s, and three HK5s. A fair four-sided die is rolled. If the outcome is {1}, a

coin is selected randomly from C1. If the outcomes belong to {2, 3, 4}, a coin is

selected randomly from C2. Find

a) )20(BP , the probability of selecting a Belgian coin (Answer: 11/24)

b) )20|1( BCP , the conditional probability that the coin selected from C1,

given that it was a Belgian coin. (Answer: 2/11)

c)

3. Given two urns, suppose urn I contains 4 black and 7 white balls. Urn II contains

3 black, 1 white, and 4 yellow balls. Select an urn and then select a ball.

a) What is the probability that you obtain a black ball? (Answer: 65/176)

b) What is the probability that you obtain a ball from urn II, given that the ball is a

black ball? (Answer: 33/65)

4. An absence minded nurse is to give Mr. Brown a pill each day. The probability

that the nurse forgets to administer the pill is 2/3. If he receives the pill, the

probability that Mr. Brown will die is 1/3. If he does not get the pill, the

probability that he will die is ¾. Mr. Brown died. What is the probability that the

nurse forgot to give Mr. Brown the pill? (Answer: 9/11)

Solution hints: A = the nurse forgets to give pill, B = do not forget, E = Mr.

Brown dies. Now P(A) = 2/3, P(E|A) = 1/3, P(E) = P(A)P(E|A)+P(B)P(E|B), find

P(A|E).

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5. A company that specializes in language tutoring lists the following information

concerning its English-speaking employees: 23 speak German, 25 speak French,

31 speak Spanish, 43 speak Spanish or French, 38 speak French or German, 46

speak German or Spanish, 8 speak Spanish, French and German, and 7 office

workers and secretaries speak English only. Make a Venn Diagram and show all

information in it. Find the following a) What percent of the employees speak at

least one language other than English? b) What percent of the employees speak at

least two languages other than English. Answer: a) 88.9% b) 23.8%

6. A box of candy hearts contains 52 hearts of which 19 are white, 10 are tan, 7 are

pink, 3 are purple, 5 are yellow, 2 are orange, and 7 are green. If you select 9

pieces candy randomly from the box, without replacement, give the probability

that a) three of the hearts are white b) Three are white, two are tan, one is pink,

one is yellow, and two are green. [a) 29.17% b) 0.87%]

7. Suppose that P(A) = 0.7 and P(B) = 0.5 and P[ )( BA ] = 0.1. Find a) )( BAp

b) )|( BAp c) )|( ABp

8. Let A and B be the events that a person is left-eye dominant or right-eye

dominant, respectively. When a person folds their hands, let C and D be the

events that their left thump and right thump, respectively are on top. A survey in

one statistics class yielded the following table:

If a student is selected randomly, find the following probabilities: a) )( CAp

b) )( CAp c) )|( CAp d) )|( DBp

Chapter 8 random Variables and Statistics

Random variables of the discrete type

In probability theory, a probability distribution is called discrete if it is characterized by a

probability mass function. Thus, the distribution of a random variable X is discrete, and X

is then called a discrete random variable, if ( ) 1p X

If a random variable is discrete, then the set of all values that it can assume with non-zero

probability is finite or countably infinite, because the sum of uncountably many positive

real numbers (which is the least upper bound of the set of all finite partial sums) always

diverges to infinity.

Given a random experiment with an outcome space S, a function X that assigns to each

element s in S one and only one real number X(s) = x is called a random variable, like a

A B

C 5 7

D 14 19

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function of s. The space of X is the set of real numbers { : ( ) , }x X s x s S , where s S

means the element s belongs to S. The probability mass function (pmf) ( )f x of a

discrete random variable X is a function that satisfies the following properties:

1. ( ) 0, ;f x x S

2. ( ) 1;x S

f x

3. ( ) ( ),x A

P X A f x A S

Example 1. Suppose that X has a discrete uniform distribution on {1,2,3,4,5,6}S and

its pmf is 1

( ) , 1,2,3,4,5,66

f x x .

As a general case we may write pmf as 1

( ) , 1,2,3,4, ,f x x mm

Example 2. Roll a 4 –sided die twice and let X equal the larger of the two outcomes if

they are different and common value if they are the same. The outcome space for this

experiment is 0 1 2 1 2{( , ); 1,2,3,4; 1,2,3,4}S d d d d , where we assume that each of

these 16 points has probability 1

16. Then

1 3 5( 1) [(1,1)] , ( 2) [{(1,2),(2,2), (2,1)] , ( 3)]

16 16 16P X P P X P P X and

7( 4)

16P X . Looking at the pattern one can easily find the pmf

2 1( ) , 1,2,3,4

16

xf x x

Exercise

1. Let the pmf of X be defined by ( ) , 2,3,49

xf x x , a) draw a bar graph and a

b) probability histogram

2. For each of the following, determine the constant c, so that ( )f x satisfies the

conditions of being a pmf for a random variable X,

a) ( ) , 1,2,3,4x

f x xc

b) ( ) , 1,2,3,4 10f x cx x

c) 1

( ) , 1,2,3,44

x

f x c x

d) 2( ) (1 ) , 0,1,2,3f x c x x

e) ( ) , 1,2,3,4, ,x

f x x nc

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Mathematical expectation

In probability theory and statistics, the expected value (or expectation value, or

mathematical expectation, or mean, or first moment) of a random variable is the

integral of the random variable with respect to its probability measure. For discrete

random variables this is equivalent to the probability-weighted sum of the possible

values.

The term "expected value" can be misleading. It must not be confused with the "most

probable value." The expected value is in general not a typical value that the random

variable can take on. It is often helpful to interpret the expected value of a random

variable as the long-run average value of the variable over many independent repetitions

of an experiment.

When it exists, mathematical expectation E satisfies the following properties:

a) If c is a constant, ( )E c c

b) If c is a constant and u is a function, [ ( )] [ ( )]E cu x cE u x

c) If 1c and 2c are constants and 1u and 2u are functions, than

1 1 2 2 1 1 2 2[ ( ) ( )] [ ( )] [ ( )]E c u X c u X c E u X c E u X

Example 1. Let X have the pmf ( ) , 1,2,3,410

xf x x . Find ( )E X

Solution: 1

( ) ( ) 310

x

i i

x

xE X x f x x , verify.

Example 2. Let X have the pmf ( ) , 1,2,36

xf x x . Find mean = ( )E X and also

2( )E X and variance 2 2 2( ) ( ( ))E X E X and also standard deviation .

Solution: Mean = 1

7( ) ( )

6 3

x

i i

x

xE X x f x x

2 2 2

1

36( ) ( ) 6

6 6

x

i i

x

xE X x f x x

Variance

2

2 2 2 7 5( ) ( ( )) 6

3 9E X E X and

5

3

Example 3. A politician can emphasize jobs or the environment in her election

campaign. The voters can be concerned about jobs or the environment. A payoff matrix

showing the utility of each possible outcome is shown.

Jobs Voters Environment

Jobs 25 10

Environment 15 30

The political analysts feel there is a 0.39 chance that the voters will emphasize

jobs. Which strategy should the candidate adopt to gain the highest utility

a) Environment b) Jobs, explain mathematically.

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Solution: For the environment the expected value is ( ) 15(.39) 30(1 .39) 12.45E x

On the other hand for jobs the expected value is ( ) 25(.39) 10(1 .39) 3.65E x . So

the preference will go for a) Environment (because of higher expected value).

Exercise

1. Find mean and standard deviation of the following:

a) 1

( ) , 5,10,15,20,255

f x x

b) ( ) 1, 5f x x

c) 4

( ) , 1,2,36

xf x x

2. Given 2( 4) 10, [( 4) ] 116E X E X , determine 2 Var ( 4)X and

mean= ( )E X

Bernoulli trials and the Binomial distribution

A Bernoulli experiment is a random experiment, the outcome of which can be classified

in but one of two mutually exclusive and exhaustive ways, say, success or failure (life or

death, head or tail, 3 or not 3 etc. The pmf of a Bernoulli trail is 1( ) (1 ) , 0,1x xf x p p x and we say that the random variable x has Bernoulli

distribution. The mean of Bernoulli trial is given as 1

1

0

( ) (1 )x x

x

E X xp p p , verify.

The variance of Bernoulli trial is 1

2 2 1

0

( ) ( ) (1 ) (1 ) , 1x x

x

Var X x p p p p p pq q p

Example 1. In the instant lottery with 20% winning tickets, if X is equal to the number of

winning tickets among n = 8 that are purchased, the probability of purchasing two

winning tickets is 2 68

(2) ( 2) (0.20) (1 0.2) 0.2936 29.36%2

f P X

One may use calculator as follows (TI)

2nd

DISTR 0 (binompdf) (8, 0.20, 2) will display 0.29360128

Example 2. In the instant lottery with 20% winning tickets, if X is equal to the number of

winning tickets among n = 8 that are purchased, the probability of purchasing at best 6

winning tickets is

7 1 88 8

( 6) 1 (7) (8) 1 (0.20) (1 0.2) (0.2) 0.999915527 8

P X f f

One may use calculator as follows (TI)

2nd

DISTR A (binomcdf) (8, 0.20, 6) will display 0.99991552

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Example 3. In the instant lottery with 20% winning tickets, if X is equal to the number of

winning tickets among n = 8 that are purchased. Find the probability of purchasing at

least 6 winning tickets.

Hint. Find ( 6) ( 6) ( 7) ( 8)P X P X P X P X or

( 6) 1 ( 5) 1 (8,.2,5)P X P X binomcdf

Example 4. A quiz consists of 24 multiple choice questions. Each question has 5 possible

answers, only one of which is correct. If you answer the questions completely based on

guessing, what is the probability that

a) You will answer exactly 4 wrong?

b) You will answer exactly 4 correctly?

c) You will answer at least 20 correctly?

d) You will answer at most 3 wrong?

e) You will answer at most 3 correctly?

Solution: The probability that you will answer one question wrong is 4

0.85

.

a) The probability of answering exactly 4 wrong is a binomial probability of

B(24, 0.8, 4), which is 4 20 1124

( 4) (24,0.8,4) (0.8) (0.2) 4.56 104

P X B ,

which is almost zero.

If you use TI calculator use binompdf (24, 0.8, 4). Check your calculator using the

following code:

2nd

DISTR 0 binompdf (24, .8, 4)

b) The probability that you will answer exactly 4 correct is B(24, 0.2, 4) = 0.196

c) At least 20 correct ( 20)P X

= 20 correct + 21 correct + 22 correct + 23 correct + 24 correct = 114.79 10 .

It is easy to use calculator with binomcdf as follows: 11( 20) 1 (24,0.2,19) 4.79 10P X binomcdf

d) At most three wrong: 12( 3) (24,0.8,3) 2.25 10P X binomcdf

e) At most three correct: ( 3) (24,0.2,3) 0.264P X binomcdf

Example 5. A computer manufacturer tests a random sample of 28 computers. The

probability that a computer is non defective is 91.3%. What is the probability that:

a) Exactly 7 computers are defective? Answer: 0.006605

b) At least two computers are defective? Answer: 0.7131689

c) At most two computers are defective? Answer: 0.555224

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Example 6. A quiz consists of 10 multiple choice questions, each with 4 possible

choices. For someone who makes random guesses for all of the questions, find the

probability of passing if the minimum passing grade is 90%.

Solution: 5( 9) 1 (10,0.25,8) 2.95639 10P X binomcdf

Example 8. A student claims that he has extrasensory perception (ESP). A coin is flipped

25 times, and a student is asked to predict the outcome in advance. He gets 20 out of 25

correct. What was the probability that he would have done at least this well if he had no

EPS?

Solution: ( 20) 1 (25,0.5,19) 0.002038658P X binomcdf

Exercise 1. Toss a fair coin 12 times. How many possible outcomes do you have? What

is the probability of getting a) exactly 7 heads, b) at least 7 heads, c) at most 7 heads?

Exercise 2. A student claims that he has extrasensory perception (ESP). A coin is flipped

30 times, and a student is asked to predict the outcome in advance. He gets 25 out of 30

correct. What was the probability that he would have done at least this well if he had no

EPS?

Exercise 3. A quiz consists of 20 multiple choice questions, each with 5 possible choices.

For someone who makes random guesses for all of the questions, find the probability of

passing if the minimum passing grade is 80%.

Example 4. A computer manufacturer tests a random sample of 30 computers. The

probability that a computer is defective is 7

78

%. What is the probability that:

a) Exactly 7 computers are defective?

b) At least two computers are defective?

c) At most two computers are defective?

Exercise 5. In the instant lottery with 10% winning tickets, if X is equal to the number of

winning tickets among 20 tickets that are purchased, find the probability of purchasing

a) at best 7 winning tickets,

b) at least 7 winning tickets,

c) no more than 6 winning tickets,

d) no less than 6 winning tickets

Exercise 6. The rates of on-time flights for commercial jets are continuously tracked by

the U.S Department of transportation. Recently, Southwest Air had the best rate with

80% of its flights arriving on time. A test is conducted by randomly selected 16

Southwest flights and observing whether they arrive on time. Find

a) the probability that exactly 4 flights arrive on time

b) The probability that at least 4 flights arrive on time

c) At best 4 flights arrive on time

14

Random variable of the continuous type

A random variable is a function X that assigns to each element s in the outcome space S

one and only one corresponding real number X(s) = x. The space of X is the set of real

numbers { : ( ) , }S x X s x s S is an interval. In discrete case the S is the set of discrete

points. In the continuous case we call the integrable function ( )f x , a probability

density function (pdf) which satisfies the following:

a) ( ) 0,f x x S b) ( ) 1S

f x dx

c) The probability of the event X A is ( ) ( )A

P X A f x dx

Probability Distribution Function: A function F is a distribution function of the

random variable X iff the following conditions are satisfied:

a) F is non decreasing i.e., 0F , or ( ) ( )F x F y for all x y

b) F is continuous c) F is normalized i.e., lim ( ) 0; lim ( ) 1x x

F x F x

Example 1. Evaluate the integral / 20

020

xedx

Solution: / 20

/ 20

00

lim 120

xb

x

b

edx e

Example 2. Show that ( ) , 0mxe

f x xm

is a probability density function.

Solution (Hint): Show that ( ) 0f x and /

0

1x me

dxm

Example 3. Let Y be a continuous random variable with pdf ( ) 2 , 0 1g y y y and the

distribution function is defined by

2

0

0 0

( ) 2 0 1

1 1

y

y

G y t dt y y

y

Find mean

1

0

2( ) ( )

3E Y yg y dy (check the integral) and

Variance

1

2 2 2 2 2

0

1( ) ( ) ( )

18Var Y E Y y g y dy (check the integral). Find

also the standard deviation .

Example 4. The probability density function of a continuous random variable x is given

as ( ) 1 1 , 0 2f x x x . Find its corresponding distribution function, mean,

variance and standard deviation, interval of one standard deviation of mean, two standard

deviation of mean and three standard deviation of mean.

15

Solution: The distribution function is the integral of the 1

pdf function over the real line.

Draw the graph of the pdf and notice that F(0) = 0,

0 1 2

F(1) = 1/2 and F(2) = 1. We also notice that distribution function is zero, i.e., (0) 0F

when x < 0.

The distribution function over the interval 0 1x is 2

( ) (1 1) (1 (1 ) , 0,as (0) 02

xF x x dx x dx x dx c c F

The distribution function over the interval 1 2x is 2 1

( ) (1 1) (1 1) (2 ) 2 , 1, (1)2 2

xF x x dx x dx x dx x c c as F

The distribution function over the interval 2 x is

( ) ( ) 0 1, as (2) 1F x f x dx dx c F

Thus we have the probability distribution function defined as follows:

2

2

0 0

, 0 12

( )

2 1, 1 22

1 2

x

xx

F xx

x x

x

You can calculate mean, standard deviation and variance. Look at example 3.

Example 5. Show the following function is a probability distribution function

2

2

0 0

, 0 12

( )

2 1 1 22

1 2

x

xx

F xx

x x

x

Solution: We need to check the following properties:

a) Check that F is non-decreasing.

0 00, 0

, 0 1( ) 1 | 1|, 0 2

2 1 20, 2

0 2

xx

x xF x x x

x xx

x

shows that F is not decreasing.

16

b) For continuity check that 0 0

lim ( ) lim ( ) (0) 0x x

F x F x F and

2 2lim ( ) lim ( ) (2) 1x x

F x F x F

c) F is normalized: lim ( ) 0; lim ( ) 1x x

F x F x

The function F(x) defined above is probability distribution function.

Exercise:

1. For each of the following functions, i) find the constant c so that f (x) is a pdf of

the random variable X, ii) find the distribution function F(x) ( )P X x and iii)

sketch f (x) and F(x), iv) find also 2, , .

a) 3

( ) , 04

xf x x c

b) 23

( ) ,16

xf x c x c

c) ( ) 4 , 0 1cf x x x

d) ( ) , 0 4f x c x x

2. Sketch the graph of the following pdf f (x), then find and sketch the probability

distribution function F(x) on the real line. Review example 4.

a) 23

( ) , 1 12

xf x x b)

1( ) , 1 1

2f x x

c) 1 , 1 0

( )1 , 0 1

x xf x

x x

The Normal Distribution

A normal distribution of a random variable X with mean and variance 2 is a statistic

distribution with probability density function (pdf)

(1)

17

on the domain . While statisticians and mathematicians uniformly use the term

"normal distribution" for this distribution, physicists sometimes call it a Gaussian

distribution and, because of its curved flaring shape, social scientists refer to it as the

"bell curve."

De Moivre developed the normal distribution as an approximation to the binomial

distribution, and it was subsequently used by Laplace in 1783 to study measurement

errors and by Gauss in 1809 in the analysis of astronomical data (Havil 2003, p. 157).

The normal distribution is an extremely important probability distribution in many

fields. It is a family of distributions of the same general form, differing in their location

and scale parameters: the mean ("average") and standard deviation ("variability"),

respectively. The standard normal distribution is the normal distribution with a mean

of zero and a standard deviation of one (the green curve in the plots below). It is often

called the bell curve because the graph of its probability density resembles a bell.

If a random variable X has this distribution, we write ~ . If and , the

distribution is called the standard normal distribution and the probability density

function reduces to

18

Area under a normal curve:

For a standard normal variate z, the normal distribution has mean zero and standard

deviation one with pdf 22 / 21 1

( ) exp / 22 2

zf x z e

The area under the standard normal distribution curve for 21

( ) exp22

zu

f Z z du . We have now the difficulty to evaluate the integral

without having the knowledge of multivariable calculus and polar coordinate form. But

this difficulty we can manage using standard values from the table 5 of Normal

distribution at page # 423 or using our calculator. Look at example 3.

Important Information: All normal density curves satisfy the following property which

is often referred to as empirical rule:

1. 68.26% of the observations fall within 1 standard deviation of mean.

2. 95.44% of the observations fall within 2 standard deviation of mean

3. 99.74% of the observations fall within 3 standard deviation of mean

Note: Within 5 standard deviation of mean we assume 100% data points.

Example 1. Find the mean and standard deviation of the normal distribution whose pdf is

given as 21 ( 7)

( ) exp128128

xf x

Solution: Compare with the standard formula of pdf for the normal distribution and find

that 8, 7 .

Example 2. Write the pdf of a normal distribution with mean 3 and variance 16.

Solution: We have 4, 3 , the pdf of the normal distribution is given as

21 ( 3)( ) exp

324 2

xf x

Example 3. Find the area under the normal curve with mean zero and standard deviation

one for the standard variate 1.24z .

Solution: From table 5a:

( 1.24) 0.8925 89.25%P z

For this value choose row with 1.2 and column 0.04.

0 1.24

Using calculator: ( 1.24) 0.8925 89.25%P z

19

The calculator code:

2nd

DISTR 2 normalcdf (-5, 1.24) =0.8925120

Example 4. Find the area under the normal curve with mean zero and standard deviation

one for the standard variate 1.24z .

Solution: From table 5a:

( 1.24) 1 0.8925 10.75%P z

For this value choose row with 1.2 and column 0.04.

0 1.24

Using calculator: ( 1.24) 0.1075 10.75%P z

The calculator code: 2nd

DISTR 2 normalcdf ( 1.24, 5) =0.1074875

Example 5. Find the area under the normal curve with mean zero and standard deviation

one for 0.12 1.24z .

Solution: From table 5a:

( 0.12 1.24) ( .12) ( 1.24) 1 44.03%P z P z P z

-0.12 0 1.24

Using calculator: ( 0.12 1.24) 0.4402707 44.03%P z

The calculator code:

2nd

DISTR 2 normalcdf (-0.12, 1.24) =0.4402707

Example 6. Suppose x is a normally distributed random variable with mean 10.2 and

standard deviation 1.5. Find each of the following probabilities.

a) (6.1 13.3)P x .

b) (9.4 13)P x

c) (15.5 13.1)P x

d) ( 11.6)P x

e) ( 14.4)P x

Draw normal curve and show the region bounded by the normal curve and the x values.

20

Solution: from calculator a)

(6.1 13.3) 6.1,13.3,10.2,1.5 0.9774824 97.75%P x normalcdf

try similar way for b), and c),

b) 11.6 10.2

( 11.6) ,5 17.53%1.5

P x normalcdf

Try for e).

6.1 10.2 13.3

Exercise Set

1. The physical fitness of an athlete is often measured by how much oxygen the

athlete takes in (which is recorded in millimeters per kilogram, ml/kg). The

maximum oxygen uptake for elite athletes has been found to be 80 with a standard

deviation 9.2. Assume that distribution is approximately normal.

a) What is the probability that an elite athlete has a maximum oxygen uptake of

at least 75 ml/kg? Answer: 70.66%

b) What is the probability that an elite athlete has a maximum oxygen uptake of

65 ml/kg or lower? Answer: 5.15%

c) Consider someone with a maximum oxygen uptake of 26 ml/kg. Is it likely

that this person is an elite athlete? Answer: No

2. The combined score of SAT – 1 test are normally distributed with mean of 998

and a standard deviation of 202. If a college includes a minimum score of 800

among its requirements, what percentage of students do not satisfy that

requirement? Answer: 16.35%

3. IQ score are normally distributed with mean of 100 and a standard deviation 15.

Mensa is an international society that has one – and only one qualification for

membership, a score in the top 2 on an IQ test.

a) What IQ score should one have in order to be eligible for Mensa?

Answer: hint: (x-100)/15 = invnorm(0.98), x = 130.81

b) In a typical region of 90,000 people, how many are eligible for Mensa?

Answer: 90,000 (0.02) = 1800

4. Using diaries for many weeks, a study on the lifestyle of visually impaired

students was conducted. The students kept track of many lifestyle variables

including how many hours of sleep obtained on a typical day. Researchers found

that visually impaired students averaged 9.6 hours of sleep, with a standard

deviation of 2.56 hours. Assume that the number of hours of sleep for these

visually impaired students is normally distributed.

a) What is the probability that a visually impaired student gets less than 6.1

hours of sleep? Answer: 8.58%

b) What is the probability that a visually impaired student gets between 6.3 and

10.35 hours of sleep? Answer: 51.65%

21

c) Forty percent of students get less than how many hours of sleep on a typical

day? Answer: 8.95 hours

5. Healthy people have body temperatures that are normally distributed with a mean

of 98.20 degree Fahrenheit and a standard deviation of 0.62 degree Fahrenheit.

a) If a healthy person is randomly selected, what is the probability that he or she

has a body temperature above 98.9 degree Fahrenheit? Answer: 12.94%

b) A hospital wants to select a minimum temperature for requiring further

medical tests. What should that temperature be, if we want only 1% of healthy

people to exceed it? Answer: hint: (x-98.2)/.62 = invnorm(0.99), 99.64

6. The heights of a large group of people are assumed to be normally distributed.

Their mean height is 68 inches, and the standard deviation is 4 inches. What

percent of these people are taller than 73 inches? Answer: 10.56%

7. Suppose a population is normally distributed with a mean of 24.6 and a standard

deviation of 1.3. What percent of the data will lie between 25.3 and 26.8?

Answer: 24.91%

Statistics (this section will be discussed briefly in the class)

Measures of Central Tendency

Mean Median and Mode of a set of data:

Measures of Dispersion: