The Binomial Distributionrchidester/Stat 211 Week Five Display Version.pdf · Binomial Experiment • We have an experiment with the following qualities : 1. A fixed number of trials.

Stat 211 Week Five The Binomial Distribution

Last Week • 𝐸 𝑥 = ∑ 𝑥 ∙ 𝑝(𝑥) • 𝐸 𝑥 = 𝑛 ∙ 𝑝

• 𝜎𝑥 = ∑ 𝑥 − 𝜇𝑥

2𝑝(𝑥)

• We will see this again soon!!

Binomial Experiment • We have an experiment with the following

qualities : 1. A fixed number of trials. 2. Each trial has a result of either success or failure. 3. P(success) is the same for every trial. 4. Each trial is independent of all others. 5. X = # of successes

So X count the successes.

Is it Binomial! • Flip a coin 30 times, the side that lands upward is

observed (we want to flip heads). 1. 30 trials 2. Success = Heads, Failure = Tails 3. P(Success) = 0.5 on every trial. 4. Trials are independent, flips don’t affect each

other. 5. X = Side that lands up (can equal an H or a T)

Does not count number of successes.

Is it Binomial! • Flip a coin 30 times, the side that lands upward is

observed (we want to flip heads). ▫ This is NOT a Binomial Experiment!!!

Is it Binomial! • Roll a die 24 times, the number of 6’s is observed.

1. 24 trials 2. Success = 6, Failure = Anything else 3. P(Success) = 1

6 on every trial.

4. Trials are independent, rolls don’t affect each other.

5. X = # of 6’s = Number of successes

Is it Binomial! • Roll a die 24 times, the number of 6’s is observed.

▫ This is a Binomial Experiment!!! ▫ We meet all of the criteria!

Binomial Distribution • We let X = # successes in a binomial experiment. • From this we say :

X ~ Binomial(n, p) X ~ B (n, p) p = P(success per trial)

Mean and St. Dev. for Binomial • Our mean for Binomial

▫ 𝐸 𝑥 = 𝜇𝑥 = 𝑛 ∙ 𝑝

• Our standard deviation for Binomial ▫ 𝑆𝑆 𝑥 = 𝜎𝑥 = 𝑛 ∙ 𝑝 ∙ (1 − 𝑝)

Example of Mean/St. Deviation • X ~ Binomial(n = 4, p = ½)

▫ 𝐸 𝑥 = 𝜇𝑥 = 𝑛 ∙ 𝑝 ▫ 𝐸 𝑥 = 𝜇𝑥 = 4 ∙ 1

2= 2

▫ 𝑆𝑆 𝑥 = 𝜎𝑥 = 𝑛 ∙ 𝑝 ∙ (1 − 𝑝)

▫ 𝑆𝑆 𝑥 = 𝜎𝑥 = 4 ∙ 12

∙ (1 − 12) = 1 = 1

Ideas from last week • 𝐸 𝑥 = 𝜇𝑥 = ∑ 𝑥 ∙ 𝑝(𝑥)

• 𝐸 𝑥 = 0 116

+ 1 416

+ 2 616

+ 3 416

+ 4 116

=3216

= 2

𝑥 𝑝(𝑥)

0 1 2 1

16

3 4 4

16

616

4

16

116

More Ideas from Last Week • 𝑆𝑆 𝑥 = 𝜎𝑥 = ∑ 𝑥 − 𝜇𝑥

2𝑝(𝑥)

• 𝑉 𝑥 = 0 − 2 2 116

+ 1 − 2 2 416

+ 2 − 2 2 616

+ 3 − 2 2 416

+ 4 − 2 2 116

• 𝑉 𝑥 = 416

+ 416

+ 016

+ 416

+ 416

= 1616

= 1

• 𝑆𝑆 𝑥 = 𝑉(𝑥) = 1 = 1

𝑥 𝑝(𝑥)

0 1 2 1

16

3 4 4

16

616

4

16

116

Notice

• We get the same results both way!!!

• But, using the Binomial equations is way way way easier!!!!

• It is a shortcut.

Another Example • X Binomial(n = 480, p = 1 6⁄ )

• 𝜇𝑥 = 𝐸 𝑥 = 𝑛 ∙ 𝑝 = 480 ∙ 1

6 = 80

• 𝜎𝑥 = 𝑛 ∙ 𝑝 ∙ (1 − 𝑝) = 480 16

56

= 66.7

𝜎𝑥 = 8.2

Binomial Probabilities • What if we have a Binomial experiment and we want

to calculate the probability of a certain number of successes (counted as X) of occurring?

• There’s a formula for that!!!

• 𝑃 𝑋 = 𝑥 = 𝑛𝑥 𝑝𝑥 1 − 𝑝 𝑛−𝑥

• 𝑋 = # successes • 𝑥 = # of successes of interest

Binomial Formula Explained

• 𝑃 𝑋 = 𝑥 = 𝑛𝑥 𝑝𝑥 1 − 𝑝 𝑛−𝑥

𝐶𝐶𝐶𝐶𝐶𝑛𝐶𝐶𝐶𝐶𝑛, ℎ𝐶𝑜 𝐶𝐶𝑛𝑚 𝑜𝐶𝑚𝑤 𝑐𝐶𝑛 𝑜𝑤 ℎ𝐶𝑎𝑤 𝑥 𝐶𝐶𝑛𝑚 𝑤𝑠𝑐𝑐𝑤𝑤𝑤𝑤𝑤

𝑝𝑝𝐶𝐶𝐶𝐶𝐶𝑝𝐶𝐶𝑚 𝐶𝑜 𝑤𝑠𝑐𝑐𝑤𝑤𝑤 𝐶𝑠𝑝𝐶𝐶𝑝𝑝𝐶𝑤𝑚 𝑥 𝐶𝐶𝐶𝑤𝑤

𝑝𝑝𝐶𝐶𝐶𝐶𝐶𝑝𝐶𝐶𝑚 𝐶𝑜 𝑜𝐶𝐶𝑝𝑠𝑝𝑤 𝐶𝑠𝑝𝐶𝐶𝑝𝑝𝐶𝑤𝑚 𝑛 − 𝑥 𝐶𝐶𝐶𝑤𝑤

Example of the Formula #1 • A coin is tossed 3 times, find P(2 Heads). ▫ Sample space :

{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} what we want!

• 𝑃 2 𝐻𝑤𝐶𝑚𝑤 = # 𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝𝑠𝑠𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑠𝑠

# 𝑝𝑝𝑝𝑡𝑝 𝑝𝑝𝑠𝑠𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑠𝑠 = 3

8

Example #1 Continues • We want P(2 Heads), this is P(Successes = 2)

or P(X = 2). • The experiment is Binomial! (Use the formula!) • X ~ B(n = 3, p = ½ )

• 𝑃 𝑋 = 2 = 32

12

2 12

1= 3 1

212

12

= 38

• The same as we saw with the other method.

Formula Example #2 • X ~ Binomial(n = 10, p = 1 5⁄ )

• Find P(X = 3)

• 𝑃 𝑋 = 3 = 103

15

3 45

7= 120 1

1251638478125

= 0.201

Something a Little Different • What if we aren’t interested in an exact number of successes,

but in a range of successes. • For instance seven or fewer sucesses. • We want P(X ≤ 7).

• One way is to add up all of the probabilities in this grouping.

• 𝑃 𝑋 ≤ 7

= 𝑃(0 ∪ 1 ∪ 2 ∪ 3 ∪ 4 ∪ 5 ∪ 6 ∪ 7)

= 𝑃 0 + 𝑃 1 + 𝑃 2 + 𝑃 3 + 𝑃 4 + 𝑃 5 + 𝑃 6 + 𝑃 7

• Each of the smaller components can be found with the formula.

Why That is Possible • Finding the probabilities of each piece of the

group comes from our probability laws.

• P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

• In our case the pieces are all mutually exclusive, so we can just add the probabilities of each component without worrying about any overlap.

Too Much Effort • Doing a problem that way is possible but not

feasible.

• There is an easier way!!!

• The table.

Using the Table Example #1 • X ~ Binomial(n=10, p=0.5) • 𝑃 𝑋 ≤ 3 = 𝑃 0 + 𝑃 1 + 𝑃 2 + 𝑃 3

𝑃 0 = 100

12

0 12

10

= 112

0 12

10

= 0.0009

𝑃 1 = 101

12

1 12

9

= 1012

1 12

9

= 0.0097

𝑃 2 = 102

12

2 12

8

= 4512

2 12

8

= 0.0439

𝑃 3 = 103

12

3 12

7

= 12012

3 12

7

= 0.117

• P(X ≤3) = 0.0009 + 0.0097 + 0.0439 + 0.117 = 0.1715

Example #1 Continued • Look in the table.

• P(X ≤ 3) = 0.1719.

• These are really close, but not perfect which is OK! • Usually they will be close but not perfectly

matched, this mainly happens due to rounding error in the first approach or due to any rounding in the table itself.

Using the Table Example #2 • X ~ Binomial(n=10, p= ½ )

• Find P(X ≤ 2)

• From the table we get 0.9730

• The tables are important, we will being using

several different tables throughout the remainder of the semester.

Forgoing the Formula • X ~ Binomial(n=10, p= ½ )

• Find P(X = 3) • Suppose I don’t want to use the formula, I can use

the table instead. • How? • P(X=3) = P(0) + P(1) + P(2) + P(3)

- P(0) - P(1) - P(2)

Forgoing the Formula • X ~ Binomial(n=10, p= ½ )

• Find P(X = 3) • Suppose I don’t want to use the formula, I can use

the table instead. • How? • P(X=3) = P(0) + P(1) + P(2) + P(3)

- P(0) - P(1) - P(2)

• This is the same as : P(X ≤ 3) – P(X ≤ 2)

Forgoing the Formula • Using the formula :

𝑃 𝑋 = 3 = 103

12

3 12

7

= 0.117

• Using the table : • P(X ≤ 3) = 0.1719 • P(X ≤ 2) = 0.0547

• P(X = 3) = 0.1719 – 0.0547 = 0.1172

• They are super duper close, YAY!!!

What about > ? • P(X > 2) (where n = 5)

• P(X > 2) = P(3 ∪ 4 ∪ 5) = P(3) + P(4) + P(5)

• We could do this in pieces, but that is a pain. (>.<)

• We need another way!!!

A better way for > • Remember the sample space, in this case it is :

{0, 1, 2, 3, 4, 5} • P(Sample Space) = 1

= P(0) + P(1) + P(2) + P(3) + P(4) + P(5) • Remember compliment. • If we want P(not E) we use 1 – P(E), we can use

that here. • P(X > 2) = 1 – P(X ≤ 2)

Proof that it works!!! • 1 – P(X ≤ 2) • P(Sample Space) – P(X ≤ 2) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) - P(0) - P(1) - P(2) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) - P(0) - P(1) - P(2) = P(3) + P(4) + P(5) = P(X > 2)

Example for > (or even ≥ *gasp*) • X ~ Binomial(n = 5, p = 0.3) • P(X ≥ 3) = P(X > 2) = P(3) + P(4) + P(5)

• 𝑃 3 = 53 0.3 3 0.7 2 = 0.132

• 𝑃 4 = 54 0.3 4 0.7 1 = 0.028

• 𝑃 5 = 55 0.3 5 0.7 0 = 0.002

• P(X ≥ 3) = 0.132 + 0.028 + 0.002 = 0.162 • P(X ≥ 3) = 1 - P(X ≤ 2) = 1 – 0.8369 = 0.1613

close

What about in between? • P 2 ≤ X ≤ 4

= P(X = 2) + P(X = 3) + P(X = 4)

• We could do this by hand but this would be difficult with larger numbers.

• We can also use the table. Let’s think!

• P(X ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4) This contains all of what we want plus some extra.

• Subtract out what we don’t want. P(0) + P(1) we don’t want. This is the same as P(X ≤ 1).

Flipping Success and Failure • The Binomial table has a maximum p of 0.5. • Males have a 0.7 chance of marrying. • We take a sample of 25 males. • What is the probability that 20 or more males

marry? • P(X ≥ 20) = ?

• X = # males that marry. • Let Y = # males that do not marry.

Flipping Continued • We want P(X ≥ 20), put it in terms of Y.

• P(X ≥ 20) = P(Y ≤ 5)

• X ~ Binomial(n = 25, p = 0.7) • Y ~ Binomial(n = 25, p = 0.3)

• Use the table!

• P(Y ≤ 5) = 0.1935 = P(X ≥ 20)

Weird Words • Probability questions are often word questions.

• More than > • Less than < • ____ or more ≥ • ____ or less ≤

• At most ≤ • At least ≥

Be Careful When Flipping • P(X > 3) = 1 – P(X ≤ 3)

• P(X ≥ 3) = 1 – P(X ≤ 2)

• Think before you flip!!!!!

Extra Example • A student is taking a multiple choice exam with 16

questions, and guessing at the answers. Each question has 4 possible answers, A, B, C, D. Each question is independent. Is this a binomial distribution?

• Fixed number of trials, n = 16 • Either the answer is right or wrong. (success/failure) • p = probability of success = ¼ • Trials are independent. • Let X = # of correct answers.

• Yes, this is binomial.

Extra Example • A student is taking a 16 question multiple choice

exam, and guessing at the answers. There are four choices per question, A, B, C,D. What is the expected number that he/she will get right?

• 16 questions – 16 trials

• Probability of success per question = ¼

• E(x) = n ∙ p = 16(¼) = 4 correct answers

Extra Example • A student is taking a multiple choice exam with 16

questions, and guessing at the answers. Each question has 4 possible answers, A, B, C, D. Each question is independent. Is this a binomial distribution?

• What is the standard deviation?

• 𝜎𝑥 = 𝑛 ∙ 𝑝 ∙ (1 − 𝑝) = 16 ∙ 14

∙ 34

= 3

• 𝜎𝑥 ≈ 1.732

Extra Example • What is the probability that the student will get 8 questions

correct?

• P(x= 8) = P(x ≤ 8) – P(x ≤ 7)

= 168

14

8

1 −14

16−8

= 1287014

8 34

8

= 0.0197

Extra Example • What is the probability that the student will get

less than 8 correct?

• P(x < 8) = P(x ≤ 7) = P(0) + P(1) + P(2) + … + P(7)

• From the table we can find that P(x ≤ 7) = 0.9729

• What is the probability that the student will get less than 8 correct?

• P(x < 8) = P(x ≤ 7) = P(0) + P (1) + P (2) + … + P (7)

• From the table we can find that P(x ≤ 7) = 0.9729