the binomial distributionrchidester/stat 211 week five display version.pdf · binomial experiment...
TRANSCRIPT
![Page 1: 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](https://reader033.vdocuments.us/reader033/viewer/2022043023/5f3f48744376b6086f2f788e/html5/thumbnails/1.jpg)
Stat 211 Week Five The Binomial Distribution
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Last Week • 𝐸 𝑥 = ∑ 𝑥 ∙ 𝑝(𝑥) • 𝐸 𝑥 = 𝑛 ∙ 𝑝
• 𝜎𝑥 = ∑ 𝑥 − 𝜇𝑥
2𝑝(𝑥)
• We will see this again soon!!
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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.
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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.
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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!!!
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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
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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!
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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)
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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)
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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
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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
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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
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Notice
• We get the same results both way!!!
• But, using the Binomial equations is way way way easier!!!!
• It is a shortcut.
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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
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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
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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
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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
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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.
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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
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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.
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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.
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Too Much Effort • Doing a problem that way is possible but not
feasible.
• There is an easier way!!!
• The table.
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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
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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.
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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.
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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)
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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)
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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!!!
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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!!!
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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)
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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)
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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
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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).
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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.
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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)
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Weird Words • Probability questions are often word questions.
• More than > • Less than < • ____ or more ≥ • ____ or less ≤
• At most ≤ • At least ≥
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Be Careful When Flipping • P(X > 3) = 1 – P(X ≤ 3)
• P(X ≥ 3) = 1 – P(X ≤ 2)
• Think before you flip!!!!!
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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.
![Page 39: 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](https://reader033.vdocuments.us/reader033/viewer/2022043023/5f3f48744376b6086f2f788e/html5/thumbnails/39.jpg)
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
![Page 40: 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](https://reader033.vdocuments.us/reader033/viewer/2022043023/5f3f48744376b6086f2f788e/html5/thumbnails/40.jpg)
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
![Page 41: 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](https://reader033.vdocuments.us/reader033/viewer/2022043023/5f3f48744376b6086f2f788e/html5/thumbnails/41.jpg)
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
![Page 42: 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](https://reader033.vdocuments.us/reader033/viewer/2022043023/5f3f48744376b6086f2f788e/html5/thumbnails/42.jpg)
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
![Page 43: 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](https://reader033.vdocuments.us/reader033/viewer/2022043023/5f3f48744376b6086f2f788e/html5/thumbnails/43.jpg)
• 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