Download - The Binomial & Geometric Distribution
![Page 1: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/1.jpg)
The Binomial & Geometric Distribution
The Binomial & Geometric Distribution
Chapter 8Chapter 8
![Page 2: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/2.jpg)
Everyone’s worst nightmare
Activity: In a multiple choice test, there will be five choices for each question: A B C D E. Select an answer to each question.
Answer:
1. C2. E3. D4. E5. A6. B7. C8. E 9. D10.A
![Page 3: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/3.jpg)
![Page 4: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/4.jpg)
![Page 5: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/5.jpg)
k 0 1 2 3 4 5 6 7 8 9 10
nCk 1 10 45 120 210 252 210 120 45 10 1
# possibleP(X=K)
# possible = 510 = 9 765 625P(X=0) = 1/9 765 625 = .0000001024
Binomial Coefficient
![Page 6: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/6.jpg)
X: The number of successes that result from the
binomial experiment.n: The number of trials in the
binomial experiment. P: The probability of success
on an individual trial.Q: The probability of failure on an individual trial. (This is equal to 1 - P.)b(x; n, P): Binomial probability -
nCr: The number of combinations of n things, taken r at a time.
NOTATION
![Page 7: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/7.jpg)
Binomial Formula
Suppose a die is tossed 5 times. What is the probability of getting exactly 2
fours?
number of trials is equal to 5
number of successes is equal to 2
the probability of success (getting a 4) on a single trial is 1/6 or about 0.167
b = (x;n,p) =
x=2n=5p=.167
![Page 8: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/8.jpg)
x=2n=5p=.167
=5C2 (.167)2 * (1-.167)5-2
=10 (.167) 2(.833) 3
=0.161
The Probability of getting exactly two 4’s after 5 trials is 16%
![Page 9: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/9.jpg)
Suppose a coin is tossed 5 times. What is the probability of getting exactly 3
heads?
n=5x=3p=.5
b=(n,x,p)
=5C3 (.5)3 * (1-.5)5-3
=10 (.5)3 * (1-.5)5-3
=.3125
![Page 10: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/10.jpg)
What is the probability of getting 8 correct answers from guessing on the pop
quiz?
The probability of getting 8 correct answers from guessing on the quiz will
be .0000737%
![Page 11: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/11.jpg)
Peter is a basketball player who makes 75% of his free throws over the course of the season. Peter shoots 12 free throws and makes only 7 of them. The fans think he failed because he is nervous. Is it unusual for Peter to perform this poorly?
P (X ≤ 7)
P (X = 0)+ P (X = 1)+ P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7)
= 0.0000 + 0.0000 + 0.0000 +0.0004 + 0.0024 + o.0115 + 0.0401 + 0.1032
= 0.1576
b(n,p)Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x
![Page 12: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/12.jpg)
Suppose we need to find out the probability of Peter making at most 9 shots during the game, how are you going to compute for the probability of this event happening?
Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x
= 0.6093The probability of Sufian making at most
9 basket out of the 12 free throws is 61
%
P (X ≤ 9) b(n,p)
![Page 13: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/13.jpg)
What if we need to find Peter’s probability of making at least 9 shots in a game?
= 1- binomcdf(n,p,x)
= 1- binomcdf(12,.75,9)
=.3907
b(n,p)P (X ≥ 9)
![Page 14: The Binomial & Geometric Distribution](https://reader036.vdocument.in/reader036/viewer/2022062309/5681357f550346895d9cdf45/html5/thumbnails/14.jpg)
75% 30%
μx =n(p)
σ2x =n(p)( 1 - p )
σx
= √n(p)( 1 -
P )
Compute for the following statistical measure for this particular event:
=9
=2.25
=1.5
=3.6
=2.52
=1.6