Download Chapter 8 binomial and geometric distributions 8.1 binomial

Chapter 8 binomial and geometric distributions
8.1 binomial distributions
Requirements for binomial distributions
1. Each observation falls into one of two categories
Success or failure
2. There is a fixed number of observations
3. The observations are all independent
4. The probability of success is the same for each observation.
Binomial distribution must meet all 4 requirements
B(n,p)
Binomial number
probability of success
Of observations
Discrete probability
Examples page 440-441
Finding binomial probabilities
Binompdf (n,p,x) calculates the binomial probability of the value x
Pdf (probability distribution function)
Where to find on calculator
2nd distribution
0: binompdf
Ex 8.6 page 443
B(12,.75) (Corrine makes 75% of her free throws and she shoots 12
baskets)
P(she makes at most 7 baskets)
P(x 7)
P(x=0) +p(x=10+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6)+p(x=7)
Binompdf o(12,.75,0)+binompdf(12,.75,1) +….+binompdf(12,.75,7)
443-444
Find probability that number of girls is 3
P(x=3)
Binompdf(3,.5,3) = .125
Cdf (cumulative distribution function)
Calculates the sum of the probabilities
For p(x 7) we can use cdf
Binomcdf(12,.75,7)
Let’s say I want to know what the probability of Corrine making more
than 5 shots
P(x
)
Easy way to find is 1 – p(x
5)
1 – binomcdf(12,.75,5)
Making at least 7 baskets
P(x
7)
1- Binomcdf(12,.75,6)
Read pages 439 – 449 do problems 1-5, 9-13
Binomial mean and standard deviation
P=probability of success
q = probability of failure
p+ q = 1 or q = 1 – p
E(x) = μx = 0(q) + 1(P) = p
Failure
Success
Mean and standard deviation of a binomial random variable
μ = np
μ=(number of observations)(probability)
σ=
back to Corrine
n = 12
p = .75
find μ and σ for Corrine
μ = (12)(.75)
μ=9
σ=
σ=
σ=
σ = 1.5
***** As the number of trials gets larger, the binomial distribution gets
close to a normal distribution******
This means that we can use a z-score to find the area under the curve.
Ex. Attitudes about shopping
N = 2500
P = 0.6
μ = 2500(0.6) = 1500
σ=
σ=
p(x
= 24.49
1520)
z=
=
= .8167
.7939
.2061
___________________________
1500 1520
20.61% are grate than 1520.
Normal approximation for binomial distribution
N(np,
***As a rule of thumb we use normal approximation when it satisfies
np
n(1-p)
10
10
You need to check this every time you do a problem!!!!!!!!!
Can we use normal approximation of Corrine?
np
10
(12)(.75)
n(1-p)
10
12(.25)
10
10
9
10
NO
3
10
NO
Page 455 -457 exploring binomial distributions on calculator
Read pages 450 – 459
HW problems 27 – 34
Quiz over binomial probability
8.2 Geometric distributions
Geometric distributions- there is no fixed number of observations
(x = the numbers of trials to reach the first success)
Ex. Roll a die until you get a 3
Shot baskets until you make a basket
Geometric setting
1.
2.
3.
4.
Each observation falls into success and failure
Probability of success, p, is the same for each observation
Observations are all independent
The variable of interest is the number of trials required to obtain
the first success.
Rules for calculation geometric probabilities
P = probability of success
1-p = probability of failure
P(x=n) = (1-p)n-1 p
Think 90% of getting it correct.
What is the probability the first one correct is the 4th?
First 3 failures then success
(.10)3 (.90)
(.001)(.90)
.0009
Mean and standard deviation of a geometric random variable
Μ=
σ² = (1-p)/p²
σ=
p(x >n) probability it takes more than n trials
p(x>n) = (1-p)n
pages 471 – 472 calculator geometric distribution
Homework problems 37-40