Transcript Chapter 6 Probability: The Study of Randomness

Chapter 17
Probability Models
The Binomial Distribution
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The Binomial Distribution
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A binomial variable is a variable whose domain
contains the number of successes observed in repeated
trials of a given experiment. The binomial setting
satisfies the following four conditions:
Each observation falls into one of just two categories,
“success” or “failure.”
The probability of success, called p, is the same for
each observation.
The n observations are all independent. That is,
knowing the result of one observation tells you
nothing about the other observations.
There is a fixed number, n, of observation (trial).
Binomial Distribution
The distribution of the X of successes in the
binomial distribution with parameters n and p. The
parameter n is the number of observations, and p
is the probability of a success on any one
observation.
The possible values of X are the whole numbers
from 0 to n.
As an abbreviation, we say that X is B(n, p); that is
to say that the number of success are defined by
our random variable X described by the binomial
distribution given n observations with each
success having a probability of p.
Example
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An archer has determined that the probability that she hits
a bull’s eye is 0.72 on every shot. What is the probability
that she will hit exactly 8 bull’s eyes in 10 shots?
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Let X = random variable whose values are the number of bull’s
eyes she could hit on her next 10 shorts.
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Does this situation satisfy our assumptions?
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Are there only two outcomes?
• Yes, either she hits or misses
Is the probability the same for each shot?
• Yes, we are told that the probability of each shot
Are her shots independent?
• Although it is possible that one shot may influence the next
(she may adapt), it is acceptable to assume that each shot is
independent unless there is a reason not to make this
assumption. When in doubt, leave a disclaimer.
Example
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An archer has determined that the probability that she hits
a bull’s eye is 0.72 on every shot. What is the probability
that she will hit exactly 8 bull’s eyes in 10 shots?
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Let X = random variable whose values are the number of bull’s
eyes she could hit on her next 10 shorts.
X is a binomial random variable and we write X: B(10, 0.72)
 In the calculator, use the function binompdf (n, p)
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If you use binompdf (10, .72), it will give you the entire probability
distribution for this event.
For a specific probability (the probability that the archer hits
exactly 8 of the 10 shots) use binompdf (n, p, x):
binompdf (10, .72, 8) = .2548
Binomial distributions allow us to find the probabilities of these
specific types of problems without having to do the tree diagrams.
They are intended to save us time and energy.
pdf - probability distribution function
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Given a discrete random variable X, the probability
distribution function assigns a probability to each value
of X. The probabilities must satisfy probability rules.
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Try to find the probability distributions for the following
scenarios:
 A basketball player hits .80 of his free-throws and he shoots
six baskets.
 A child desires a certain toy that is randomly distributed
within cereal boxes with a 20% chance of obtaining the
particular toy and eight boxes are opened.
Try to find the specific probabilities of the following:
 A basketball player hits .80 of his free-throws and he shoots
six baskets. Find the probability that he makes exactly 4
baskets.
cdf-cumulative distribution function
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In applications we frequently want to find the
probability that a random variable takes a range of
values. The cumulative binomial probability is useful in
these cases.
cdf - Given a random variable X, the cumulative
distribution function (cdf) of X calculates the sum of
probabilities for 0, 1, 2, ….., up to the value X. That is, it
calculates the probability of obtaining at most X
successes in n trials.
 It is important to note that the binomcdf (n, p, X)
function adds up all the probabilities from 0 to X.
The Binomial Coefficient
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A Binomial Coefficient counts the number of ways of
arranging k successes among n observations and is
given by the formula:
n
n!
  
 k  k!(n  k )!
For k = 0, 1, 2, ……., n.
In other words, this gives the number of combinations
for each event.
For example, it can count the number of ways you can
pick a group of three people from this class. The
number of students would be n and k = 3.
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The formula for binomial coefficients uses the factorial
notation.
n! = n x (n-1) x (n - 2) x …….x 3 x 2 x 1.
(note: 0! = 1)
4! = 4 x 3 x 2 x 1 = 24
For example, let’s say you want to pick three people
from this class and there are 20 students present. Then
n = 20 and k = 3.
 20 20! 20191817! 201918
  
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 1140
17! 3  2 1
3  2 1
 3  17!3!
There are 1140 different groups of three that I can
choose from a group of 20 people.
Binomial Probability
If X has the binomial distribution with n observations
and probability p of success on each observation, the
possible values of X are 0, 1, 2, …, n. If k is any one of
these values,
n k
P(X = k) =  p (1 - p)n-k
k 
Example:
Lets say an archer hits the “bulls-eye” 72% of the time
that she shoots. Using the Binomial distribution,
determine the number ways that she can hit the bulls-eye
exactly 4 out of 6 times.
6
P(X = 4) =  (.72)4(.28)2 = 0.3160
 4
Or you can you binompdf (n, p, x) = binompdf (6, .72, 4)
Mean and Standard Deviation of a
Binomial Random Variable
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If a count X has the binomial distribution with number of
observations n and probability of success p, the mean and
standard deviation of X are
µ = np
 = np1 p
These formulas are good only for binomial distributions.
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Back to the bulls-eye: The archer hits the “bullseye” 72% of the time that she shoots. She takes 6 shots.
What is her Expected Value or the mean number of
bulls-eyes we expect her to get? What is the standard
deviation?
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Formulas:
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So, the E(X) = µX = (6)(.72) = 4.32
σX = (6)(.72)(.28)  1.0998
The mean number of bulls-eyes in six shots is
4.32 with a standard deviation of 1.0998.
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Mean: µ = np
Standard deviation: σ = np1 p
Binomial Distributions
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Suppose an Olympic archer is able to hit the bullseye 87% of the time. If she shoots eight arrows,
what is the probability that she will get:
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(Assume that each shot is independent.)
4 bulls-eyes binompdf (8, .87, 4) ≈ 0.0115
 at most 4 bulls-eyes binomcdf (8, .87, 4) ≈ 0.0129
 at least 4 bulls-eyes 1 – binomcdf (8, .87, 3) ≈ 0.9985
 What is the Expected Value? np = 8(0.87) ≈ 6.96
 What is the Mean and standard deviation?
Since E(X) = μ=np = 8(0.87) ≈ 6.96
 exactly
  (8)(0.87)(0.13)  0.9512
Normal Approximation for Binomial Distributions
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Suppose that a count X has a binomial distribution with n
trials and success probability p. When n is large, the
distribution of X is approximately normal,
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N ( X , X )  N np, np(1  p)
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The Success/Failure Condition: we will use the normal
approximation when n and p satisfy:
np 10 and n(1- p)  10.
Basically, we can use the normal model if we expect to see
at least 10 successes and 10 failures, based on the
probabilities, of course.
Example
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Use a normal distribution to estimate the
following: A baseball player has a .300 batting
average. What is the probability that he will hit
at least 40 of the next 100 at bats?
First determine if the binomial setting is
appropriate. Yes, all of the conditions are met.
Now find E(X) E(X) = np = 100(.300) = 30
Find σX σ  npq  100(.3)(.7 )  4.58
X
40  30
 2.18
Find the z-score for 40 z 
4.58
Determine the P(X ≥ 40) normalcdf(2.18,E99)
≈ 0.01463
Assignment
Chapter 17
Lesson:
Geometric and
Binomial Distributions
Read:
Chapter 17
Problems:
1 – 37 (odd)
Ch. 17 WS