DOES THIS QUESTION RESULT IN A
BERNOULLI TRIAL?
1.
2.
3.
Yes
No
Who was Bernoulli again?
33%
33%
33%
Slide
1- 1
1
2
3
BERNOULLI, GEOMETRIC, BINOMIAL

Bernoulli – single trial

Geometric – number of trials until success

Binomial – number of success in n trials

Poisson – number of success with large n and
very small p
Slide
1- 2
THE GEOMETRIC MODEL (CONT.)
Geometric probability model for Bernoulli trials:
Geom(p)
p = probability of success
q = 1 – p = probability of failure
X = number of trials until the first success occurs
x-1
P(X = x) = q p
1
E(X)   
p

q
p2
Slide
1- 3
THE BINOMIAL MODEL (CONT.)
Binomial probability model for Bernoulli trials:
Binom(n,p)
n = number of trials
p = probability of success
q = 1 – p = probability of failure
X = number of successes in n trials
n!
 n  x n x
n
P( X  x)    p q where  
 x
 x  x !(n  x)!
  np
  npq
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1- 4
THE NORMAL MODEL TO THE RESCUE!
When dealing with a large number of trials in a
Binomial situation, making direct calculations of
the probabilities becomes tedious (or outright
impossible).
 Fortunately, the Normal model comes to the
rescue…

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1- 5
THE NORMAL MODEL TO THE RESCUE
(CONT.)

As long as the Success/Failure Condition holds,
we can use the Normal model to approximate
Binomial probabilities.

Success/failure condition: A Binomial model is
approximately Normal if we expect at least 10
successes and 10 failures:
np ≥ 10 and nq ≥ 10.
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1- 6
CONTINUOUS RANDOM VARIABLES
When we use the Normal model to approximate
the Binomial model, we are using a continuous
random variable to approximate a discrete
random variable.
 So, when we use the Normal model, we no longer
calculate the probability that the random
variable equals a particular value, but only that
it lies between two values.

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1- 7
APPLYING THE NORMAL MODEL




A lecture hall has 190 seats with folding arm
tablets, 25 of which are designed for left-handers.
The average size of classes that meet there is
177.
We can assume that about 12% of students are
left-handed.
What’s the probability that a right-handed
student in one of these classes is forced to use a
lefty arm tablet?
CAN WE USE THE NORMAL MODEL?
1.
2.
3.
4.
Yes, n*p ≥ 10 and n*(1-p) ≥ 10 and we are examining
a range of possible outcomes.
Yes, n*p ≥ 10 and n*(1-p) ≥ 10 and we are examining
one possible outcome.
No, n*p ≥ 10 and n*(1-p) ≥ 10 and we are examining a
range of possible outcomes.
No, n*p ≥ 10 and n*(1-p) ≥ 10 and we are examining
one possible outcome.
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1- 9
WHAT IS THE EXPECTED NUMBER OF RIGHT
HANDED STUDENTS IN THE AVERAGE CLASS?
1.
2.
3.
4.
177*(.12)
177*(.88)
190*(.12)
190*(.88)
0%
0%
0%
0%Slide
1- 10
1
2
3
4
WHAT IS THE STANDARD DEVIATION OF RIGHT
HANDED STUDENTS IN THE AVERAGE CLASS?
25%
1.
2.
3.
4.
25%
25%
25%
177*(.12)*(.88)
sqrt(177*(.12)*(.88))
190*(.12)*(.88)
sqrt(190*(.12)*(.88))
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1
2
3
4
WHAT HAS TO HAPPEN FOR A RIGHT-HANDED
STUDENT TO HAVE TO SIT IN A LEFTY CHAIR?
1.
2.
3.
4.
X ≥ 190
X<190
X ≥ 166
X<166
0%
1
0%
2
0%
3
Slide
0%
1- 12
4
WHAT’S THE PROBABILITY THAT A RIGHTHANDED STUDENT HAS TO SIT IN A CHAIR WITH
A LEFT-HANDED TABLET?
1.
2.
3.
4.
.0089
.9911
.0166
.9834
0%
0%
0%
0%Slide
1- 13
1.
2.
3.
4.
PROCESS
Determine X
 Check Normality Assumptions
 Find E(X) and SD(X) based on the Binomial
model
 Determine z-score based on a fixed level or fixed
standard given by the problem.
 Apply to z-tables to find probability

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1- 14
USING STATISTICS TO TEST THE VALIDITY
OF CLAIMS.
A newly hired telemarketer is told he will
probably make a sale on about 12% of his phone
calls.
 The first week he called 300 people, but only
made 12 sales.


Should he suspect he was misled about the true
success rate? Explain.
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1- 15
SHOULD HE SUSPECT HE WAS MISLED?
0%
1.
2.
0%
3.
0%
0%
4.
Twelve sales is less than 1 SD below the mean. He
was probably misled.
Twelve sales is less than 1 SD below the mean. He
was probably NOT misled.
Twelve sales is less than 5 SD below the mean. He
was probably NOT misled.
Twelve sales is more than 3 SD below the mean. He
was probably misled.
THE POISSON MODEL
The Poisson probability model was originally
derived to approximate the Binomial model when
the probability of success, p, is very small and the
number of trials, n, is very large.
 The parameter for the Poisson model is λ. To
approximate a Binomial model with a Poisson
model, just make their means match: λ = np.

Slide
1- 17
THE POISSON MODEL (CONT.)
Poisson probability model for successes:
Poisson(λ)
λ = mean number of successes or n*p
X = number of successes
e is an important mathematical constant
(approximately 2.71828)
 x
e 
P  X  x 
x!
EX   
SD  X   
Slide
1- 18
APPLYING THE POISSON MODEL
The probability of contracting a disease is small,
with p about 0.0005 for a new case in a given
year.
 In a town of 6,000 people, what is the expected
number of new cases?

Slide
1- 19
WHAT IS THE EXPECTED NUMBER OF NEW
CASES IN A TOWN OF 6,000?
1.
2.
3.
4.
6,000*0.0005
6,000*0.9995
6,000*0.0005*0.9995
We need more information.
0%
0%
0%
0%Slide
1- 20
1
2
3
4
USE THE POISSON MODEL TO APPROXIMATE
THE PROBABILITY THAT THERE WILL BE AT
LEAST ONE NEW CASE OF THE DISEASE NEXT
YEAR
1.
2.
3.
4.
Exp(-3)*31/1!
Exp(-3)*30/0!
1-Exp(-3)*31/1!
1-Exp(-3)*30/0!
0%
1
0%
0%
2
3
0%
4
Slide
0%
1- 21
5
WHAT’S THE PROBABILITY THAT THERE
WILL BE MORE THAN TEN NEW CASES NEXT
YEAR?
1.
2.
3.
4.
Exp(-3)*310/10!
0.9901
1-Exp(-3)*310/10!
1-0.9901
0%
1
0%
0%
2
3
0%
4
Slide
0%
1- 22
5
APPLYING THE POISSON MODEL


Hurricanes in a particular place arrive with a
mean of 2.35 per year.
Suppose the number of hurricanes can be
modeled by a Poisson distribution with this
mean.
Slide
1- 23
WHAT’S THE PROBABILITY OF NO
HURRICANES NEXT YEAR?
1.
2.
3.
4.
Exp(2.35)*2.35
Exp(-2.35)*2.35
Exp(2.35)*2.350
Exp(-2.35)*2.350
0%
0%
0%
0%Slide
1- 24
1
2
3
4
WHAT’S THE PROBABILITY THAT DURING THE
NEXT TWO YEARS, THERE IS EXACTLY ONE
HURRICANE?
1.
2.
3.
4.
Exp(-2.35)*2.35
Exp(-2.35)*2.350
Exp(-2.35)*2.35*Exp(-2.35)*2.350
Exp(-2.35)*2.35*Exp(-2.35)*2.350
+Exp(-2.35)*2.35*Exp(-2.35)*2.350
0%
0%
0%
0%Slide
1- 25
1
2
3
4
UPCOMING
Homework 7 due Sunday
 Review for Exam on Tuesday
 Exam 1 on Thursday

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