Transcript Chapter 6: Modeling Random Events... Normal

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Chapter 6: Modeling
Random Events... Normal
& Binomial Models
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Models...
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Help us predict what is likely to happen

Remember LSRLs (a model for linear, bi-variate data); not
exact (some points above line, some below); but best model
we have
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Two more models discussed in this chapter are for Normal
(or ≈ Normal) and for Binomial distributions
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Both of these describe/model numeric, uni-variate data
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Again, models are not perfect... But in many cases, they are
good, effective, or “good enough”
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Two types of numeric data...
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Discrete random variables/data (we will use binomial
distribution with discrete data)
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Continuous random variables/data (we will often use Normal
distribution with continuous data)
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Let’s discuss discrete random variables/data first
+ Discrete Random Variables
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Discrete random variables have a “countable” number of possible
positive outcomes and must satisfy two requirements. Note:
‘countable’ is not the same as finite.
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(1) Every probability is a # between 0 and 1;
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(2) The sum of the probabilities is 1
+ World-Wide 2015 High School AP Statistics
Score Distribution
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Is this a discrete random variable probability distribution?
Why or why not?
1
2
3
4
5
.238
.189
.252
.189
.132
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Discrete Random Variables...
Discuss
other examples of discrete
random variables. 1 minute.
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+ Other Examples of Discrete Random
Variables...
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Number of times people have seen Shawn Mendes in concert
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Number of gifts we get on our birthday
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Number of burgers sold at In-N-Out Burger per day
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Number of stars in the sky
All are whole, countable numbers; all vary; usually represented by a
table or probability histogram
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Non-examples of Discrete Random
Variables...
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Your height
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Weight of a candy bar
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Time it takes to run a mile
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Continuous Random Variables...
are
usually measurements
heights, weights, time
amount
of sugar in a granny smith
apple, time to finish the New York
marathon, height of Mt. Whitney
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How can we distinguish between
continuous and discrete?
 Discuss
in your groups for a few minutes.
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How can we distinguish between
continuous and discrete?
 Discuss
in your groups for a few minutes.
 Ask
yourself ‘How many? How much? Are
you sure?’
 For
example, # of children, pounds of
Captain Crunch produced each year, # of
skittles, ounces in a bag of skittles
+ Continuous Random Variables ...
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take on all values in an interval of numbers
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probability distribution is described by a density curve
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probability of event is area under the density curve and
above the values of X that make up the event
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total area under (density) curve = 1
+ Continuous Random Variables...
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Probability distribution is area under the density curve, within an
interval, above x-axis
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Continuous Random Variables ...
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+ Random Variables...
Consider a six-sided die... What is the probability...
P ( roll less than or equal to a 2) is
P ( roll less than a 2) is
Different probabilities; discrete random variable
Note: possible outcomes are 1, 2, 3, 4, 5, 6; but
probabilities of those outcomes are (often)
fractions/decimals
+ Continuous random variables ...
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All continuous random variables assign probabilities to intervals
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All continuous random
variables assign a probability of zero to every
individual outcome. Why?
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Continuous Random Variables...
 There
is no area under a vertical line (sketch)
 Consider...
0.7900 to 0.8100
P = 0.02
0.7990 to 0.8010
P = 0.002
0.7999 to 0.8001
P = 0.0002
P (an exact value –vs. an interval--) = 0
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Density Curves & Continuous
RV’s..
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Can use ANY density curve to assign probabilities/model
continuous distributions/RV’s; many models
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Most familiar density curves are the Normal (bell) density
curves
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Based on Empirical Rule, 68-95-99.7, symmetric, uni-modal,
chapter 3)
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Many distributions/events are considered Normal & can be
modeled by Normal density curves, such as ... cholesterol
levels in young boys, heights of 3-year-old females, Tiger
Woods’ distance golf ball travels on driving range, basic
skills vocabulary test scores for 7th graders, etc.
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Mean & Standard Deviation of Normal
Distributions...
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μx for continuous random variables lies at the center of a
symmetrical (or fairly symmetrical) density curve (Normal or
approximately Normal)
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N (μ, σ)
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Remember... μand σ are population parameters;
sample statistics.
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
Calculating σ and/or σ2 for continuous random
variables….
beyond the scope of this course… will be given this
information if needed
xand
s are
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Normal distributions/density
curves...
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Four possibilities that we need to
know how to calculate...
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When calculating probabilities for Normal distributions,
there are four possibilities that we might be asked to
calculate.
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Let’s draw some pictures and match up some questions to
each drawing
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Remember for continuous random variables (like Normal
distributions) there is no difference between ‘less than’ and
‘less than or equal to;’ likewise no difference between
‘greater than’ or ‘greater than or equal to.’
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Female heights... N(64.5, 2.5)
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Female heights... N(64.5, 2.5)
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Calculate the probability that a randomly chosen female is:
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shorter than 63 inches
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no more than 61 inches
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taller than 66 inches
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68 inches or taller
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between 63.5 inches and 64 inches
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between 5 inches and 72 inches
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Female heights... N(64.5, 2.5)
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Calculate the probability that a randomly chosen female is:
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shorter than 63 inches or taller than 70 inches
as short or shorter than 65 inches or as tall or taller than 67 inches
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Female heights... N(64.5, 2.5)
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Sometimes we want to turn this around... Sometimes we are
given a probability & we need to find the value that
corresponds to that probability.
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What is the height of a randomly chosen woman if she is in
the 20th percentile? Let’s draw a picture...
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What is the height of a randomly chosen woman if she is in
the 85th percentile?
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Normal model...
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Very helpful, but one size does not fit all
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Good first choice if data is continuous, uni-modal, symmetric,
68-95-99.7
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Another important, “special” type of
distribution...
 If
certain criteria is met, easier to calculate
probabilities in specific situations
 Next
types of distributions we will examine are
situations where there are only two outcomes
 Win
or lose; make a basket or not; boy or girl ...
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Discuss situations where there are only
two outcomes...
 Yes
or no
 Open
or closed
 Patient
has a disease or doesn’t
 Something
 Person
A
is alive or dead
has a job or doesn’t
part is defective or not
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that is what this section is all about...
 ... a
class of distributions that are concerned about
events that can only have 2 outcomes
 The
Binomial Distribution
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Binary; Independent; fixed Number;
probability of Successes
The Binomial setting is:
1.
Each observation is either a success or a failure (i.e., it’s
binary)
2.
All n observations are independent
3.
Fixed # (n) of observations
4.
Probability of success, p, is the same for each observation
“BINS”
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Binomial Distribution: practice…
I roll a die 3 times and observe each roll to see if it is
even or odd. Is:
1.
each observation is either a success or a failure?
2.
all n observations are independent?
3.
fixed # (n) of observations?
4.
probability of success, p, is the same for each
observation ?
BINS
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Binomial Distribution
 If
BINS is satisfied, then the distribution can be
described as B (n, p)
B
binomial
n
the fixed number of observations
p
probability of success
 Note:
This is a discrete probability distribution.
 Remember
N (μ, σ)… is that discrete??
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Binomial Distribution
 Most
important: being able to recognize situations
and then use appropriate tools for that situation
 Let’s
practice...
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Are these binomial distributions? Why
or why not?
 Toss
a coin 20 times to see how many tails occur.
 Asking
200 people if they watch ABC news
 Rolling
a die until a 6 appears
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Are these binomial distributions or
not? Why or why not?
 Asking
20 people how old they are
 Drawing
 Rolling
5 cards from a deck for a poker hand
a die until a 5 appears
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How could we change the situations
to make these/force these to be
binomial distributions?
 Asking
20 people how old they are
 Drawing
 Rolling
5 cards from a deck for a poker hand
a die until a 5 appears
+ Side note: Binomial Distribution... Is
the situation ‘independent enough’?
An engineer chooses a SRS of 20 switches from a shipment of 10,000
switches. Suppose (unknown to the engineer) 12% of switches in
the shipment are bad.
Not quite a binomial setting. Why?
For practical purposes, this behaves like a binomial setting; ‘close
enough’ to independence; as long as sample size is small
compared to population. Rule of thumb: sample ≤ 10% of
population size
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Binomial probabilities...in Minitab
use Probability Density Function
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Two different ‘flavors’ of binomial calculations... First one is
use for finding the probability of an exact value, a particular
point (i.e., probability that a family has exactly 2 boys)
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Remember, these are discrete random variables
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Can calculate probabilities of exact values (unlike
continuous random variables)
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Minitab: go to Probability, Probability Density Function, x =,
binomial, trials, event probability
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Binomial probabilities: in Minitab
use Cumulative Distribution
Function...
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Use for cumulative values (i.e., probability that a family has at
most 2 boys; at least 3 girls; no more than 4 boys, etc.)
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Cumulative to left; at that value or less
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If want area to right, need to do “1 minus ....”
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Pay attention to < vs. ≤; and > vs. ≥
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Minitab go to Probability, Cumulative Distribution Function,
value (what x = ), binomial, trials, event probability
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
probability that exactly 2 of them have type
O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=2
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
probability that exactly 2 of them have type
O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=2
= 0.2636; context, always!
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
probability that exactly 4 of them have type
O blood? Binomial setting? Check for
BINS.
p = 0.25
n=5
X=4
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type
O. If these parents have 5 children, what is
the probability that exactly 4 of them have
type O blood? Binomial setting? Check
for BINS.
p = 0.25
n=5
X=4
= 0.0146; context, always
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
probability that exactly 1 of them have type
O blood? Binomial setting? Check for
BINS.
p = 0.25
n=5
X=1
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
probability that exactly 1 of them have type
O blood? Binomial setting? Check for
BINS.
p = 0.25
n=5
= 0.3955; context, always
X=1
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type
O. If these parents have 5 children, what is
the probability that at most 2 of them have
type O blood? Binomial setting? Check for
BINS.
p = 0.25
n=5
X=2
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type
O. If these parents have 5 children, what is
the probability that at most 2 of them have
type O blood? Binomial setting? Check for
BINS.
p = 0.25
n=5
X=2
= 0.8965; context, always
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type
O. If these parents have 5 children, what is
the probability that at most 4 of them have
type O blood? Binomial setting? Check
for BINS.
p = 0.25
n=5
X=4
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
probability that at most 4 of them have type
O blood? Binomial setting? Check for
BINS.
p = 0.25
n=5
X=4
= 0.9990, context, always
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
probability that at most 1 of them have type
O blood? Binomial setting? Check for
BINS.
p = 0.25
n=5
X=1
+ Practice...
Each child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
probability that at most 1 of them have type
O blood? Binomial setting? Check for
BINS.
p = 0.25
n=5
X=1
= 0.6328, context, always
+ Practice...
... probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that at least 2 (meaning 2, 3, 4,
or 5) of them have type O blood?
p = 0.25
n=5
X = 2, 3, 4, or 5
+ Practice...
... probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that at least 2 (meaning 2, 3, 4,
or 5) of them have type O blood?
p = 0.25
n=5
X = 2, 3, 4, or 5
1 - 0.6328 = 0.3672; context, always
+ Practice...
... probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that at least 3 (meaning 3, 4,
or 5) of them have type O blood?
p = 0.25
n=5
X = 3, 4, or 5
+ Practice...
... probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that at least 3 (meaning 3, 4, or
5) of them have type O blood?
p = 0.25
n=5
X = 3, 4, or 5
1 - 0.8965 = 0.1035; context, always
+ Practice...
... probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that more than 3 (meaning 4
or or 5) of them have type O blood?
p = 0.25
n=5
X = 4 or 5
+ Practice...
... probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that more than 3 (meaning 4
or or 5) of them have type O blood?
p = 0.25
n=5
X = 4 or 5
1 - 0.9844 = 0.0156; context, always
+ Practice...
... probability 0.25 of having blood type O. If
these parents have 5 children, what is the
probability that … of them have type O
blood?
p = 0.25
a) …at most 3 of them…
b) …at least 4 of them…
c) …more than 1 of them…
d) …exceeds 3 of them…
e) … below 2 of them…
f) … 0 of them…
n=5
+
Binomial Distribution: Mean and
Standard Deviation
If
a basketball player makes 75% (“p”
of her free throws, what do you think
the mean number of baskets made
will be in 12 tries?
x

+
Binomial Distribution: Mean and
Standard Deviation

If a basketball player makes 75% (“p” of her free throws,
what do you think the mean number of baskets made will be
in 12 tries?
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(0.75) (12) = 9; we expect she should make 9 baskets in 12
tries.

Her mean number of baskets made should be 9.

We expect her

x
= 9 baskets; E(x) = 9
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Binomial Mean & Standard Deviation
 If
a count X is a binomial distribution with number
of observations n and probability of success p,
then
 μ=
np
σ=
and
 Only
np(1  p)
for use with binomial distributions;
remember criteria... BINS

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Practice...
 If
a basketball player makes 75% (“p”) of her free
throws, we expect her to make 9 baskets in 12
tries.
 What
is the SD of this distribution?

+
Practice...
 If
a basketball player makes 75% (“p”) of her free
throws, we expect her to make 9 baskets in 12
tries.
 What
 SD
=
is the SD of this distribution?
np(1  p)
=
(12)(0.75)(1 0.75)
= 1.5

+ Practice...
 Each
child born to a particular set of parents
has probability 0.25 of having blood type O.
If these parents have 5 children, what is the
mean and standard deviation for this
distribution?
+ Practice...
 Each
child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the mean and
standard deviation for this distribution?
 mean
= np = (5)(.25) = 1.25; the family should
expect to have 1.25 children with type O blood
SD
= np(1  p)
= 0.9682
=
(5)(0.25)(1 0.25)
+ Remember & Caution...
 Binomial
distribution is a special case of a
probability distribution for a discrete random
variable
 All
binomials distributions are discrete random
variable distributions BUT not all discrete random
variable distributions are binomial distributions
 Don’t
assume; don’t apply binomial tools to all
discrete RV distributions; must meet binomial
distribution criteria (BINS)
+
Next test... Exam #2 ...
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Chapters 4, 5, & 6