Transcript AP Statistics

AP Statistics
Chapter 8
Section 1
A gaggle of girls
• The Ferrell family have 3 children: Jennifer,
Jessica, and Jaclyn.
• If we assume that a couple is equally likely to
have a girl or a boy, then how unusual is it for a
family to have 3 children who are all girls?
• If success = girl (of course) and failure = boy
(sorry), then p(success) = .5
• Define random variable X as the number of girls
• Simulate families with 3 children to determine
the long-term relative frequency of a family with
3 girls, P(X=3)
1. Let even digits represent girls: 0,2,4,6,8
Let odd digits represent boys: 1,3,5,7,9
2. RandInt (0,9,3)
If all three digits are even it represents a family with three girls.
Any other combination of digits represents a family with a mixture of
genders or all boys.
3 girls
Not 3 girls
Perform 40 simulations. Combine with classmates.
Calculate the relative frequency of the event {3 girls}.
X
P(X)
Determine the sample space for this simulation.
What are the possible outcomes?
X = number of girls
bbg
bgg
bgb
gbg
bbb
gbb
ggb
ggg
0 girls
1 girl
2 girls
3 girls
Vocab
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Binomial Setting
Binomial random variable
Binomial distribution – B(n,p)
pdf
cdf
factorial
Binomial coefficient
Binomial probability
Mean of a binomial random variable
Standard deviation of a binomial random variable
Binomial Setting
1.
2.
3.
4.
two categories: Success, Failure
fixed number of observations, n
all observations are independent
probability of success (p) is the same for
all observations
Binomial?
• Blood Type is inherited. If both parents
carry genes for the O and A blood types,
each child has probability 0.25 of getting
two O genes and so of having blood type
O. Different children inherit independently
of each other. These parents have 5
children.
B(5, .25)
Binomial?
• Deal cards from a shuffled deck and count
the number X of red cards. There are 10
observations, and each gives either a red
or black card.
• Not a binomial distribution – not
independent events
A quality engineer selects an SRS of 10 switches from
a large shipment for detailed inspection. Unknown
to the engineer, 10% of the switches in the shipment
fail to meet the specifications. What is the
probability that no more than 1 of the 10 switches in
the sample fail inspection?
Binomial?
1. Pass/Fail
2. N = 10
3. Independent events? Yes, each switch inspected will
succeed or fail independently of the other switches
chosen.
4. P = .1 for all switches
Yes, this is a binomial distribution. B(10, 0.1)
Calculator
2nd vars
Binomial pdf (10, .1) sto L2
In L1 put 0 - 10
This means that in this inspection of 10 switches the
probability that 0 of the switches fails inspection is
approx. 35%. The probability that one of the switches
fails is approx. 38%, etc.
So…
What is the probability that no more than 1 of the
switches fails inspection?
P X  0  P X  1 
.3487 .3874 .7361
So, in this inspection the probability that no more
than 1 of the switches fails inspection is
approx. 74%.
Histogram
• Turn axes off – format – axes off
• 2nd y =
• Set viewing window
Outcomes larger than 6 do not have probability exactly 0 but their
probabilities are so small that the rounded values are 0.0000.
You can use the 1-Var stats to verify that the sum of the
probabilities =1.
Corinne is a basketball player who makes 75% of her
free throws over the course of a season. In a key
game, Corinne shoots 12 free throws and makes
only 7 of them. The fans think that she failed
because she was nervous. Is it unusual for Corinne
to perform this poorly?
Binomial?
1. miss/make
2. N =12
3. The outcome of each shot is considered
independent of the success/failure of the other shots.
4. P = .75 for all shots
B(12, .75)
Was the outcome determined by
her nervousness?
The probability of Corinne making at most 7 free throw
shots out of 12 is approx. 16%. This means that approx.
1 out of every 6 games Corinne could be expected to
make 7 or less free throw shots out of a possible 12.
pdf and cdf tables
X
0
1
2
3
4
5
6
7
8
9
10
11
12
pdf
.000
.000
.000
.000
.002
.011
.040
.103
.194
.258
.232
.127
.032
cdf
.000
.000
.000
.000
.003
.014
.054
.158
.351
.609
.842
.968
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Binomial Formulas
n!
n
 
 k  k!n  k !
n k
nk
P X  k     p 1  p 
k
Simulation
• Simulate 12 free throws with 75% success
• Let 0 represent missed and 1 represent
made
• randBin(1, .75, 12)
• Out of the 12 shots in
the simulation 10 were
made – approx. 83%
Binomial mean and
standard deviation
  np
  np1  p 
Find the mean & s.d. for the free
throw situation
  np
 12.75
9
  np1  p 


12.751  .75
9.25
 2.25
 1 .5