038­037B_6.3_binomial­distr_103634_s12.notebook
Class #38
STA 2023 ­­
_103634__
MON.7/16/12
July 16, 2012
Sect. 6.3: How Can We Find Probabilities
When Each Observation Has Two Possible
Outcomes? (p. 291).
The BINOMIAL DISTRIBUTION
Criteria
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(p.292)
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038­037B_6.3_binomial­distr_103634_s12.notebook
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The Formula, p. 393.
n n!
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038­037B_6.3_binomial­distr_103634_s12.notebook
binompdf(n,p,s)
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n = # of trials
p = prob. of success in any one trial.
S = # of successes.
I prefer to think "r" for # right. Then I can easily remember "n,p,r."
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038­037B_6.3_binomial­distr_103634_s12.notebook
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0.125
0.375
0.375
0.125
P(2)=binompdf(3,0.5,2)= 0.375,
for example.
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Continued on the next page.
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038­037B_6.3_binomial­distr_103634_s12.notebook
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4 children = 4 trials (in this problem).
The desired outcome is "has a girl," and
the probability of a girl is 0.46.
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