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Probability flow chart
Are there only two outcomes possible
(success/fail)?
Bernoulli trials:
Are we finding quantity
of successes or location
of first success?
location
Geometric Probability
1
    =

=

2
quantity
Binomial Probability
    = 
=

Yes
No
Continued
on next slide
Are there only two outcomes possible (success/fail)?
No
Is it a question involving an “and/or”?
No
No
Simple
Probability
Is it a question
involving a “given”?
Yes
Are the events disjoint?
No
Yes
Draw a tree
diagram if possible
( ∩ )
  =
()
Draw a Venn diagram
Yes
  ∪  =   +   − ( ∩ )
  ∩  = () ∙ (|)
  ∪  =   + ()
  ∩  = () ∙ (|)
Practice FR #1
You are up for your annual job performance review. You
estimate that there’s a 30% chance you’ll get a
promotion, a 40% chance of a raise, and a 25% chance
of getting both a promotion and a raise.
a)Find the probability that you get a raise or a
promotion.
b)Find the probability that you got neither.
c)Are the raise and the promotion independent events?
d)Are the raise and the promotion disjoint events?
a) 0.45 or 45%
Promotion
Raise
b) 0.55 or 55%
c) The events are not
0.05
0.25
independent because
0.15
P(Pr|R)=0.25/0.4
0.55 which isn’t equal to
P(Pr) of 0.3.
d) The events are not disjoint since it is possible to get
both a promotion and a raise.
Practice FR #2
Assume that 75% of the AP Stat students
studied for this test. If 40% of those who study
get an A, but only 10% of those who don’t
study get an A, what is the probability that
someone who gets an A actually studied for
the test?
A
  ∩  = 0.75 0.4 = 0.3
0.6
Not A
  ∩  = 0.75 0.6 = 0.45
0.1
A
  ∩  = 0.25 0.1 = 0.025
0.9
Not A
  ∩  = 0.25 0.9 = 0.225
0.4
Yes
0.75
0.25
No
( ∩ )
0.3
   =
=
= .923
()
0.3 + 0.025
Practice FR #3
Assume the heights of high school basketball players are
normally distributed. For girls the mean is 70 inches with a
standard deviation of 0 inches, while boy players have a mean
height of 74 inches with a standard deviation of 4.5 inches.
a) On average, how much taller is a boy player than a girl
player?
b) What will be the standard deviation of the difference in
heights between a boy player and a girl player?
c) On what percentage of the team would you expect the girl
to be taller than the boy?
a) E(B-G) = E(B)-E(G) = 74-70 = 4 inches
b) SD B − G =
(4.5)2 (3)2 = 5.4 inches
c) Let D = B-G and we want the probability that a girl is
taller than a boy, so B-G<0.
P(D<0) = P( <
0−4
)
5.4
= P(z < -0.7407) = 0.2294
We would expect about 23% of the time, a girl will be
taller than a boy.
Practice FR #4
The National Association of Retailers reports that 62% of all
purchases are now made by credit card. You think this is true
at your store as well. On a typical day you make 20 sales.
a) Explain why your sales can be considered Bernoulli trials.
b) What is the probability that your 4th customer is the first
one who uses a credit card?
c) What is the probability model for the number of customers
who use a credit card on a typical day?
d) What is the probability that on a typical day at least half of
your customers use a credit card?
a)Bernoulli trials have only two possible outcomes
(success = used credit card; failure = doesn’t use
credit card). Trials are independent (one transaction
does not influence the next transaction). The
probability of success stays the same on every trial
(62% of all purchases).
b)P(X=4) =0.034
c)Model: Binom(20, 0.62) with mean = 12.4 and
standard deviation = 2.17
d)P X ≥ 10 = 1 − P X = 0 + P X = 1 +
Practice FR #5
New York public health officials report that currently 22% of
adults smoke. They hope that newly increased state cigarette
taxes will reduce this rate. They plan to check in December by
selecting a random sample of 1200 New Yorkers to estimate
again the percentage of adults who smoke.
a)Verify that a Normal model is a useful approximation for the
binomial in this situation.
b)In that December sample, how many smokers would it take to
convince you that the percentage of NY adults who smoke had
decreased significantly? Explain.
a)np = 1200 0.22 = 264 ≥ 10
nq = (1200)(0.78) = 936 ≥ 10
Also, 1200 < 10% of all New Yorkers. (The population size
isn’t given, but it’s BIG! Much bigger than 12,000.)
b)It would be unusual to see the number of adult smokers be
lower than 2 standard deviations below the mean.
The standard deviation is 14.35 and the mean is 264, so
264 – 2(14.35) = 235.3. I would conclude that the estimate
of 22% of adults smoke had decreased significantly if there
were fewer than 235 smokers in the December sample of
1200 NY adults.

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