Transcript 4.3 PowerPoint

Sta220 - Statistics
Mr. Smith
Room 310
Class #12
Section 4.3
4.3- The Binomial Random Variable
Many experiments result in responses for which
there exist two possible alternatives, such as YesNo, Pass-Fail, Defective-Nondefective or MaleFemale.
These experiments are equivalent to tossing a coin
a fixed number of times and observing the number
of times that one of the two possible outcomes
occurs. Random variables that possess these
characteristics are called Binomial random
variables.
Definition
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Example 4.10
The Heart Association claims that only 10% of
U.S. adults over 30 years of age meet the
minimum requirements established by
President’s Council of Fitness, Sports and
Nutrition. Suppose four adults are randomly
selected and each is given the fitness test.
a) Find the probability that none of the four adults
passes the test.
b) Find the probability that three of the four adults
pass the test.
c) Let x represent the number of four adults who
pass the fitness test. Explain why x is a binomial
random variable.
d) Use the answers to parts a and b to derive a
formula for p(x), the probability distribution of
the binomial random variable x.*
a)
1) First step is define the experiment.
We are observing the fitness test results of each
of the four adults: Pass(S) or Fail(F)
2) Next we need to list the sample points
associated with the experiment.
Example: FSSS represents the sample point
denoting that adult 1 fails, while adult 2, 3, and
4 pass the test.
There are 16 sample points.
Table 4.2
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3) Now we have to assign probabilities to the
sample points.
Since sample points can be viewed as the
intersection of four adults’ test results, and
assuming that the results are independent, we
can obtain the probability of each sample point
by the multiplicative rule.
Let find the probability that all four adults passed the fitness test.
P(SSSS)
= 𝑃[ 𝑎𝑑𝑢𝑙𝑡 1 𝑝𝑎𝑠𝑠𝑒𝑠 ∩ 𝑎𝑑𝑢𝑙𝑡 2 𝑝𝑎𝑠𝑠𝑒𝑠 ∩ 𝑎𝑑𝑢𝑙𝑡 3 𝑝𝑎𝑠𝑠𝑒𝑠 ∩
(𝑎𝑑𝑢𝑙𝑡 4 𝑝𝑎𝑠𝑠𝑒𝑠)]
= 𝑃(𝑎𝑑𝑢𝑙𝑡 1 𝑝𝑎𝑠𝑠𝑒𝑠) ∗ 𝑃(𝑎𝑑𝑢𝑙𝑡 2 𝑝𝑎𝑠𝑠𝑒𝑠) ∗ 𝑃(𝑎𝑑𝑢𝑙𝑡 3 𝑝𝑎𝑠𝑠𝑒𝑠) ∗
𝑃(𝑎𝑑𝑢𝑙𝑡 4 𝑝𝑎𝑠𝑠𝑒𝑠)
= (.1)(.1)(.1)(.1) = .1
So, P(SSSS) = .0001
4
= .0001
Another sample point.
P(FSSS)
= 𝑃[ 𝑎𝑑𝑢𝑙𝑡 1 𝑓𝑎𝑖𝑙𝑠 ∩ 𝑎𝑑𝑢𝑙𝑡 2 𝑝𝑎𝑠𝑠𝑒𝑠 ∩ 𝑎𝑑𝑢𝑙𝑡 3 𝑝𝑎𝑠𝑠𝑒𝑠
∩ (𝑎𝑑𝑢𝑙𝑡 4 𝑝𝑎𝑠𝑠𝑒𝑠)]
= 𝑃(𝑎𝑑𝑢𝑙𝑡 1 𝑓𝑎𝑖𝑙𝑠) ∗ 𝑃(𝑎𝑑𝑢𝑙𝑡 2 𝑝𝑎𝑠𝑠𝑒𝑠) ∗ 𝑃(𝑎𝑑𝑢𝑙𝑡 3 𝑝𝑎𝑠𝑠𝑒𝑠)
∗ 𝑃(𝑎𝑑𝑢𝑙𝑡 4 𝑝𝑎𝑠𝑠𝑒𝑠)
= .9 .1 .1 .1 = .9 .1
So, P(FSSS) = .0009
3
= .0009
4)
We wanted to obtain the sample point that none of the
four adults’ passed.
P(FFFF)
= 𝑃 𝑎𝑑𝑢𝑙𝑡 1 𝑓𝑎𝑖𝑙𝑠 ∗ 𝑃 𝑎𝑑𝑢𝑙𝑡 2 𝑓𝑎𝑖𝑙𝑠
∗ 𝑃 𝑎𝑑𝑢𝑙𝑡 3 𝑓𝑎𝑖𝑙𝑠 ∗ 𝑃 𝑎𝑑𝑢𝑙𝑡 4 𝑓𝑎𝑖𝑙𝑠
= (.9)(.9)(.9)(.9) = .9
So , P(FFFF) = .6561
4
= .6561
b) Probability that three of the four adults pass the
test. Looking at the sample points, we have four
sample points: FSSS, SFSS, SSFS, and SSSF.
P(3 of 4 adults pass)
= P(FSSS) + P(SFSS) + P(SSFS) + P(SSSF)
= .1 3 (.9) + .1 3 (.9) + .1 3 (.9) + .1 3 (.9)
= 4 .1 3 (.9)
= .0036
c.
I. We can characterize this experiment as a
consisting on four identical trials: the four test
results
II. There are two possible outcomes to each trail, S
of F.
III. The probability of passing , p = .1, is the same
for each trial.
IV. Assuming that each adult’s test result is
independent of all others, so that the four trials
are independent.
d. *Since time is limited in this class, we will not
derive a formula for p(x) in class. However, I
have attached it to the end of the PowerPoint.
Procedure
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Example 4.11
Refer to Example 4.10. use the formula for a
binomial random variable to find the probability
distribution of x, where x is the number of
adults who pass the fitness test. Graph the
distribution.
For this application, we have n = 4 trails.
Since a success S is defined as an adult who passes
the test,
p = P(S) = .1
and
q = 1- p = .9
So we have the following:
n=4
p = .1
q = .9
p(0) =
4
0
.1
0
.9
4−0
= 1 .1
0
.9
4
= .6561
p(1) =
4
1
.1
1
.9
4−1
= 4 .1
1
.9
3
= .2916
p(2) =
4
2
.1
2
.9
4−2
= 6 .1
2
.9
2
= .0486
p(3) =
4
3
.1
3
.9
4−3
= 4 .1
3
.9
1
= .0036
p(4) =
4
4
.1
4
.9
4−4
= 1 .1
4
.9
0
= .0001
Table 4.3
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Education, Inc.. All rights
reserved.
Figure 4.8 Probability distribution for physical
fitness example: graphical form
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Example 4.12
Refer to Examples 4.10 and 4.11 . Calculate
𝜇 and 𝜎, respectively, of the number of the four
adults who pass the test. Interpret the results.
We first need to find the mean of a discrete probability distribution
𝜇 = ∑𝑥𝑝(𝑥)
We must refer back to the table
𝜇 = 0(.6561) + 1(.2916) + 2(.0486) + 3(.0036) + 4(.0001)
= .4
Thus, in the long run, the average number of adults (out of four) who
pass the test is only .4.
The variance:
𝜎 2 = ∑ 𝑥 − 𝜇 2 𝑝 𝑥 = ∑ 𝑥 − .4 2 𝑝 𝑥
= 0 − .4 2 (0.6561) + 1 − .4 2 (.2916)
+ 2 – . 4 2 (.0486) + 3 − .4 2 (.0036)
+ 4 − .4 2 (.0001)
= .36
𝜎 = .36 = .6
Since the distribution is skewed right, we should
apply Chevyshev’s rule to describe where most of
the x-values fall. According to the rule, at least 75%
of the x values will fall into the interval 𝜇 ± 2𝜎 =
.4 ± 2 .6 which is −.8, 1.6 . Since x cannot be
negative, we expect the number of adults out of
four who pass the fitness test to be less than 1.6.
If we look at the 𝜇 and 𝜎 2 in relation to binomial
probability distribution ( that using n, p, and q),
you need not use the expectation summation
rule to calculate 𝜇 and 𝜎 2 for a binomial random
variable.
𝜇 = .4 = 4(.1) = 𝑛𝑝
𝜎 2 = .36 = 4(.1)(.9) = 𝑛𝑝𝑞
Definition
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Using Binomial Tables
Calculating binomial probabilities becomes
tedious when n is large. For some values of n
and p, the binomial probabilities have been
tabulated to Table II of Appendix A. The entries
in the table represent cumulative binomial
probabilities.
Example (put this on you table)
Let p = .10 and x = 2.
𝑃 𝑥 ≤ 2 = .930
Figure 4.9 Binomial probability distribution for
n=10 and p=.10, with P ( x  2) highlighted
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𝑃 𝑥=2
=𝑃 𝑥 ≤2 −𝑃 𝑥 ≤1
= .930 − .736 = .194
𝑃 𝑥>2
=1−𝑃 𝑥 ≤2
= 1 − .930 = .070
Example 4.13
Suppose a poll of 20 voters taken in a large city.
The purpose is to determine x, the number who
favor a certain candidate for mayor. Suppose
that 60% of all the city’s voters favor the
candidate.
a) Find the mean and the standard deviation of x.
b) Use Table II of Appendix A to find the probability
that x less than equal 10
c) Use Table II to find the probability x > 12
[P(x >12)]
d) Use Table II to find the probability that x = 11
[P(x =11)]
e) Graph the probability distribution of x, and
locate the interval 𝜇 ± 2𝜎 on the graph.
First,
n = 20
p = .60
q = .40
a)
𝜇 = 𝑛𝑝 = 20(.6) = 12
𝜎 2 = 𝑛𝑝𝑞 = 20(.6)(.4) = 4.8
𝜎 = 2.19
The 𝜇 = 12 and 𝜎 = 2.19
b)Find the probability that 𝑥 ≤ 10 [𝑃 𝑥 ≤ 10 ]
k = 10
p = .60
n = 20
𝑃 𝑥 > 10 = .245
c)Find the probability that 𝑥 > 12 [𝑃 𝑥 > 12 ]
𝑃 𝑥 > 12 = 1 − 𝑃 𝑥 ≤ 12
= 1 − .584
= .416
d)Find the probability that 𝑥 = 11 [𝑃 𝑥 = 11 ]
𝑃 𝑥 = 11
= 𝑃 𝑥 ≤ 11 − 𝑃 𝑥 ≤ 10
= .404 − .245 = .159
E. The probability distribution for x,
𝜇– 2𝜎 = 12 − 2(2.2) = 7.62
𝜇 + 2𝜎 = 12 + 2(2.2) = 16.38
Figure 4.10 The binomial probability
distribution for x in Example 4.13; n=20 and p=.6
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The probability that x falls into the interval
P(x = 8, 9, 10,…, 16)
= 𝑃 𝑥 ≤ 16 − 𝑃 𝑥 ≤ 7
=.984-.021
=.963
Note that his probability is very close to the .95
given by the empirical rule. Thus, we expect the
number of voters in the same of 20 who favor
mayoral candidate to between 8 and 16.