Section 4.2 (cont.) Binomial Random Variables

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Binomial Random Variables
Binomial Probability Distributions
Binomial Random
Variables
Through 2/25/2014 NC State’s free-throw
percentage is 65.1% (315th out 351 in Div. 1).
 If in the 2/26/2014 game with UNC, NCSU
shoots 11 free-throws, what is the probability
that:

NCSU makes exactly 8 free-throws?
NCSU makes at most 8 free throws?
NCSU makes at least 8 free-throws?
“2-outcome” situations are very
common
Heads/tails
 Democrat/Republican
 Male/Female
 Win/Loss
 Success/Failure
 Defective/Nondefective

Probability Model for this Common
Situation

Common characteristics
◦ repeated “trials”
◦ 2 outcomes on each trial

Leads to Binomial Experiment
Binomial Experiments

n identical trials
◦ n specified in advance

2 outcomes on each trial
◦ usually referred to as “success” and “failure”
p “success” probability; q=1-p “failure”
probability; remain constant from trial to
trial
 trials are independent

Classic binomial experiment: tossing a
coin a pre-specified number of times
Toss a coin 10 times
 Result of each toss: head or tail (designate one
of the outcomes as a success, the other as a
failure; makes no difference)
 P(head) and P(tail) are the same on each toss
 trials are independent

◦ if you obtained 9 heads in a row, P(head) and P(tail)
on toss 10 are same as P(head) and P(tail) on any
other toss (not due for a tail on toss 10)
Binomial Random Variable
The binomial random variable X is
the number of “successes” in the n
trials
 Notation: X has a B(n, p)
distribution, where n is the number
of trials and p is the success
probability on each trial.

Examples
a.
b.
c.
d.
Yes; n=10; success=“major repairs within
3 months”; p=.05
No; n not specified in advance
No; p changes
Yes; n=1500; success=“chip is defective”;
p=.10
Binomial Probability Distribution
n trials, p  success probability on each tri al
probability distribution:
p(x)  n C x p q
x
nx
, x  0,1, 2,
n
E ( x) 

x0
n
xp ( x ) 

,n
xx p q
n
x0


V ar ( x )  E  ( x      npq
x
nx
 np
Rationale for the Binomial
Probability Formula
P(x) =
n!
•
(n – x )!x!
Number of
outcomes with
exactly x
successes
among n trials
px •
n-x
q
Binomial Probability
Formula
P(x) =
n!
•
(n – x )!x!
Number of
outcomes with
exactly x
successes
among n trials
px •
n-x
q
Probability of x
successes
among n trials
for any one
particular order
Graph of p(x); x binomial n=10 p=.5;
p(0)+p(1)+ … +p(10)=1
The sum of all the
areas is 1
Think of p(x) as the area
of rectangle above x
p(5)=.246 is the area
of the rectangle above 5
Binomial Probability Histogram: n=100, p=.5
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
70
68
66
64
62
60
58
56
54
52
50
48
46
44
42
40
38
36
34
32
30
0
Binomial Probability Histogram: n=100, p=.95
0.18
0.17
0.16
0.15
0.14
0.13
0.12
0.11
0.1
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
70
72
74
76
78
80
82
84
86
88
90
92
94
96
98
100
Example
A production line produces motor
housings, 5% of which have cosmetic
defects. A quality control manager
randomly selects 4 housings from the
production line. Let x=the number of
housings that have a cosmetic defect.
Tabulate the probability distribution for x.
Solution
(i) D=defective, G=good
outcome
x
P(outcome)
GGGG
0
(.95)(.95)(.95)(.95)
DGGG
1
(.05)(.95)(.95)(.95)
GDGG
1
(.95)(.05)(.95)(.95)
:
:
:
DDDD
4
(.05)4

Solution
( ii ) x is a binom ial random variable
p(x)  n C x p q
x
nx
, x  0,1, 2,
,n
n  4, p  .05 ( q  .95)
p (0)  4 C 0 (.05) (.95)  .815
0
4
p (1)  4 C 1 (.05) (.95)  .171475
1
3
p (2)  4 C 2 (.05) (.95)  .01354
2
2
p (3)  4 C 3 (.05) (.95)  .00048
3
1
p (4)  4 C 4 (.05) (.9 5)  .00000625
4
0
Solution
x 0
p(x) .815
1
2
.171475 .01354
3
4
.00048 .00000625
Example (cont.)
x
0
p(x) .815
1
2
.171475 .01354
3
.00048
4
.00000625
What is the probability that at least 2 of
the housings will have a cosmetic defect?
P(x  p(2)+p(3)+p(4)=.01402625

Example (cont.)
x
0
p(x) .815
1
2
.171475
.01354
3
4
.00048 .00000625
What is the probability that at most 1
housing will not have a cosmetic defect?
(at most 1 failure=at least 3 successes)
P(x  3)=p(3) + p(4) = .00048+.00000625 =
.00048625

Using binomial tables; n=20, p=.3
9, 10, 11, … , 20
P(x  5) = .416
 P(x > 8) = 1- P(x  8)= 1- .887=.113
=P(x 8)
 P(x < 9) = ? 8, 7, 6, … , 0
 P(x  10) = ? 1- P(x  9) = 1- .952
 P(3  x  7)=P(x  7) - P(x  2)
.772 - .035 = .737

Binomial n = 20, p = .3 (cont.)
P(2 < x  9) = P(x  9) - P(x  2)
= .952 - .035 = .917
 P(x = 8) = P(x  8) - P(x  7)
= .887 - .772 = .115

Color blindness
The frequency of color blindness (dyschromatopsia) in the
Caucasian American male population is estimated to be
about 8%. We take a random sample of size 25 from this population.
We can model this situation with a B(n = 25, p = 0.08) distribution.
 What
is the probability that five individuals or fewer in the sample are color blind?
Use Excel’s “=BINOMDIST(number_s,trials,probability_s,cumulative)”
P(x ≤ 5) = BINOMDIST(5, 25, .08, 1) = 0.9877
 What
is the probability that more than five will be color blind?
P(x > 5) = 1  P(x ≤ 5) =1  0.9877 = 0.0123
 What
is the probability that exactly five will be color blind?
P(x = 5) = BINOMDIST(5, 25, .08, 0) = 0.0329
x
P (X = x) P (X < = x)
0
1
2
1 2 .4 4 %
2 7 .0 4 %
2 8 .2 1 %
1 2 .4 4 %
3 9 .4 7 %
6 7 .6 8 %
30%
3
4
1 8 .8 1 %
9 .0 0 %
8 6 .4 9 %
9 5 .4 9 %
25%
5
6
7
8
3 .2 9 %
0 .9 5 %
0 .2 3 %
0 .0 4 %
9 8 .7 7 %
9 9 .7 2 %
9 9 .9 5 %
9 9 .9 9 %
9
10
11
0 .0 1 %
0 .0 0 %
0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
12
13
0 .0 0 %
0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
5%
14
15
0 .0 0 %
0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
0%
16
17
18
0 .0 0 %
0 .0 0 %
0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
19
20
0 .0 0 %
0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
21
22
0 .0 0 %
0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
Probability distribution and histogram for
23
24
0 .0 0 %
0 .0 0 %
1 0 0 .0 0 %
1 0 0 .0 0 %
the number of color blind individuals among
25
0 .0 0 %
1 0 0 .0 0 %
B(n = 25, p = 0.08)
15%
24
22
20
18
16
14
12
10
8
6
4
2
10%
0
P(X = x)
20%
Number of color blind individuals (x)
25 Caucasian males.
What are the expected value and standard deviation of the
count X of color blind individuals in the SRS of 25 Caucasian
American males?
E(X) = np = 25*0.08 = 2
SD(X) = √np(1  p) = √(25*0.08*0.92) = 1.36
What if we take an SRS of size 10? Of size 75?
E(X) = 10*0.08 = 0.8
E(X) = 75*0.08 = 6
SD(X) = (75*0.08*0.92)=2.35
0.5
0.2
0.4
0.15
0.3
p = .08
n = 10
0.2
0.1
P(X=x)
P(X=x)
SD(X) = √(10*0.08*0.92) = 0.86
p = .08
n = 75
0.1
0.05
0
0
0
1
2
3
4
5
Number of successes
6
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Number of successes
Recall Free-throw
question


Through 2/25/14 NC State’s
free-throw percentage was
65.1% (315th in Div. 1).
If in the 2/26/14 game with
UNC, NCSU shoots 11 freethrows, what is the probability
that:
1.
2.
3.
NCSU makes exactly 8
free-throws?
NCSU makes at most 8
free throws?
NCSU makes at least 8
free-throws?
1.
n=11; X=# of made
free-throws; p=.651
p(8)= 11C8 (.651)8(.349)3
=.226
2. P(x ≤ 8)=.798
3.
P(x ≥ 8)=1-P(x ≤7)
=1-.5717 = .4283
Recall from beginning of Lecture
Unit 4: Hardee’s vs The Colonel
Out of 100 taste-testers, 63 preferred
Hardee’s fried chicken, 37 preferred KFC
 Evidence that Hardee’s is better? A
landslide?
 What if there is no difference in the
chicken? (p=1/2, flip a fair coin)
 Is 63 heads out of 100 tosses that unusual?

Use binomial rv to analyze
n=100 taste testers
 x=# who prefer Hardees chicken
 p=probability a taste tester chooses
Hardees
 If p=.5, P(x  63) = .0061 (since the
probability is so small, p is probably NOT .5;
p is probably greater than .5, that is,
Hardee’s chicken is probably better).

Recall: Mothers Identify
Newborns





After spending 1 hour with their newborns,
blindfolded and nose-covered mothers were asked
to choose their child from 3 sleeping babies by
feeling the backs of the babies’ hands
22 of 32 women (69%) selected their own newborn
“far better than 33% one would expect…”
Is it possible the mothers are guessing?
Can we quantify “far better”?
Use binomial rv to analyze
n=32 mothers
 x=# who correctly identify their own baby
 p= probability a mother chooses her own baby
 If p=.33, P(x  22)=.000044 (since the probability
is so small, p is probably NOT .33; p is probably
greater than .33, that is, mothers are probably not
guessing.

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