Download Chapter 17 Notes filled in

Ch. 17: Binomials
Example: Computer chips have a 25% chance of being defective. Create the probability distribution for X, if X is
the # of defective chips in a sample of 3. What is the probability of having 2 or more defective chips?
D
.25
.25
.25
.75
D
.75
D
.25
c
.25
.75
D
.25
D
c
D
D
.75
.25
.75
.75
Dc
.25
P(X)
Dc
D
Dc
D
.75
X
Dc
Dc
0
1
.4219
.4218
2
.1407
3
.0156
(a) What is the probability of having 2 or more defective chips?
P( X  2)  .1563
(b) What is the probability of having 1 or less defective chips?
P( X  1)  .8437
(c) What is the probability of having exactly 2 defective chips?
P( X  2)  .1407
BINOMIAL MODELS:
 Interested in the number of successes in a set number of trials
 4 conditions that must apply:
o Only 2 possible outcomes (success/failure)
o Probability of success remains constant (called p)
o Number of trials is set/known (called n)
o Independent trials
 10% Condition: If we cannot assume independence, we can proceed as long as the
sample is smaller than 10% of the population (pop  10n)
 If these 4 conditions apply, we have a Bernoulli trial
Notation: B(n, p)
probability of success
µX= n * p = E(x)
σX=
n * p *(1  p)  n * p * q
(1-p)
Example: It is known that only 15% of the population is left handed. Create a probability distribution for
the number of left handed people in a sample of 3.
X
P(X)
0
.6141
1
.3251
2
.0574
3
.0034
Quicker way to get probabilities:
n
k 
Formula: P(X = k) =   p k (1  p)( n k )
combinations n Ck
X
P(X)
C0 (.15)0 (.85)3  .6141
0
3
1
3
2
3
C2 (.15) 2 (.85)1  .0574
3
3
C3 (.15)3 (.85)0  .0034
C1 (.15)1 (.85) 2  .3251
Example: I am playing a game in which I have a 39% chance of winning each time I play. Create the probability
distribution for the number of wins out of 5 plays of the game.
STEP 1: Check if the problem is binomial
1. Two outcomes: Winning/Not Winning
2. P = .39 and remains constant (q = .61)
3. Number of trials is set at 5
4. Independence is not stated but we can assume independence since more than 50 games will be
played.
STEP 2: Create the probability distribution
X
P(X)
0
5
C0 (.39) (.61)5  .0845
1
5
2
5
C2 (.39)2 (.61)3  .3452
3
5
C3 (.39)3 (.61) 2  .2207
4
5
C4 (.39)4 (.61)1  .0706
5
5
C5 (.39)5 (.61)0  .0090
0
C1 (.39)1 (.61) 4  .2700
STEP 3: answer questions
P(X=2) = .3452
P(X<2) = .3545
P(X≥3) = .3003
P(2≤X≤4) = .6365
Now let’s look at changing the sample size to 10, and answering similar questions:
X
P(X)
0
10
P(X=9) = .0013
 .0071
10 C0 (.39) (.61)
0
C1 (.39)1 (.61)9  .0456
1
10
2
10
C2 (.39)2 (.61)8  .1312
3
10
C3 (.39)3 (.61)7  .2237
4
10
C4 (.39)4 (.61)6  .2503
5
10
C5 (.39)5 (.61)5  .1920
6
6
4
10 C6 (.39) (.61)  .1023
7
10
C7 (.39)7 (.61)3  .0374
8
10
C8 (.39)8 (.61) 2  .0090
9
10
10
P(X≥6) = .15008
P(5≤X≤7) = .3317
C9 (.39) (.61)  .0013
9
10
P(X<4) = .4076
1
C0 (.39)10 (.61)0  .00008
Would you want to answer these questions for a sample size of 50? Of 100? NO! So we can use the calculator!
For P(X=k)
 Use binompdf (n, p, k)

k = the number you are looking for…
Example: P(X = 5) …k = 5

pdf = probability distribution function
(gives each individual outcome’s probability)
For P(X≤k)
 Use binomcdf (n, p, k)

k = the number you are looking for…
Example: P(X  5)…k = 5

Notice that is ONLY GIVES YOU: _LESS THAN OR EQUAL TO____

cdf = cumulative distribution function
(adds up all the probabilities below and equal to an outcome)
Ex: P( X  3)  P( X  0)  P( X  1)  P( X  2)  P( X  3)
Example: John is taking archery. He has a 30% chance of hitting the target each time he shoots. He shoots 8
times
1. Two outcomes: Hit/Miss
2. p = .30 and is constant (q=.70)
3. Number of trials is set at 8
4. Independence can be assumed because we can assume that he could take more than 80 shots so we
are sampling less than 10 % of the population
B(8, .30)
1) What is the probability that he hits the target 4 times?
P( X  4)  .1361
2) What is the probability that he hits the target 2 times or less?
P( X  2)  .5518
3) What is the probability that he hits the target at least 3 times?
1  P( X  2)  .4482
4) What is the probability that he hits the target less than 5 times?
P( X  4)  .9420
5) What is the probability that he hits the target more than 6 times?
1  P( X  6)  .0013
6) How many times do we expect him to hit the target? (average!)
E ( x)   x  (8)(.3)  2.4 hits
7) What is the standard deviation of the number of times he hits the target?
 x  (8)(.3)(.7)  1.296 hits
Try this example on your own:
150 businesses are sent mailings asking them to answer a survey question and send the mailing back. The
probability of nonresponse is 55%.
1. Two outcomes: Response/Non-Response
2. p = .45 and is constant (q=.55)
3. Number of trials is set at 150
4. Independence can be assumed because we can assume that there are more than 1500 businesses so
we are sampling less than 10% of the population
B(150, .45)
1) What is the average number of businesses that WILL respond?
E ( x)  x  (150)(.45)  67.5 businesses
2) What is the standard deviation of the number of businesses that WILL respond?
 x  (150)(.45)(.55)  6.093 businesses
3) What is the probability that 75 businesses will respond?
P( X  75)  .0306
4) What is the probability that 60 businesses or less will respond?
P( X  60)  .1251
5) What is the probability that 60 businesses or more will respond?
1  P( X  59)  .9058
6) What is the probability that less than 60 businesses will respond?
P( X  59)  .0942
7) What is the probability that greater than 60 businesses will respond?
1  P( X  60)  .8749
8) What number of surveys would you have to send out if you wanted to be able to expect to get 90 back?
90  n(.45)
n  200
9) What is the probability that between 50 and 70 businesses will respond?
P( X  50)  .0024
P( X  69)  .6295
.6295-.0024=.6271
EXPERIMENT:
I have a 35% chance of winning a game. Let X be the # of wins if I play 4 times.
Create the Probability Distribution
X
P(X)
0
.1785
1
.3845
2
.3105
3
.1115
4
.0150
Sketch a histogram (rough sketch!):
Prob. Distribution Directions:
 Put the x-values in L1
 Go to the TOP of L2
 Type in binompdf(n, p, L1) = binompdf(4, 0.35, L1)
 Be sure that you don’t type in L1, but you put in the
name of L1 from your list names.
 Hit ENTER
 This will give you each probability for the table at left
Histogram Directions:
 Go to STAT PLOT
 Go into Plot 1, and make it a histogram
 Under X-list, put L1
 Under Freq, put L2
 Hit Zoom 9
 You will probably need to adjust your
Window. Make your min the lowest X
value (0); make your max 1 more than
that highest X value (5); make your bar
width 1.
Describe the shape only of the histogram:
Right skewed
Now let’s change the sample size to 10. Create the distribution, then the histogram. (Use the same directions as
above, but with n=10)
Distribution:
Histogram:
X
P(X)
0
.0135
1
.0725
2
.1757
3
.2522
4
.2377
5
.1536
6
.0689
7
.0212
8
.0043
9
.0005
10
.00003
Describe the shape of the histogram:
Roughly symetric
Now let’s change the sample size to 20. Create the distribution in your calculator, then the histogram.
Histogram:
Shape:
Roughly symmetric
Try it with a sample size of 30. Just create the distribution and histogram in your calculator. What is the shape
of this histogram??
Roughly symmetric
What do you notice about the shape of the distribution as the sample size increases? We have a specific NAME
for these types of distributions…. (unimodal, bell shaped, symmetric… starts with an “N”)
As the sample size increase, the distribution becomes more Normal.
Ch. 17: Probability Models: Binomial Random Variables: LARGE SAMPLE SIZE
What happens to the shape of the Binomial Random Variable when n is large?
The shape goes from being skewed (small sample size) to being Normal (large sample size)
What is considered a “large enough” n (for the shape to look normal)??
n * p  10
n * q  10
So if the check passes…
 We can say that the distribution is _apprx. Normal______, and can use _normalcdf(__!

Calculator: normalcdf(lower bound, upper bound, mean, std. dev)

Same mean and std dev. that we learned before:
N ( , )
 X  np
 X  np(1  p)
Example 1:
It is said that 75% of people pay their credit card bill on time. If we take a sample of 125 adults, what is the
chance that over 80 of them paid their bill on time this past month?
Bernoulli:
1. Two outcomes: On time/Not on time
2. p = .75 and is constant (q = .25)
3. Number of trials is set at 125
4. 10% condition: We can assume independence since there are more than 1250 adults with credit cards
Check:
(125)(.75)  10
(125)(.25)  10
B(125, .75)
N(93.75, 4.841)
1  P( X  80)  .9977
Example 2:
Assume 15% of adults jog regularly. We survey 1,000 adults what is the probability that between 120 and 160
people in our sample jogs?
Bernoulli:
1. Two outcomes: Jog regurlary/Don’t jog regurlarly
2. p = .15 and is constant (q = .85)
3. Number of trials is set at 1,000
4. 10% condition: We can assume independence since there are more than 10,000 adults that jog
Check:
(1000)(.15)  10
(1000)(.85)  10
B(1000, .15)
N(150, 11.292)
P(120  X  160)  .8081