>Output
inomial distribution
. 0 0.0 56 375
2 38
75
0000
00
rmal distribution
z dev
,400
X mean 44
7,340 istributed Random Numbers
1 462
Uniformly Distributed Random Numbers Uniform 615
.37234 037
072
(n = 6, p = ) 0.0 1.0 2.0 3.0 0.015625 0.09375 0.234375 0.3125 0.234375 0.09375 0.0156 25 X 3 -1 -2 -3 f(z) z -1.28 3 ounting Probability – combinations order does not count- in permutations order counts. This following shows how many combinations one permutations there are of 2 given 4 things.
A-B A-C A-D B-C B-D C-D 6 12 ions, how many arrangements of new businesses locate there?
27 43 36 79 es shown here correspond to your experience? (d) Why might
ii. P(W) iii. P(S | W)
30 120 24 50 6 n cumulative 1 0.09375 0.10938 3 0.65625 expected value The mean annual income of workers in a particular industry is $49,400 with a standard deviation of $7340.
P(lower) P(upper) z X mean std.dev .0743 1.44 60,000 49,400 7,340 .9257 .0743 1.44 60,000 49,400 7,340 .7256 -0.60 45,000 49,400 7,340 one is research exactly. Then one must determine the appropriate sample size for the research. The sampling must be designed so everyone in the population as an equal chance of being selected (the procedures are Random, Cluster, Stratified, and Systematic – we will do an exercise using random selection)
9 dog 2.3877465129 grizzly 1 cat 29.6666839123 bird 2 panda 77.6992698908 elephant grizzly cat dog &P of &N
Binomial distribution (n = 6, p = 0.5)
2
B
6
n
0
5
p
cumulative
X
P(X)
probability
1
3
0.015
63
1
0.0
9
0.10938
0.23
4
0.3
43
3
0.3
12
50
0.65625
4
0.23438
0.89063
5 0.09375
0.98438
6
0.01563
1.0
1.00000
3.0
expected value
1.500
variance
1.225
standard deviation
3.0
No
P(lower)
P(upper)
X
mean
st
d.
.92
57
.0743
1.44
60,000
49
7,340
2.0
Normal distribution
P(lower) P(upper)
z
std.dev
.
27
.7256
-0.60
45,000
49,400
Uniformly
D
Minimum
Maximum
100
Uniform
70.84920
36
Minimum 1
Maximum 100
53.80897
79
58
30
29.6666839123
30.8928530812
77.6992698908
2.3877465129
76.3116354942
81.6345
120
71.1947520971
Binomial distribution
0.5
4.0
5.0
6.0
P(X)
2
1
0
1.44
2
1
0
-1
-2
-3
f(z)
z
-0.60Sheet1
Week 3 – Probabilities
First we cover counting probabilities
C
Things
Combinations
Permutations
A
A-B
B-A
B
A-C
C-A
C
A-D
D-A
D
B-C
C-B
B-D
D-B
C-D
D-C
totals
1.0
If a new shopping center has seven lo
cat
Use Factorial to do.
use factorial
2.0
If seven business want to locate in a new shopping center but there are only five locations, how many
arrangements could there be?
use permutations
3.0
If seven bossiness want to locate in a new shopping center but there are only five locations, how many
different combinations could be selected?
use combinations
Discrete Probabilities: Concept and Binomial
Probability must between 0 and 1; total of all probabilities is 1
Concept is simply “The Specific Question” divided by the total possibilities (Excel for these)
Example
Married
Not Married
B
-B
Sport car
14
41
Not sport car
57 63 120
Question: What is probability of being married = 57/120
Question: What is the probability of having a sports car and being married = 14/120
Question: What is the probability of having a spots car or being married =(41+57-14)/120
Another old example: Refer to the contingency table shown below. (a) Calculate each probability (i–vi) and explain in
words what it means. (b) Do you see evidence that smoking and race are not independent?
Explain. (c) Do the smoking
rat
public health officials be interested in this type of data? (Data are from Statistical Abstract of the
United States, 2001, pp. 12 and 16. Note: Actual statistics are applied to a hypothetical sample of
1,000.) Smoking2
i. P
(S)
iv. P(S | B) v. P(S and W) vi. P(N and B)
Smoking by Race for Males Age 18-
24
Smoker
Nonsmoker
Row Total
(S)
(N)
White (W)
290
560
850
Black (B)
150
Column Total
320
680
1000
a.-i
320/1000 =
probability of a smoker
a.-ii
850/1000=
probability of a White
a.iii
290/850=
probability of a While being a smoker
a.iv
30/150=
probability of a Black being a smoker
a.v
290/1000=
probability of selecting a White and being a smoker
a.vi
120/1000=
probability of selecting a Black and being a nonsmokeroer
b.
The P of smoking and white is .29 and the P(S)*P(W) =.32*.85=
The P of smoking and Black is .03 and the P(B)*P(S)=.32*.15 =
d.
If smoking is dependent on race, then health officials might target special
programs based on race.
4.0
Your exercise to do
Size of company
Small
Medium
Large
Plan MBA?
Yes
105
179
No 49 58 100
207
73
108
205
386
P(Yes) =
P(Yes | Medium) =
P(Yes and Large) =
P(Small and Yes) =
P(Medium or Yes) =
P(Small) =
P(Medium | Yes) =
Now Binomial (MegaStat for these)
Example
What is the probability of observing two heads is a coin is tossed six times?
n=6
P/head = .5
Use MegaStat to compute
Results
Binomial distribution
0.5 p
X P(X) probability
0 0.01563 0.01563
2
0.34375
Note the
Probability of throwing 2 heads is .23438
0.31250
4 0.23438 0.89063
5 0.09375 0.98438
6 0.01563 1.00000
1.000003.000
1.500 variance
1.225 standard deviation 5.0
Your turn:
A basketball team that shoots free throws at 80.5% shoots 14 free throws in their next game.
What is the Probability they will be successful in 11 of them?
and what is the probability they will be successful in at least 9 of them?
Now we do what is the very basis of much of statistics – six sigma -lean. The Normal Distribution
Example:
Assume a normal distribution.
A. What is the probability a worker would have an income below $60,000?
Use MegaStat for this
Normal distribution
.9257
B. What is the probability a worker would have an income above $60,000?
.0743
C. What is the probability a worker would make between $45,000 and $60,000
.2744
0.6512
6.0
Your turn to do the exercise
Assume a college entrance exam has a mean of 1000 and a standard deviation of 100. Assume a normal distribution.
A. What is the probability the student’s score would be below 1000?
B. What is the probability the student’s score would be below 850?
C. What is the probability the student’s score would b above 1250″
D. What it the probability the student’s score would be between 850 and 1250?
E. What would be the student’s score to have only 5% of the students taking the test have higher >
Now sampling. Sampling is the basis of nearly all medical improvements and used in all industries
Sampling requires that one Identifies the
Population
Example – what is the required sample size for a customer with a .035 error and 95% confidence.
Answer:
784
7.0
Your turn:
A. Customer states he wants a 3% error with 95% confidence. What is the require sample size
B. What is the require sample size for a study have an error of 1,000 whose standard deviation is 7,800
and a 90% confidence level.
Now we randomly select
Example – random selection
Random select 5 from the following “population”.
Population
dog
N count of population =
cat
bird
lion
elephant
grizzly
polar bear
panda
turtle
Sort on Uniform
Uniform Population Uniform Population
53.8089779615
58.3723430037
29.6666839123 bird 30.8928530812 lion 3
30.8928530812 lion 53.8089779615 dog 4
77.6992698908 elephant 58.3723430037 cat 5
2.3877465129 grizzly 71.1947520971 turtle
76.3116354942 polar bear 76.3116354942 polar bear
81.6345120072
71.1947520971 turtle 81.6345120072 panda
select top 5
bird
lion
dog
8.0
Your turn.
Randomly select 7 from the following – show all your steps as in the example
cat
bird
lion
elephant
grizzly
polar bear
panda
turtle
mouse
rat
squirrel
rabbit
skunk
And we end the week with confidence interval
Example:
A local retailer wants to know the 95% confidence interval of the proportion of customers who use a credit card. A
random sample of 142 receipts showed that 83 used a credit card.
95% confidence interval .503 to .666
9.0
Your Turn
A. Compute the 95% confidence interval of those people choosing chocolate ice cream. 12 of 27 customer
chose a favor other than chocolate.
B. Compute the 90% confidence interval when the mean is 12,000, the standard deviation is 500 and the N count is 86
0.0 1.0 2.0 3.0 4.0 5.0 6.0 0.015625 0.09375 0.234375 0.3125 0.234375 0.09375 0.0156 25 X
P(X)Sheet2
Sheet3