>Output B

inomial

distribution 6 n 0

.

5 p cumulative X P(X) probability0 0.0

156

3 0.015 63 1 0.0 9375 0.10938

2

0.23 438 0.3

4375 3

0.3 12 50 0.65625 4 0.23438 0.89063 5 0.09375 0.98438 6 0.01563 1.00000 1.00000 3.0

00
expected value
1.500
variance
1.225
standard deviation
3.0

No

rmal distribution P(lower) P(upper)

z

X mean st d.dev .92

57 .0743 1.44 60,000 49,400 7,340 2.0 Normal distribution P(lower) P(upper)

zX mean

std.dev . 2744 .7256 -0.60 45,000 49,400

7,340

Uniformly Distributed Random Numbers Minimum

1

Maximum
100
Uniform
70.84920

462

Uniformly Distributed Random Numbers

Minimum 1

Maximum 100

Uniform

53.80897 79615 58

.37234

30037 29.6666839123 30.8928530812 77.6992698908 2.3877465129 76.3116354942 81.6345

120072 71.1947520971

Binomial distribution

(n = 6, p =

0.5)

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)

3

2

1

0

-1

-2

-3

f(z)

z

1.44

-1.28

3

2

1

0

-1

-2

-3

f(z)

z

-0.60

## Sheet1

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

BA-C

CA-D

DB-C

B-D

C-D

6 12

ions, how many arrangements of new businesses locate there?

2.0 3.0 B27

43 36 79

es shown here correspond to your experience? (d) Why might

ii. P(W) iii. P(S | W)

(S)30 120

d. 4.024 50

No 49 58 100 ExampleBinomial distribution

6 n

0.5 pcumulative

X P(X) probability0 0.01563 0.01563

1 0.09375 0.10938

23

0.65625

4 0.23438 0.89063

5 0.09375 0.98438

6 0.01563 1.00000

1.00000

expected value

1.500 variance

1.225 standard deviation

The mean annual income of workers in a particular industry is $49,400 with a standard deviation of $7340.

Normal distributionP(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)

Population 9

Uniform Population Uniform Population

dog 2.3877465129 grizzly 1

cat 29.6666839123 bird 2

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

panda 77.6992698908 elephant

grizzly

bird

lion

cat

dog

cat

bird

lion

elephant

grizzly

polar bear

panda

turtle

Example:

Week 3 – Probabilities | |||||

First we cover counting probabilities | |||||

C | |||||

Things | Combinations | Permutations | |||

A | A-B | B-A | |||

A-C | C-A | ||||

A-D | D-A | ||||

B-C | C-B | ||||

B-D | D-B | ||||

C-D | D-C | ||||

totals | |||||

If a new shopping center has seven lo | cat | ||||

Use Factorial to do. | use factorial | ||||

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 | ||||

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 | |||||

Sport car | 14 | 41 | |||

Not sport car | |||||

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 | |||

(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 = | |||||

If smoking is dependent on race, then health officials might target special | |||||

programs based on race. | |||||

Your exercise to do | |||||

Size of company | |||||

Small | Medium | Large | |||

Plan MBA? | Yes | 105 | 179 | ||

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) | |||||

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 | |||||

0.34375 | Note the | Probability of throwing 2 heads is .23438 | |||

0.31250 | 3.000 | ||||

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 | |||||

.9257 | |||||

B. What is the probability a worker would have an income above $60,000? | |||||

C. What is the probability a worker would make between $45,000 and $60,000 | |||||

.2744 | |||||

0.6512 | |||||

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”. | ||||

dog | N count of population = | ||||

bird | |||||

lion | |||||

elephant | |||||

grizzly | |||||

polar bear | |||||

panda | |||||

turtle | |||||

Sort on Uniform | |||||

53.8089779615 | |||||

58.3723430037 | |||||

81.6345120072 | |||||

select top 5 | |||||

8.0 | Your turn. | Randomly select 7 from the following – show all your steps as in the example | |||

mouse | |||||

squirrel | |||||

rabbit | |||||

skunk | |||||

And we end the week with confidence interval | |||||

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 |

&P of &N

Binomial distribution (n = 6, p = 0.5)

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)