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Statistics Quick Overview

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Statistics Quick Overview

Class #2

1

2

3

- Given that a consumer has looked at a green shirt 5 times
- What is the probability that they will buy?
- What is the probability that they will buy if you give them 5% off?
- What is the probability that they will buy if you give them 10% off?
- What is the probability given that they are a new customer/existing customer?

- Given that a consumer has bought a flashlight
- What is the probability that they will buy batteries?
- What is the probability that they will buy a 2nd flashlight?

Solution space = 1

A and B

Solution space = 1

A and B

(75%)

= 90% + 80% - 75%=95%

1-95% = 5% chance neither will start

Acura

Starts Doesn’t

Starts

Doesn’t

80%

75%

5%

BMW

20%

15%

5%

10%

90%

Solution space = 1

A and B

(75%)

= 75% / 80% = 93.75%

Acura

Starts Doesn’t

Normalize this row:

75% / 80%

Starts

Doesn’t

80%

75%

5%

BMW

20%

15%

5%

10%

90%

Right

Center

Left

- Let’s start with hockey…
- Here, order matters
- The person on the left must stay on the left
- The person on the right must stay on the right

- So, how many different potential line-ups does Mr. Brown have to consider?
- Choices are: Mr Blonde, Mr White, Mr Orange, Mr Pink, and Mr Blue

5 x 4 x 3 = 60

Where n is the number of choices, and k is the number picked. In Excel, this is PERMUT(n,k)

Note: Each column is a unique combination of players

Note: The entries within each row are different permutations of the players. This is our same problem again where n= 3 and k = 3

==6

- Here, order does not matter
- He just needs a front line

- All that matters is the number of unique combinations
- What observation from the permutation table helps us determine the unique combinations

When calculating the permutations, we naturally determined the unique combinations (the columns) and then ran the permutations for each combination. If we divide out that last step, we will have just the combinations:

= 60 / 6=10

- Sample Size of 10
- Case #1: Assume that the lot is good with 5% defectives
- When will you reject because you find 3 or more defectives

- Case #2: Assume that the lot has 40% defectives
- When will you accept because you find 2 or less defectives

- Let’s assume:
- s is the probability of success
- f is the probability of failure

- Example of getting 3 Failures
- fssfsssssf
- Probability of this is (5%)3(95%)7

- Example of getting 4 Failures
- fssfsfsssf
- Probability of this is (5%)4(95%)6

- What are we missing?

The number of combinations

- A1 uses drugs P(A1) = 5%
- A2 does not use drugs P(A2) = 95%
- B tests shows drug use
- P(B | A1) = 98%
- P(B | A2) = 2%

What we want….

=

= 5%*(98%) / ((5%*98%)+(95%*2%)) =72%