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Hypergeometric DistributionPowerPoint Presentation

Hypergeometric Distribution

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Hypergeometric Distribution

- Example*:
Automobiles arrive in a dealership in lots of 10. Five out of each 10 are inspected. For one lot, it is know that 2 out of 10 do not meet prescribed safety standards.

What is probability that at least 1 out of the 5 tested from that lot will be found not meeting safety standards?

*from Complete Business Statistics, 4th ed (McGraw-Hill)

- This example follows a hypergeometric distribution:
- A random sample of size n is selected without replacement from N items.
- k of the N items may be classified as “successes” and N-k are “failures.”

- The probability associated with getting x successes in the sample (given k successes in the lot.)
Where,

k = number of “successes” = 2 n = number in sample = 5

N = the lot size = 10 x = number found

= 1 or 2

Hypergeometric Distribution

- In our example,
= _____________________________

Expectations of the Hypergeometric Distribution

- The mean and variance of the hypergeometric distribution are given by
- What are the expected number of cars that fail inspection in our example? What is the standard deviation?
μ =___________

σ2 =__________ , σ =__________

Your turn …

A worn machine tool produced defective parts for a period of time before the problem was discovered. Normal sampling of each lot of 20 parts involves testing 6 parts and rejecting the lot if 2 or more are defective. If a lot from the worn tool contains 3 defective parts:

- What is the expected number of defective parts in a sample of six from the lot?
- What is the expected variance?
- What is the probability that the lot will be rejected?

Binomial Approximation

- Note, if N >> n, then we can approximate this with the binomial distribution. For example:
Automobiles arrive in a dealership in lots of 100. 5 out of each 100 are inspected. 2 /10 (p=0.2) are indeed below safety standards.

What is probability that at least 1 out of 5 will be found not meeting safety standards?

- Recall: P(X≥ 1) = 1 – P(X< 1) = 1 – P(X = 0)

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