Multinomial Experiments. What if there are more than 2 possible outcomes? (e.g., acceptable, scrap, rework) That is, suppose we have: n independent trials k outcomes that are mutually exclusive (e.g., ♠, ♣, ♥, ♦) exhaustive (i.e., ∑ all k p i = 1) Then
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f(x1, x2, …, xk; p1, p2, …, pk, n) =
EGR 252.001 Spring 2011
EGR 252.001 Spring 2011
EGR 252.001 Spring 2011
EGR 252.001 Spring 2011
k = number of “successes” = 2 n = number in sample = 5
N = the lot size = 10 x = number found = 1 or 2
P(X > 1) = 0.556 + 0.222 = 0.778
EGR 252.001 Spring 2011
μ = nk/N = 5*2/10 = 1
σ2 =(5/9)(5*2/10)(1-2/10) = 0.444
σ =0.667
EGR 252.001 Spring 2011
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:
P(X>2) = 1 – [P(0)+P(1)]
EGR 252.001 Spring 2011
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?
(Compare to example 5.14, pg. 129)
EGR 252.001 Spring 2011
Historical data indicates that 30% of all bits transmitted through a digital transmission channel are received in error. An engineer is running an experiment to try to classify these errors, and will start by gathering data on the first 10 errors encountered.
What is the probability that the 10th error will occur on the 25th trial?
EGR 252.001 Spring 2011
Where,
k = “success number” = 10
x = trial number on which k occurs = 25
p = probability of success (error) = 0.3
q = 1 – p= 0.7
EGR 252.001 Spring 2011
k = “success number” = 10
x = trial number on which k occurs = 25
p = probability of success (error) = 0.3
q = 1 – p= 0.7
EGR 252.001 Spring 2011
In our example, what is the probability that the 1st bit received in error will occur on the 5th trial?
= (0.3)(0.7)4 = 0.072
EGR 252.001 Spring 2011
A worn machine tool produces 1% defective parts. If we assume that parts produced are independent:
EGR 252.001 Spring 2011
EGR 252.001 Spring 2011
EGR 252.001 Spring 2011
An average of 2.7 service calls per minute are received at a particular maintenance center. The calls correspond to a Poisson process. To determine personnel and equipment needs to maintain a desired level of service, the plant manager needs to be able to determine the probabilities associated with numbers of service calls.
EGR 252.001 Spring 2011
Where,
λ= average number of outcomes per unit time or region
t = time interval or region
EGR 252.001 Spring 2011
P(X < 2) =P(X = 0) + P(X = 1)
μ = λt =________________
EGR 252.001 Spring 2011
P(X > 6) = 1 – P(X <6) = 1 - 0.3134 = 0.6866
EGR 252.001 Spring 2011
EGR 252.001 Spring 2011
The effect of λ on the Poisson distribution
EGR 252.001 Spring 2011