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# STAT 10-17-16

### Geometrics (Test 1)

TermDefinition
Continuity correction factor use P(x<(number .5))
Let x~b(50, .3) np = 15, n(1-p) = 35, both are large so normal approximation should work well
X~b, P(x < 18) P(x <= 17.5)
X~b, P(x > 10) P(x >= 10.5)
X~b, P(9 < x <= 16) P(9.5 <= x <= 16.5)
X~b, P(x <= 17) P(x <= 17.5)
X~b, P(x >= 11) P(x >= 10.5)
X~b, P(11 <= x < 18) P(11.5 <= x <= 17.5)
Geometric distribution binomial experiment independently repeating Bernoulli trials where we don’t know how many we perform before starting
Geometric random variable number of trials it takes to get first success
Possible values of geometric distribution 1,2,3,...
X~g(p), P(x = 1) P
X~g(p), P(x = 2) (1-p)p
X~g(p), P(x = 3) (1-p)^2 * p
X~g(p), P(x = k) (1-p)^k-1 * p
Excel geometric distribution no specific function, but there is a generalization
Negative binomial random variable number of independent Bernoulli trials needed to get the rth success
Geometric binomial negative binomial with r = 1
X~g(.8), P(x <= 1) P(x = 1) = negbinom.dist(0,1,.8,0)
X~g(.8), P(x < 3) P(x <= 2) = negbinom.dist(1,1,.8,1)
X~g(.8), P(x > 3) 1 - P(x <= 3) = negbinom.dist(2,1,.8,1)
X~g(.8), P(4 <= x <= 6) P(x <= 6) - P(x <= 3) = negbinom.dist(5,1,.8,1) - negbinom.dist(2,1,.8,1)
X~g(.6), P(x >= 2) 1 - P(x < 2) = 1 - P(x <= 1) = 1 - .6 = .4
X~g(.6), P(x < 5) P(x <= 4) = negbinom.dist(3,1,.6,1)
X~g(.6), P(x <= 4) negbinom.dist(3,1,.6,1)
X~g(.6), P(3 <= x < 9) P(x <= 8) - P(x <= 2) = negbinom.dist(7,1,.6,1) - negbinom.dist(1,1,.6,1)
Mean: If x~g(p), then μ = 1 / p
Variance: If x~g(p), then σ = (1-p)/p^2
Created by: Spencer Gowey