HW 26 Suppose that random variable X has a geometric distrib

HW 2.6. Suppose that random variable X has a geometric distribution with a mean of 2.5. Determined the following: (a) P(X = 1) (b) P(X = 4) (c) P(X = 5) (d) P(X 3) (f) V(X)

Solution

For a geomtric distribution,

p = 1/mean = 1/2.5 = 0.4

Also,

P(x) = (1 - p)^(x- 1) p

a)

P(x =1) = (1-0.4)^(1-1) (0.4) = 0.4 [answer]

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b)

P(x = 4) = (1-0.4)^(4-1) (0.4) = 0.0864 [answer]

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c)

P(x = 5) = (1-0.4)^(5-1) (0.4)^5 = 0.05184 [answer]

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d)

P(x <=3) = P(1) + P(2) + P(3)

= (1-0.4)^(1-1) (0.4) + (1-0.4)^(2-1) (0.4) + (1-0.4)^(3-1) (0.4)

= 0.784 [answer]

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