The probability that a wildcat well will be productive is 1/13. Assume that a gr

Question

The probability that a wildcat well will be productive is 1/13. Assume that a group is drilling wells in various parts of the country so that the status of one well has no bearing on that of any other. Let X denote the number of wells drilled to obtain the first strike. Verify that X is geometric, and identify the value of the parameter p. What is the exact expression for the density for X? What is the exact expression for the moment generating function for X? What are the numerical values of E[X], E[X^2], sigma^2, and sigma? Find P[X greaterthanorequalto 2].

Explanation / Answer

(a) X = number of wells to be drilled to obtain the first strike.

so X has a geometric here as it has a geomtric pattern of probability.

Value of parameter p = 1/13

(b) Density expression for X; f(x) = (1-p)x-1(p) = (12)X-1 (1/13)X

(c) MOment genrating function Mx (t) = p et / [ 1 - (1-p)et] where p = 1/13

(d) E(X) = 1/p = 1/(1/13) = 13

E [X2 ] = (2-p)/p2 = (2-1/13) / (1/13)2 = 325

2 = E [X2 ] - [ E(X)]2 = 325 - 169 = 156

= sqrt (156) = 12.49

(e) P( X >=2) = 1 - P(1) = 1 - (12)0 /13 = 1 - 1/13 = 12/13 = 0.923

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