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 strikes.

a. Verify that X is geometric, and identify the value of the parameter p.

b. What is the exact expression for the density for X?

c. what is the exact expression for the moment generating function for X?

d. What are the numerical values of E[x], E[x2], \sigma 2, and \sigma ?

e. Find P[X>=2]

a) p = 1 / 13

b) f(x) = ( 12 / 13 ) ^(n-1) * 1 / 13

c) M(x) =  1/13 / ( 1 - (12/13)*e^t)  

d) E(X) = 13 ,  E(X^2) =  325 , Var (X) = 156 , S.d = 12.49

e)  P(X >= 2) = 12/13

Step-by-step explanation:

Given:

- The probability that a wildcat well is productive p = 1/13

Find:

- 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[x2], \sigma 2, and \sigma ?

- Find P[X>=2]

Solution:

- Declaring a random variable X is the number of wells drilled to obtain the first strikes.

                                     X ~ Geo ( 1 / 13 )

- The probability of success is independent from successive trials. Where X denotes the number of successive trials till there is a success. Hence, the parameter p = 1 / 13.

- The probability density function of the geometric distribution for number f trails till first success is given by:

                               f(x) = ( 1 - p ) ^(n-1) * p

                               f(x) = ( 12 / 13 )^(n-1) * 1 / 13

- The moment generating expression for a Geometric distribution is given by:

                              M(x) =  p / ( 1 - (1-p)*e^t)  

                              M(x) =  1/13 / ( 1 - (12/13)*e^t)  

- The expected value E(X) of a geometric function is given by:

                              E(X) = 1 / p

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

                              E(X) = 13

Where,

                              Var(X) = ( 1 - p ) / p^2

                              Var(X) = ( 12/13 )*13^2

                              Var(X) = 156  

                               S.d = sqrt(156) = 12.49

We know,

                              Var(X) = E(X^2) - [ E(X) ]^2

                               E(X^2) =  Var(X) + [ E(X) ]^2

                               E(X^2) =  156 + 13^2

                              E(X^2) =  325

- The required probability of P(X >= 2 ) can be computed using f(x)

                              P(X >= 2 ) = 1 - f(1)

                             P(X >= 2 ) = 1 - ( 12 / 13 ) ^(1-1) * 1 / 13

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

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl