Answer:
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
. Now to solve, we will, plug it's value in the second equation.
