Respuesta :
Answer:
(a) The distribution of X is Binomial distribution.
(b) The probability of obtaining at least 3 oil strikes if 8 wells are dug is 0.0381.
(c) The probability that 8 or fewer wells will strike oil is 0.9876.
Step-by-step explanation:
Let random variable X = number of exploratory oil well drilled in a certain region should strike oil.
(a)
The probability of successful strike is, P (X) = p = 0.10.
The number of wells dug is, n = 8.
The outcome of the random variable X are:
- Successful strike.
- Unsuccessful strike.
The event of striking oil at one drilling location is independent of success at another location.
The random variables satisfies all the properties of a binomial random variable.
Thus, the distribution of X is Binomial distribution.
(b)
The probability function of a Binomial distribution is:
[tex]P(X=x) ={n\choose x}p^{x}(1-p)^{n-x};\ x=01,2,3,...[/tex]
Compute the probability of obtaining at least 3 oil strikes if 8 wells are dug as follows:
P (X ≥ 3) = 1 - P (X < 3)
= 1 - P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
[tex]=1-{8\choose 0}(0.10)^{0}(1-0.10)^{8-0}-{8\choose 1}(0.10)^{1}(1-0.10)^{8-1}\\-{8\choose 2}(0.10)^{2}(1-0.10)^{8-2}\\=1-0.4305-0.3826-0.1488\\=0.0381[/tex]
Thus, the probability of obtaining at least 3 oil strikes if 8 wells are dug is 0.0381.
(c)
A Poisson distribution is used to approximate the Binomial distribution when the following conditions are satisfied:
[tex]np\leq 10\\n\geq 20\\P(Success)\ is\ small[/tex]
The sample size is, n = 40.
The P (Success) = p = 0.10 (small)
Check the conditions as follows:
[tex]np=40\times0.10=4<10[/tex]
n = 40 > 20.
Thus, a Poisson distribution can be used to approximate the Binomial distribution of the random variable X.
The random variable X thus follows a Poisson distribution with parameter [tex]\lambda=np=4[/tex].
The probability function of a Poisson distribution is:
[tex]P(X=x)=\frac{e^{-\lamda}\lambda^{x}}{x!};\ x=0,1,2,...[/tex]
Compute the probability that 8 or fewer wells will strike oil as follows:
P (X ≤ 8) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 8)
[tex]=\frac{e^{-4}(-4)^{0}}{0!}+\frac{e^{-4}(-4)^{1}}{1!}+\frac{e^{-4}(-4)^{2}}{2!}+...+\frac{e^{-4}(-4)^{8}}{8!}\\=0.0183+0.0733+0.1465+...+0.0298\\=0.9786[/tex]
Thus, the probability that 8 or fewer wells will strike oil is 0.9876.