Respuesta :
Answer:
a) [tex]\alpha = 3, \beta = 2[/tex]
b) 0.0620
Step-by-step explanation:
We are given the following in the question:
Population mean, [tex]\mu[/tex] = 6
Variance, [tex]\sigma^2[/tex] = 12
a) Value of [tex]\alpha, \beta[/tex]
We know that
[tex]\alpha \beta = \mu = 6\\\alpha \beta^2 = \sigma^2 = 12[/tex]
Dividing the two equations, we get,
[tex]\dfrac{\alpha\beta^2}{\alpha\beta} = \dfrac{12}{6}\\\\\Rightarrow \beta = 2\\\alpha \beta = 6\\\Rightarrow \alpha = 3[/tex]
b) probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.
We can write the probability density function as:
[tex]f(x,3,2) = \dfrac{1}{2^{3}(3-1)!}x^{3-1}e^{-\frac{x}{2}}, x > 0\\\\f(x,3,2) = \dfrac{1}{16}x^{2}e^{-\frac{x}{2}}, x > 0[/tex]
We have to evaluate:
[tex]P(x >12)\\\\= \dfrac{1}{16}\displaystyle\int^{\infty}_{12}f(x)dx\\\\=\dfrac{1}{16}\bigg[-2x^2e^{-\frac{x}{2}}-2\displaystyle\int xe^{-\frac{x}{2}}dx}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[x^2e^{-\frac{x}{2}}+4xe^{-\frac{x}{2}}+8e^{-\frac{x}{2}}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[(\infty)^2e^{-\frac{\infty}{2}}+4(\infty)e^{-\frac{\infty}{2}}+8e^{-\frac{\infty}{2}} -( (12)^2e^{-\frac{12}{2}}+4(12)e^{-\frac{12}{2}}+8e^{-\frac{12}{2}})\bigg]\\\\=0.0620[/tex]
0.0620 is the required probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.
