Answer :
Answer:
The expected cost is 152
Step-by-step explanation:
Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y
[tex] f_Y(y ) = \frac{1}{5-1} = \frac{1}{4} [/tex] if [tex]1\leq y \leq 5[/tex] and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)
Recall that, by definition
[tex]E(Y^k) = \int_{-\infty}^{\infty} y^kf_Y(y)dy[/tex]
Also, we are given that [tex] C = 50+3Y+9Y^2[/tex]. Recall the following properties of the expected value. If X,Y are random variables, then
[tex]E(a+bX+cY) = a+bE(X)+cE(Y)[/tex]
Then, using this property we have that [tex]E[C] = 50+3E[Y]+ 9E[Y^2][/tex].
Thus, we must calculate E[Y] and E[Y^2].
Using the definition, we get that
[tex]E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3[/tex]
[tex]E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}[/tex]
Then
[tex] E(C) = 50 + 3\cdot 3 + 9 \cdot \frac{31}{3} = 152[/tex]