Probability Question - Random Variable

An appliance dealer sells three different models of upright freezers having 13.5, 15.9 and 19.1 cubic feet of storage space, respectively. Let X = the amount of storage space purchased by the next customer to by a freezer. Suppose that X has pdf

X.............13.5 ......... 15.9 ........ 19.1
P(X)........ 0.2 ......... 0.5 ........ 0.3

i- Compute E(X), E(X2), and VAR(X)

ii- If the price of a freezer having a capacity X cubic feet is 2.5X - 8.5, what is the expected price paid by the next customer to buy a freezer?

iii. What is the variance of the price 2.5X – 8.5?

iv. Suppose that while the rated capacity of a freezer is X, the actual capacity is
h(X)=X – 0.01X2. What is the expected actual capacity of the freezer purchased by the next customer?


Thanks in advance ;***

just use formulas appropriately
1.
E(X) = 13.5*0.2 + 15.9*0.5 + 19.1*0.3 = 16.38
E(X2) = 13.5^2 * 0.2 + 15.9^2*0.5 + 19.1^2 * 0.3 = 272.298
V(X) = E(X2)‐E(X)2= 272.298 – 16.382= 3.993

2.
E[2.5X‐8.5] = 2.5*E[X]‐8.5 = 2.5*16.38‐8.5 = 7.45

3.
V(2.5X‐8.5) = 2.5^2 * V(X) = 24.95625

4.
E[X‐0.01X2] = E[X] – 0.01*E[X^2] = 16.38 – 0.01*272.298 = 13.66

i) E(x)=𝚺xP(x)

=(13.5)(0.2)+(15.9)(0.5)+(19.1)(0.3)=16…

Add X^2 to the table

X^2............182.25 ........252.81 .....364.81
X^2P(X)...... 36.45 .....126.404 .......109.443

E(x^2)=𝚺X^2P(X)=272.297

Var(x)=𝚺X^2P(X)- [E(x)]^2

=272.297 - (16.38)^2=3.993

ii)E(y) =2.5*E(x) -8.5=2.5(16.38)- 8.5=32.45

iii) Addition does not affect variance,so

Var(y)=2.5*var(x)=(2.5)^2*3.993=24.956

iv)
X.............13.5 ......... 15.9 .......1519.1

h(X)........ 11.7 .........13.6 ........ 15.5


E(x)=𝚺h(x)P(x)

=(11.7)(0.2)+(13.6)(0.5)+(15.5)(0.3)
=13.79

i. I'm assuming "E(x)" is the expected value, or the sample mean as I would refer to it. If so, you can find this by taking the sum of the product of each individual x-value and it's respective probability. With that said, let's take a look: