Answer:
(a) The probability distribution is shown in the attachment.
(b) The value of E (<em>Y</em>) is 7.85.
(c) The value of E (X) and E (X²) are 1.45 and 3.25 respectively.
(d) The value of P (Y ≤ 2) is 0.60.
(e) Verified that the value of E (Y) is 7.85.
Step-by-step explanation:
(a)
The random variable <em>Y</em> is defined as: ![Y=3X^{2}-2X+1](https://tex.z-dn.net/?f=Y%3D3X%5E%7B2%7D-2X%2B1)
For <em>X</em> = {0, 1, 2, 3} the value of <em>Y</em> are:
![X=0;\ Y=3\times(0)^{2}-2\times(0)+1 =1](https://tex.z-dn.net/?f=X%3D0%3B%5C%20Y%3D3%5Ctimes%280%29%5E%7B2%7D-2%5Ctimes%280%29%2B1%20%3D1)
![X=1;\ Y=3\times(1)^{2}-2\times(1)+1 =2](https://tex.z-dn.net/?f=X%3D1%3B%5C%20Y%3D3%5Ctimes%281%29%5E%7B2%7D-2%5Ctimes%281%29%2B1%20%3D2)
![X=2;\ Y=3\times(2)^{2}-2\times(2)+1 =9](https://tex.z-dn.net/?f=X%3D2%3B%5C%20Y%3D3%5Ctimes%282%29%5E%7B2%7D-2%5Ctimes%282%29%2B1%20%3D9)
![X=3;\ Y=3\times(3)^{2}-2\times(3)+1 =22](https://tex.z-dn.net/?f=X%3D3%3B%5C%20Y%3D3%5Ctimes%283%29%5E%7B2%7D-2%5Ctimes%283%29%2B1%20%3D22)
The probability of <em>Y</em> for different values are as follows:
P (Y = 1) = P (X = 0) = 0.20
P (Y = 2) = P (X = 1) = 0.40
P (Y = 9) = P (X = 2) = 0.15
P (Y = 22) = P (X = 3) = 0.25
The probability distribution of <em>Y</em> is shown below.
(b)
The expected value of a random variable using the probability distribution table is:
![E(U)=\sum[u\times P(U=u)]](https://tex.z-dn.net/?f=E%28U%29%3D%5Csum%5Bu%5Ctimes%20P%28U%3Du%29%5D)
Compute the expected value of <em>Y</em> as follows:
![E(Y)=\sum [y\times P(Y=y)]\\=(1\times0.20)+(2\times0.40)+(9\times0.15)+(22\times0.25)\\=7.85](https://tex.z-dn.net/?f=E%28Y%29%3D%5Csum%20%5By%5Ctimes%20P%28Y%3Dy%29%5D%5C%5C%3D%281%5Ctimes0.20%29%2B%282%5Ctimes0.40%29%2B%289%5Ctimes0.15%29%2B%2822%5Ctimes0.25%29%5C%5C%3D7.85)
Thus, the value of E (<em>Y</em>) is 7.85.
(c)
Compute the expected value of <em>X</em> as follows:
![E(X)=\sum [x\times P(X=x)]\\=(0\times0.20)+(1\times0.40)+(2\times0.15)+(3\times0.25)\\=1.45](https://tex.z-dn.net/?f=E%28X%29%3D%5Csum%20%5Bx%5Ctimes%20P%28X%3Dx%29%5D%5C%5C%3D%280%5Ctimes0.20%29%2B%281%5Ctimes0.40%29%2B%282%5Ctimes0.15%29%2B%283%5Ctimes0.25%29%5C%5C%3D1.45)
Compute the expected value of <em>X</em>² as follows:
![E(X^{2})=\sum [x^{2}\times P(X=x)]\\=(0^{2}\times0.20)+(1^{2}\times0.40)+(2^{2}\times0.15)+(3^{2}\times0.25)\\=3.25](https://tex.z-dn.net/?f=E%28X%5E%7B2%7D%29%3D%5Csum%20%5Bx%5E%7B2%7D%5Ctimes%20P%28X%3Dx%29%5D%5C%5C%3D%280%5E%7B2%7D%5Ctimes0.20%29%2B%281%5E%7B2%7D%5Ctimes0.40%29%2B%282%5E%7B2%7D%5Ctimes0.15%29%2B%283%5E%7B2%7D%5Ctimes0.25%29%5C%5C%3D3.25)
Thus, the value of E (X) and E (X²) are 1.45 and 3.25 respectively.
(d)
Compute the value of P (Y ≤ 2) as follows:
![P (Y\leq 2)=P(Y=1)+P(Y=2)=0.20+0.40=0.60](https://tex.z-dn.net/?f=P%20%28Y%5Cleq%202%29%3DP%28Y%3D1%29%2BP%28Y%3D2%29%3D0.20%2B0.40%3D0.60)
Thus, the value of P (Y ≤ 2) is 0.60.
(e)
The value of E (Y) is 7.85.
![E(Y)=E(3X^{2}-2X+1)=3E(X^{2})-2E(X)+1](https://tex.z-dn.net/?f=E%28Y%29%3DE%283X%5E%7B2%7D-2X%2B1%29%3D3E%28X%5E%7B2%7D%29-2E%28X%29%2B1)
Use the values of E (X) and E (X²) computed in part (c) to compute the value of E (Y).
![E(Y)=3E(X^{2})-2E(X)+1\\=(3\times 3.25)-(2\times1.45)+1\\=7.85](https://tex.z-dn.net/?f=E%28Y%29%3D3E%28X%5E%7B2%7D%29-2E%28X%29%2B1%5C%5C%3D%283%5Ctimes%203.25%29-%282%5Ctimes1.45%29%2B1%5C%5C%3D7.85)
Hence verified.