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: 
For <em>X</em> = {0, 1, 2, 3} the value of <em>Y</em> are:
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:
Thus, the value of P (Y ≤ 2) is 0.60.
(e)
The value of E (Y) is 7.85.
Use the values of E (X) and E (X²) computed in part (c) to compute the value of E (Y).
Hence verified.
