(a)
a is got by noting that the Total Probability = 1
Thus,
a + 0.25 + 0.10 + 0.05 = 1
So,
a = 1 - 0.40 = 0.60
Thus, pdf of Y is given by:
Y | P(Y) |
0 | 0.6 |
1 | 0.25 |
2 | 0.10 |
3 | 0.05 |
The cdf of Y is given by:
F(Y) = 0 for y <0
= 0.6 for 0 y < 1
= 0.85 for 1 y < 2
= 0.95 for 2 y < 3
= 1 for y 3
(c)
y | p | yp | y2 p |
0 | 0.60 | 0 | 0 |
1 | 0.25 | 0.25 | 0.25 |
2 | 0.10 | 0.20 | 0.40 |
3 | 0.05 | 0.15 | 0.45 |
Total | 0.60 | 1.10 |
So,
E(Y) = 0.60
E(Y2) = 1.10
So,
E(Y + 100 Y2) = E(Y) + 100 E(Y2)
=0.60 + (100 X 1.10) = 110.60
So,
Answer is:
110.60
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