Answer:
We can conclude that on this case we have identical processes but excersise 17 use another way to present the probability distribution and as we can see the expected value can be viewed as a dot product of two vectors with one vector containing the outcomes and the other the probabilities for each possible outcome.
Step-by-step explanation:
Assuming this previous info:
Exercise 17. Suppose that we convert the table on the previous page displaying the discrete distribution for the number of heads occurring when two coins are flipped to two vectors.
Let vector
F(x) = 2^x; h(x) = x^3 + x + 8
Table
x f(x) = 2^x h(x) = x^3 + x + 8
0 2^0 = 1 0 + 0 + 8 = 8
1 2^1 = 2 1^3 + 1 + 8 = 10
2 2^2 = 4 2^3 + 2 + 8 = 8 + 2 + 8 = 18
3 2^3 = 8 3^3 + 3 + 8 = 27 + 3 + 8 = 38
4 2^2 = 16 4^3 + 4 + 8 = 76
10 2^10 = 1024 10^3 +10 + 8 = 1018
9 2^9 = 512 9^3 + 9 + 8 = 729 + 9 + 8 = 746
Answer: an approximate value of 10
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