Homework Sourse7 If a discrete random variable X has the following probabil
7. If a discrete random variable X has the following probability distribution:
X -2 - 1 0 1 2
P(X) 0.2 0.3 0.15 0.2 0.15
Use this to find the following:
(a) The mean of X and E[X^2].
(b) The probability distribution for Y = 2X^2 + 2 (i.e, all values of Y and P(Y )).
(c) Using part (b) (i.e, the probability distribution forY ), find E[Y ].
(d) Using part (a), verify your answer in part (c) for E[Y ].
Solution
d) verified that in both ways E(y) is the same.
| X -2 - 1 0 1 2 | ||||||
| P(X) 0.2 0.3 0.15 0.2 0.15 | ||||||
| x | -2 | -1 | 0 | 1 | 2 | Total |
| p(x) | 0.2 | 0.3 | 0.15 | 0.2 | 0.15 | 1 |
| px | -0.4 | -0.3 | 0 | 0.2 | 0.3 | -0.2 |
| x^2p | 0.8 | 0.3 | 0 | 0.2 | 0.6 | 1.9 |
| E(X) = | -0.2 | |||||
| Var(x) = | 1.9-0.04= | 1.86 | ||||
| y = 2x^2+2 | ||||||
| x | -2 | -1 | 0 | 1 | 2 | Total |
| Y | 10 | 4 | 2 | 4 | 10 | |
| p(y) | 0.2 | 0.3 | 0.15 | 0.2 | 0.15 | 1 |
| yp(y) | 2 | 1.2 | 0.3 | 0.8 | 1.5 | 5.8 |
| E(Y) | 5.8 | (from table) | ||||
| E(Y) = | E(2x^2+2) = | 2E(x^2)+2 | ||||
| 5.8 |
