b:head_first_statistics:using_discrete_probability_distributions
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| b:head_first_statistics:using_discrete_probability_distributions [2023/10/04 10:18] – [Theorem 5: Var(X+Y)] hkimscil | b:head_first_statistics:using_discrete_probability_distributions [2024/10/01 21:44] (current) – [Fat Dan changed his prices] hkimscil | ||
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| Line 281: | Line 281: | ||
| __Pool puzzle__ | __Pool puzzle__ | ||
| + | |||
| + | \begin{eqnarray*} | ||
| + | \text {original win} & = & \text {c(0, 5, 10, 15, 20)} \\ | ||
| + | \text {original cost} & = & \text {c(1)} \\ | ||
| + | \text {X} & = & \text {the amount of money you receive} \\ | ||
| + | \end{eqnarray*} | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| Line 295: | Line 301: | ||
| E(X) = -.77 and E(Y) = -.85. What is 5 * E(X) + 3? | E(X) = -.77 and E(Y) = -.85. What is 5 * E(X) + 3? | ||
| - | $ 5 * E(X) = -3.85 $ | + | \begin{eqnarray*} |
| - | $ 5 * E(X) + 3 = -0.85 $ | + | E(X) & = & -.77 \\ |
| - | $ E(Y) = 5 * E(X) + 3 $ | + | E(Y) & = & E(5X+3) \; |
| + | & = & E(5X) + 3 \\ | ||
| + | & = & 5 E(X) + 3 \\ | ||
| + | & = & -0.85 \\ | ||
| + | \\ | ||
| + | \\ | ||
| + | Var(Y) & = & Var(5X+3) \\ | ||
| + | & = & Var(5X) + Var(3) \\ | ||
| + | & = & Var(5X) \\ | ||
| + | & = & 5^2 Var(X) \\ | ||
| + | & = & 67.4275 | ||
| + | \end{eqnarray*} | ||
| - | $ 5 * Var(X) = 13.4855 $ | ||
| - | $ 5^2 * Var(X) = 67.4275 $ | ||
| - | $ Var(Y) = 5^2 * Var(X) $ | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| Line 375: | Line 388: | ||
| </ | </ | ||
| - | 그런데 | + | 위는 각각 계산한 것. 아래는 Y = X - 0.5를 적용해서 계산한 것 |
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| E(X-a) & = & E(X) - a \\ | E(X-a) & = & E(X) - a \\ | ||
b/head_first_statistics/using_discrete_probability_distributions.1696382326.txt.gz · Last modified: 2023/10/04 10:18 by hkimscil