b:head_first_statistics:using_discrete_probability_distributions
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b:head_first_statistics:using_discrete_probability_distributions [2024/09/29 23:30] – [Fat Dan changed his prices] 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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\begin{eqnarray*} | \begin{eqnarray*} | ||
- | \text {original win} & = & text {c(0, 5, 10, 15, 20)} \\ | + | \text {original win} & = & \text {c(0, 5, 10, 15, 20)} \\ |
- | \text {original cost} & = & text {c(1)} \\ | + | \text {original cost} & = & \text {c(1)} \\ |
- | \text {X} & = & text {the amount of money you receive} \\ | + | \text {X} & = & \text {the amount of money you receive} \\ |
\end{eqnarray*} | \end{eqnarray*} | ||
Line 301: | 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 381: | 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.1727620246.txt.gz · Last modified: 2024/09/29 23:30 by hkimscil