b:head_first_statistics:poisson_distribution
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| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| b:head_first_statistics:poisson_distribution [2025/10/06 21:57] – [Exercise] hkimscil | b:head_first_statistics:poisson_distribution [2025/10/06 23:42] (current) – [Poisson Distribution] hkimscil | ||
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| Line 46: | Line 46: | ||
| \end{eqnarray*} | \end{eqnarray*} | ||
| + | < | ||
| + | > dpois(3, 2) | ||
| + | [1] 0.180447 | ||
| + | > | ||
| + | |||
| + | </ | ||
| ====== What does the Poisson distribution look like? ====== | ====== What does the Poisson distribution look like? ====== | ||
| Line 394: | Line 400: | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| - | P(X = 0) & = & {10 \choose 0} * 0.3^0 * 0.7^10 \\ | + | P(X = 0) & = & {10 \choose 0} * 0.3^0 * 0.7^{10} \\ |
| & = & 1 * 1 * 0.028 \\ | & = & 1 * 1 * 0.028 \\ | ||
| & = & 0.028 | & = & 0.028 | ||
| Line 425: | Line 431: | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| P(X=0) & = & \frac {e^{-1}{1^0}}{0!} \\ | P(X=0) & = & \frac {e^{-1}{1^0}}{0!} \\ | ||
| - | & = & \frac {e^-1 * 1}{1} \\ | + | & = & \frac {e^{-1} * 1}{1} \\ |
| & = & .368 | & = & .368 | ||
| \end{eqnarray*} | \end{eqnarray*} | ||
| </ | </ | ||
| + | |||
| + | < | ||
| + | > dpois(0, 1) | ||
| + | [1] 0.3678794 | ||
| + | > | ||
| + | > ppois(0, 1) | ||
| + | [1] 0.3678794 | ||
| + | > | ||
| + | </ | ||
| <WRAP box> | <WRAP box> | ||
| Line 445: | Line 460: | ||
| & = & 0.488 | & = & 0.488 | ||
| \end{eqnarray*} | \end{eqnarray*} | ||
| + | |||
| + | < | ||
| + | > sum(dgeom(0: | ||
| + | [1] 0.488 | ||
| + | > | ||
| + | > pgeom(2, 0.2) | ||
| + | [1] 0.488 | ||
| + | > | ||
| + | |||
| + | </ | ||
| + | |||
| 기대값과 분산은 각각 $1/p$, $q/p^2$ 이므로 $5$와 $20$. | 기대값과 분산은 각각 $1/p$, $q/p^2$ 이므로 $5$와 $20$. | ||
| </ | </ | ||
b/head_first_statistics/poisson_distribution.1759787833.txt.gz · Last modified: by hkimscil