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--- b:head_first_statistics:poisson_distribution [2025/10/07 06:52] – [Broken Cookies case] hkimscil
+++ b:head_first_statistics:poisson_distribution [2025/10/07 08:42] (current) – [Poisson Distribution] hkimscil
@@ Line 46: / Line 46: @@
 \end{eqnarray*}
+<code>
+> dpois(3, 2)
+[1] 0.180447
+>
+</code>
 ====== What does the Poisson distribution look like? ======
@@ Line 353: / Line 359: @@
 </WRAP>
+====== Exercise ======
+<WRAP box>
+Here are some scenarios. Your job is to say which distribution each of them follows, say what the expectation and variance are, and find any required probabilities.
+. A man is bowling. The probability of him knocking all the pins over is 0.3. If he has 10 shots, what’s the probability he’ll knock all the pins over less than three times?
+Binomial distribution 을 이용한다면,
+\begin{eqnarray*}
+X & \sim & B(n, p) \\
+X & \sim & B(10, 0.3)
+\end{eqnarray*}
+\begin{eqnarray*}
+E(X) & = & np \\
+& = & 10 * 0.3 \\
+& = & 3
+\end{eqnarray*}
+\begin{eqnarray*}
+Var(X) & = & npq \\
+& = & 10 * 0.3 * 0.7 \\
+& = & 2.1
+\end{eqnarray*}
+r을 이용한다면 ''pbinom'' 혹은 ''dbinom'' 을 이용한다.
+<code>
+> pbinom(q=2, 10, 0.3)
+[1] 0.3827828
+>
+> sum(dbinom(0:2, 10, 0.3))
+[1] 0.3827828
+>
+</code>
+손으로 계산을 한다고 하면,
+$P(X=0), P(X=1), P(X=2)$를 구한 후 모두 더하여 P(X < 3)을 구한다.
+\begin{eqnarray*}
+P(X = 0) & = & {10 \choose 0} * 0.3^0 * 0.7^{10} \\
+& = & 1 * 1 * 0.028 \\
+& = & 0.028
+\end{eqnarray*}
+\begin{eqnarray*}
+P(X = 1) & = & {10 \choose 1} *0.3^1 * 0.7^9 \\
+& = & 10 * 0.3 * 0.04035 \\
+& = & 0.121
+\end{eqnarray*}
+\begin{eqnarray*}
+P(X = 2) & = & {10 \choose 2} * 0.3^2 * 0.7^8 \\
+& = & 45 * 0.09 * 0.0576 \\
+& = & 0.233
+\end{eqnarray*}
+\begin{eqnarray*}
+P(X<3) & = & P(X=0) + P(X=1) + P(X=2) \\
+& = & 0.028 + 0.121 + 0.233 \\
+& = & 0.382
+\end{eqnarray*}
+</WRAP>
+<WRAP box>
+. On average, 1 bus stops at a certain point every 15 minutes. What’s the probability that __<fc #ff0000>no buses</fc>__ will turn up in a single 15 minute interval?
+위는 Poisson distribution 문제이므로 기대값과 분산값은 각각 lambda 값인 1 (15분마다 1대씩 버스가 온다고 한다)
+\begin{eqnarray*}
+P(X=0) & = & \frac {e^{-1}{1^0}}{0!} \\
+& = & \frac {e^{-1} * 1}{1} \\
+& = & .368
+\end{eqnarray*}
+</WRAP>
+<code>
+> dpois(0, 1)
+[1] 0.3678794
+>
+> ppois(0, 1)
+[1] 0.3678794
+>
+</code>
+<WRAP box>
+. 20% of cereal packets contain a free toy. What’s the probability you’ll need to open fewer than 4 cereal packets before finding your first toy?
+이는 geometric distribution 문제이므로,
+$$X \sim Geo(.2)$$
+$P(X \le 3)$ 을 구하는 문제이므로
+\begin{eqnarray*}
+P(X \le 3) & = & 1 - q^r \\
+& = & 1 - 0.8^{3} \\
+& = & 1 - 0.512 \\
+& = & 0.488
+\end{eqnarray*}
+<code>
+> sum(dgeom(0:2,0.2))
+[1] 0.488
+>
+> pgeom(2, 0.2)
+[1] 0.488
+>
+</code>
+기대값과 분산은 각각 $1/p$, $q/p^2$ 이므로 $5$와 $20$.
+</WRAP>