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--- b:head_first_statistics:measuring_central_tendency [2020/09/15 13:48] – [e.g.] hkimscil
+++ b:head_first_statistics:measuring_central_tendency [2024/09/09 09:50] (current) – [mean] hkimscil
@@ Line 21: / Line 21: @@
 이것을 우리는 흔히 "무"라고 부른다.
 \begin{equation*}
-\mu = \frac{\sum\limits_{i=1}^{n} X_i}{n}
+\mu = \frac{\sum\limits_{i=1}^{N} X_i}{N}
 \end{equation*}
@@ Line 31: / Line 31: @@
 \begin{equation*}
 \begin{split}
-\mu = \frac{\sum\limits_{}^{} \text{fx}}{\sum{\text{f}}}
+\mu = \frac{\sum\limits_{}^{} \text{fx}}{\sum{\text{f.nothing}}}
     = \frac{1 \text{x} 19 + 3 \text{x} 20 + 1 \text{x} 21}{5}
 \end{split}
@@ Line 43: / Line 43: @@
 ===== skewedness =====
 <WRAP info>
-[{{skewedness.jpg}}]
+skewedness: 왜도
+{{skewedness.jpg}}
 </WRAP>
@@ Line 64: / Line 65: @@
 two <- c(1,4,6,6,8,8,8,9,9,9,9,10,10,10,10,11,11,11,11,11,12,12,12,12,12)
 </code>
+<code>
+> median(one)
+[1] 3
+> mean(one)
+[1] 3.44
+> median(two)
+[1] 10
+> mean(two)
+[1] 9.28
+</code>
 <code>
@@ Line 80: / Line 94: @@
 abline(v=mean(one), col="blue", lty=1)
 abline(v=median(one), col="red", lty=2)
-legend(6.5,9.5, legend=c("mean", "median"), col=c("blue", "red"), lty=1:2, cex=.8)
+legend(3.8,9.5, legend=c("mean", "median"), col=c("blue", "red"), lty=1:2, cex=.8)
 hist(two)
 abline(v=mean(two), col="blue", lty=1)
@@ Line 87: / Line 101: @@
 par(mfrow=c(1,1))
 </code>
+[{{eg.finding.central.tendency.with.skewedness.data.histogram2.png}}]
-<code>
-> median(one)
-[1] 3
-> mean(one)
-[1] 3.44
-> median(two)
-[1] 10
-> mean(two)
-[1] 9.28
-</code>
 ==== e.g. 1 ====
-quartile 명령어는 r에서는 없다. 대신 quantile을 사용합니다. 둘은 비슷하나 quantile은 4분위 이상의 것을 한다.
+quartile 명령어는 r에서는 없다. 대신 quantile을 사용한다. 둘은 비슷하나 quantile은 4분위 이상의 것을 한다.
 R에서
@@ Line 200: / Line 203: @@
 >
 </code>
-|value | 1  | 4  | 6  | 8  | 9  | 10  | 11  | 12  |
-|freq  | 1  | 1  | 2  | 3  | 4  | 4   | 5  | 5   |
 <code>
@@ Line 214: / Line 213: @@
 {1, 1, 1, 2, 2, 2, 2, 2, 3, 31, 31, 31, 32, 32, 32, 32, 33, 33, 33}
 </code>
+<code>
+a <- c(1, 1, 1, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33)
+b <- c(1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33)
+c <- c(1, 1, 1, 2, 2, 2, 2, 2, 3, 31, 31, 31, 32, 32, 32, 32, 33, 33, 33)
+a
+b
+c
+length(a)
+length(b)
+length(c)
+mean(a)
+mean(b)
+mean(c)
+median(a)
+median(b)
+median(c)
+</code>
+<code>
+# output
+> a <- c(1, 1, 1, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33)
+> b <- c(1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33)
+> c <- c(1, 1, 1, 2, 2, 2, 2, 2, 3, 31, 31, 31, 32, 32, 32, 32, 33, 33, 33)
+> a
+ [1]  1  1  1  2  2  2  2  3  3 31 31 32 32 32 32 33 33 33
+> b
+ [1]  1  1  1  2  2  2  2  2  3  3 31 31 32 32 32 32 33 33 33
+> c
+ [1]  1  1  1  2  2  2  2  2  3 31 31 31 32 32 32 32 33 33 33
+> length(a)
+[1] 18
+> length(b)
+[1] 19
+> length(c)
+[1] 19
+> mean(a)
+[1] 17
+> mean(b)
+[1] 16.21053
+> mean(c)
+[1] 17.68421
+> median(a)
+[1] 17
+> median(b)
+[1] 3
+> median(c)
+[1] 31
+>
+</code>
+===== Ugliness of mean and median =====
+[{{hf.mean.median.improper.data.jpg}}]
+<code>
+## answer
+n <- 3+2+2 +3+4+4
+n
+mean.duckc <- 17
+median.duckc <- 17
+tot <- mean.duckc * n
+# 2 :: 4 인것은 이미 알고 있음
+known.sum <- (1*3)+(2*4)+(3*2)+(31*2)
+tot - known.sum
+# 227 = (32*4) + (33*3) 이므로
+duckc <- c(1,1,1,2,2,2,2,3,3,31,31,32,32,32,32,33,33,33)
+mean(duckc)
+median(duckc)
+</code>
+<code>
+## how do they change with 2 added and 31 added
+duckc.2.added <- c(1,1,1,2,2,2,2,2,3,3,31,31,32,32,32,32,33,33,33)
+mean(duckc.2.added)
+median(duckc.2.added)
+duckc.31.added <- c(1,1,1,2,2,2,2,3,3,31,31,31,32,32,32,32,33,33,33)
+mean(duckc.31.added)
+median(duckc.31.added)
+</code>
 ===== mode =====
 see [[:mode]] document
@@ Line 236: / Line 315: @@
   - 최빈값과 동일한 급여를 받고 있다면 어느쪽 인상을 선호할까요?
 </WRAP>
+Mean
+\begin{eqnarray*}
+\mu & = &  \frac {\Sigma {X_{i}}} {N}  \\
+& = & 50000 \\
+\end{eqnarray*}
+\begin{eqnarray*}
+\text {2000 raised to the mean} & = & \frac {\sum{(X_{i} + 2000)}} {N} \\
+& = & \frac {\sum{(X_{i})} + \sum{(2000)}} {N} \\
+& = & \frac {\sum{(X_{i})}} {N} + \frac {N * (2000)} {N} \\
+& = & 50000 + 2000 \\
+& = & 52000
+\end{eqnarray*}
+Median
+모든 연봉에 2000불을 더하는 것이므로 median 값의 위치는 변하지 않는다. 따라서
+\begin{eqnarray*}
+\text{Original Median} (20000) + 2000 & = & 22000
+\end{eqnarray*}
+Mode
+모든 연봉에 2000불을 더하는 것이므로 mode 값은 원래 모드값인 10000 불에 2000불을 더한 값이다. 따라서
+\begin{eqnarray*}
+\text{Original Mode} (10000) + 2000 & = & 12000
+\end{eqnarray*}
+Mean
+\begin{eqnarray*}
+\text {10% raised to the mean} & = & \frac {\sum{(1.1 * X_{i})}} {N} \\
+& = & \frac {1.1 * \sum{(X_{i})}} {N} \\
+& = & 1.1 * \frac {\sum{(X_{i})}} {N} \\
+& = & 1.1 * 50000 \\
+& = & 55000
+\end{eqnarray*}
+Median
+모든 연봉에 10%를 더하는 것이므로 median 값의 위치는 변하지 않는다. 따라서
+\begin{eqnarray*}
+\text{Original Median} (20000) * 1.1 & = & 22000
+\end{eqnarray*}
+Mode
+모든 연봉에 10%를 더하는 것이므로 mode 값은 원래 모드값인 10000 불에 1.1을 곱한 값이다. 따라서
+\begin{eqnarray*}
+\text{Original Mode} (10000) * 1.1 & = & 11000
+\end{eqnarray*}