<div data-bbox="165 102 243 116" data-label="Page-Header"><p>제14981호</p></div>
<div data-bbox="435 102 456 116" data-label="Page-Header"><p>관</p></div>
<div data-bbox="541 102 562 116" data-label="Page-Header"><p>보</p></div>
<div data-bbox="675 102 830 116" data-label="Page-Header"><p>2001.12.19. (수요일)</p></div>
<div data-bbox="129 512 173 875" data-label="Text"><p>통합된 교정기복의 중앙값 OOTR을 구하기 위해서는 먼저 OOTR에 의해 정렬한다. 정렬 결과 다음 결과를 얻는다.</p></div>
<div data-bbox="179 525 333 753" data-label="Table">
<table border="1">
<thead>
<tr>
<th>기술자</th>
<th>사용자</th>
<th>교정 시점</th>
<th>OOTR</th>
</tr>
</thead>
<tbody>
<tr>
<td>Eddie Zitzlaff</td>
<td>G. Gordon Gurgle</td>
<td>Bob's Cal Service</td>
<td>0.074</td>
</tr>
<tr>
<td>Eddie Zitzlaff</td>
<td>Gordwana Park</td>
<td>Bob's Cal Service</td>
<td>0.075</td>
</tr>
<tr>
<td>Hap Halvah</td>
<td>G. Gordon Gurgle</td>
<td>Bob's Cal Service</td>
<td>0.076</td>
</tr>
<tr>
<td>Mel Fernmeyer</td>
<td>Jack(Ru) Huggaboorn</td>
<td>SWAG Tech. Inc.</td>
<td>0.077</td>
</tr>
<tr>
<td>Wanda Swoose</td>
<td>Jack(Ru) Huggaboorn</td>
<td>Wendy Finger Labs</td>
<td>0.078</td>
</tr>
<tr>
<td>Mel Fernmeyer</td>
<td>Gordwana Park</td>
<td>SWAG Tech. Inc.</td>
<td>0.082</td>
</tr>
<tr>
<td>Guy Gitchemoli</td>
<td>Wally Ballou</td>
<td>OOTs-R-Us</td>
<td>0.925</td>
</tr>
<tr>
<td>Guy Gitchemoli</td>
<td>Gordwana Park</td>
<td>OOTs-R-Us</td>
<td>1.031</td>
</tr>
<tr>
<td>Guy Gitchemoli</td>
<td>G. Gordon Gurgle</td>
<td>OOTs-R-Us</td>
<td>1.151</td>
</tr>
</tbody>
</table>
</div>
<div data-bbox="343 602 361 875" data-label="Text"><p>이 표에서, <math>N=9</math> 이고 중앙값 값은 0.078이다. 즉, <math>n_s=4</math>.</p></div>
<div data-bbox="382 754 401 875" data-label="Section-Header"><p>9) 기술자 이상치 파악</p></div>
<div data-bbox="402 583 421 875" data-label="Text"><p>기술자 이상치 파악을 위해서, 관련된 값들은 다음과 같다.</p></div>
<div data-bbox="432 512 526 875" data-label="Table">
<table border="1">
<thead>
<tr>
<th>기술자</th>
<th><math>n_{su}</math></th>
<th><math>n_{suu}</math></th>
</tr>
</thead>
<tbody>
<tr>
<td>Eddie Zitzlaff</td>
<td>2</td>
<td>0</td>
</tr>
<tr>
<td>Hap Halvah</td>
<td>1</td>
<td>0</td>
</tr>
<tr>
<td>Wanda Swoose</td>
<td>2</td>
<td>0</td>
</tr>
<tr>
<td>Mel Fernmeyer</td>
<td>2</td>
<td>1</td>
</tr>
<tr>
<td>Guy Gitchemoli</td>
<td>3</td>
<td>3</td>
</tr>
</tbody>
</table>
</div>
<div data-bbox="527 512 577 875" data-label="Text"><p>중앙값 보다 큰 <math>n</math>개의 OOTR을 관찰할 확률인 <math>\rho(n)</math>을 구하기 위해, 확률 밀도함수 <math>\rho(n)</math>을 다음과 같이 정의한다.</p></div>
<div data-bbox="582 587 614 783" data-label="Equation-Block"><math display="block">F(n) = \frac{n_{su}!}{n!(n_{su} - n)!} \frac{n_{su}^n}{N^{n_{su}}} (N - n_s)^{n_{su} - n}.</math></div>
<div data-bbox="625 512 670 875" data-label="Text"><p>기술자 이상치를 90% 신뢰수준에서 파악하기를 원한다고 가정하자. 즉 <math>\alpha=0.10</math>, 그리고 다음과 같은 결과를 얻는다.</p></div>
<div data-bbox="682 796 701 875" data-label="Text"><p>Eddie Zitzlaff:</p></div>
<div data-bbox="706 703 725 783" data-label="Equation-Block"><math display="block">n_{su} = 2, n_{suu} = 0</math></div>
<div data-bbox="735 663 767 790" data-label="Equation-Block"><math display="block">\sum_{n=0}^{n_{su}-1} \rho(n) = \sum_{n=0}^0 \rho(n) = 1</math></div>
<div data-bbox="783 806 802 875" data-label="Text"><p>Hap Halvah:</p></div>
<div data-bbox="807 696 826 773" data-label="Equation-Block"><math display="block">n_{su} = 1, n_{suu} = 0</math></div>
<div data-bbox="836 656 868 783" data-label="Equation-Block"><math display="block">\sum_{n=0}^{n_{su}-1} \rho(n) = \sum_{n=0}^0 \rho(n) = 1</math></div>
<div data-bbox="129 406 146 487" data-label="Section-Header"><p>Wanda Swoose:</p></div>
<div data-bbox="151 306 170 383" data-label="Equation-Block"><math display="block">n_{su} = 1, n_{suu} = 0</math></div>
<div data-bbox="179 266 211 393" data-label="Equation-Block"><math display="block">\sum_{n=0}^{n_{su}-1} \rho(n) = \sum_{n=0}^0 \rho(n) = 1</math></div>
<div data-bbox="229 406 246 487" data-label="Section-Header"><p>Mel Fernmeyer:</p></div>
<div data-bbox="251 312 270 389" data-label="Equation-Block"><math display="block">n_{su} = 2, n_{suu} = 1</math></div>
<div data-bbox="279 271 343 429" data-label="Equation-Block"><math display="block">\rho(n) = \frac{2!}{n!(2-n)!} \frac{4^n}{9^n} (9-4)^{2-n} \\ = \frac{2!}{n!(2-1)!} \frac{4^n}{8!} 5^{2-n}</math></div>
<div data-bbox="361 208 393 487" data-label="Equation-Block"><math display="block">\rho(1) = \frac{2!}{1!(2-1)!} \frac{4^1}{9^1} (9-4)^{2-1} = (2)(4/8)(5) = 40/81</math></div>
<div data-bbox="398 206 430 429" data-label="Equation-Block"><math display="block">\rho(2) = \frac{2!}{2!(2-2)!} \frac{4^2}{9^2} (9-4)^{2-2} = 16/81</math></div>
<div data-bbox="440 454 457 487" data-label="Text"><p>그리고</p></div>
<div data-bbox="457 236 489 429" data-label="Equation-Block"><math display="block">\sum_{n=0}^{n_{su}-1} \rho(n) = \sum_{n=0}^2 \rho(n) = 56/81 = 0.691</math></div>
<div data-bbox="494 396 513 487" data-label="Section-Header"><p>Guy Gitchemoli:</p></div>
<div data-bbox="518 295 537 389" data-label="Equation-Block"><math display="block">n_{su} = 3, n_{suu} = 3</math></div>
<div data-bbox="546 293 578 429" data-label="Equation-Block"><math display="block">\rho(n) = \frac{3!}{n!(3-n)!} \frac{4^n}{9^n} 5^{3-n}</math></div>
<div data-bbox="597 147 629 403" data-label="Equation-Block"><math display="block">\rho(3) = \frac{3!}{3!(3-3)!} \frac{4^3}{9^3} = \left(\frac{4}{9}\right)^3 = 64/729 = 0.0878</math></div>
<div data-bbox="639 454 656 487" data-label="Text"><p>그리고</p></div>
<div data-bbox="661 252 693 429" data-label="Equation-Block"><math display="block">\sum_{n=0}^{n_{su}-1} \rho(n) = \sum_{n=0}^2 \rho(n) = 0.0878</math></div>
<div data-bbox="698 208 717 487" data-label="Text"><p>유의수준 <math>\alpha=0.10</math>에서는, Guy Gitchemoli가 이상치이다.</p></div>
<div data-bbox="732 367 751 487" data-label="Section-Header"><p>3) 사용자 이상치 파악</p></div>
<div data-bbox="752 200 771 487" data-label="Text"><p>사용자 이상치 파악을 위해서, 관련된 값들은 다음과 같다.</p></div>
<div data-bbox="783 340 802 429" data-label="Equation-Block"><math display="block">n_{su} = n_{suu}</math></div>
<div data-bbox="807 340 863 487" data-label="Table">
<table border="1">
<tbody>
<tr>
<td>G. Gordon Gurgle</td>
<td>3</td>
<td>1</td>
</tr>
<tr>
<td>Gordwana Park</td>
<td>3</td>
<td>2</td>
</tr>
<tr>
<td>Jack(Ru) Huggaboorn</td>
<td>2</td>
<td>0</td>
</tr>
<tr>
<td>Wally Ballou</td>
<td>1</td>
<td>1</td>
</tr>
</tbody>
</table>
</div>