{"id":40553,"date":"2026-05-16T09:00:56","date_gmt":"2026-05-16T00:00:56","guid":{"rendered":"https:\/\/blog.agentsoft.co.kr\/index.php\/2026\/05\/16\/40553\/"},"modified":"2026-05-16T09:00:56","modified_gmt":"2026-05-16T00:00:56","slug":"%ed%95%9c%ec%96%91%eb%8c%80-%ea%b3%84%ec%b8%a1%ea%b3%b5%ed%95%99-2%eb%b2%88%ec%a7%b8-%ea%b3%bc%ec%a0%9c%ec%b2%b4%ea%b7%b8-%eb%b0%8f-%ea%b5%90%ec%88%98%eb%8b%98-%ed%92%80%ec%9d%b4-%ec%b0%b8%ea%b3%a0","status":"publish","type":"post","link":"https:\/\/blog.agentsoft.co.kr\/index.php\/2026\/05\/16\/40553\/","title":{"rendered":"[\ud55c\uc591\ub300] \uacc4\uce21\uacf5\ud559 2\ubc88\uc9f8 \uacfc\uc81c(\uccb4\uadf8 \ubc0f \uad50\uc218\ub2d8 \ud480\uc774 \ucc38\uace0\ud55c \ub9cc\uc810 \ub808\ud3ec\ud2b8)"},"content":{"rendered":"

<\/p>\n

\ubaa9\ucc28<\/strong><\/p>\n

\uc5c6\uc74c<\/p>\n

\ubcf8\ubb38\ub0b4\uc6a9<\/strong><\/p>\n

3.1. For a very large set of data, the measured mean is found to be 200 with a standard deviation of 20. Assuming the data to be normally distributed, determine the range within which 60% of the data are expected to fall.<\/p>\n

\uce21\uc815\ud55c \ud3c9\uade0\uc774 x^’=200, \ud45c\uc900\ud3b8\ucc28\uac00 \u03c3=20\uc774\ubbc0\ub85c \ud45c\ubcf8 \uac1c\uc218\ub294 \ucda9\ubd84\ud558\ub2e4. \uc815\uaddc\ubd84\ud3ec 60%\uc5d0 \ud574\ub2f9\ud558\ub294 x\uac12\uc744 \uad6c\ud558\uae30 \uc704\ud574\uc11c z\ubd84\ud3ec\uc640 \ub300\uce6d\uc131\uc744 \ud65c\uc6a9\ud55c\ub2e4\uba74 \ud655\ub960 P\uac00 30%, \uc989 0.3\uc774 \ub418\ub294 z\uac12\uc744 \uad6c\ud574\uc57c \ud558\ub294\ub370 \ud45c\uc5d0 \uc815\ud655\ud55c \uac12\uc744 \ucc3e\uc744 \uc218 \uc5c6\ub2e4. \ub530\ub77c\uc11c 0.3\uacfc \uac00\uc7a5 \uac00\uae4c\uc6b4 \uac12\uc73c\ub85c \ucc3e\uc73c\uba74 z\u22480.84\uc774\ub2e4.
\nx=x^’\u00b1\u03c3z\uc774\ubbc0\ub85c \ub300\uc785\ud558\uc5ec \uacc4\uc0b0\ud558\uba74 x=216.8 ,183.2\uc774\ub2e4. \ub530\ub77c\uc11c \ubc94\uc704\ub294 183.2\u2264x\u2264216.8 \uc774\ub2e4.<\/p>\n

3.3
\n(a) A 68-k\u03a9 resistor is paralleled with a 12-k\u03a9 resistor. Each resistor has a \u00b110% tolerance. What will be the nominal resistance and the uncertainty of the combination?<\/p>\n

\ubb38\uc81c\uc5d0\uc11c R_0=1\/(1\/(1\/R_1 +1\/R_2 ))=(R_1 R_2)\/(R_1+R_2 ) \ub85c \uc2dd\uc774 \uc8fc\uc5b4\uc84c\uace0 \u00b110% \uc758 \uc624\ucc28\uc728\uc744 \uac00\uc9c4\ub2e4\uace0 \ud588\uc73c\ubbc0\ub85c R_1=(68\u00b16.8) k\u03a9, R_2=(12\u00b11.2) k\u03a9\ub85c \ud45c\ud604\ud560 \uc218 \uc788\ub2e4. \ub530\ub77c\uc11c R_(0=) (68\u00d712)\/(68+12)=10.2 k\u03a9\uc774\ub2e4.
\n\uc800\ud56d\uc758 \ubd88\ud655\ub3c4\ub97c \uad6c\ud558\uae30 \uc704\ud574\uc11c\ub294 R_0\uc2dd\uc758 \uad6c\uc131\uc131\ubd84\uc758 \uac01 \ud3b8\ubbf8\ubd84 \uac12\uc774 \ud544\uc694\ud558\ubbc0\ub85c \uacc4\uc0b0\ud558\uba74 (\u2202R_0)\/(\u2202R_1 )=(R_2^2)\/\u3016(R_1+R_2)\u3017^2 =0.0225, (\u2202R_0)\/(\u2202R_2 )=(R_1^2)\/\u3016(R_1+R_2)\u3017^2 =0.7225\uc774\ub2e4. \ub530\ub77c\uc11c \ubd88\ud655\ub3c4 \uacf5\uc2dd\uc5d0 \uc758\ud574 u_(R_0 )=\u00b10.8804 k\u03a9 \uc774\ub2e4.<\/p>\n

(b) If the values remain the same expect that the tolerance on the 68-k\u03a9 resistor is dropped to \u00b15%, what will be the uncertainty of the combination?<\/p>\n

68 k\u03a9\uc758 \uc800\ud56d \uc624\ucc28\uc728\ub9cc \u00b15%\ub85c \ub5a8\uc5b4\uc84c\uc73c\ubbc0\ub85c u_(R_1 )=3.4, u_(R_2 )=1.2 \ub85c \ub2e4\uc2dc \ud55c\ubc88 \ubd88\ud655\ub3c4 \uacf5\uc2dd\uc5d0 \ub300\uc785\ud558\uac8c \ub418\uba74 u_(R_0 )=\u00b10.8704 k\u03a9 \uc774\ub2e4.<\/p>\n

\ucd9c\ucc98 : \ud574\ud53c\ucea0\ud37c\uc2a4<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

\ubaa9\ucc28 \uc5c6\uc74c \ubcf8\ubb38\ub0b4\uc6a9 3.1. For a very large set of data, the measured mean is found to be 200 with a standard deviation of 20. Assuming the data to be normally distributed, determine the range within which 60% of the data are expected to fall. \uce21\uc815\ud55c \ud3c9\uade0\uc774 x^’=200, \ud45c\uc900\ud3b8\ucc28\uac00 \u03c3=20\uc774\ubbc0\ub85c \ud45c\ubcf8 \uac1c\uc218\ub294 \ucda9\ubd84\ud558\ub2e4. \uc815\uaddc\ubd84\ud3ec 60%\uc5d0 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[48663,12445],"class_list":["post-40553","post","type-post","status-publish","format-standard","hentry","category-uncategorized","tag-48663","tag--erica"],"_links":{"self":[{"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/posts\/40553","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/comments?post=40553"}],"version-history":[{"count":0,"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/posts\/40553\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/media?parent=40553"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/categories?post=40553"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.agentsoft.co.kr\/index.php\/wp-json\/wp\/v2\/tags?post=40553"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}