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\nThe width of a line etched on an integrated circuit chip is normally distributed with mean \\(\\var{m1}\\)nm and standard deviation \\(\\var{sd1}\\)nm.
\nA chip is rejected if its line width is found to be greater than \\(\\var{z_u}\\)nm or less than \\(\\var{z_l}\\)nm.
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If we want to find upper z-value where that prob that z is between lower z-value and upper z-value is pr% then (pr/2+50)/100 is the total prob less than upper z-value
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\n", "scripts": {}, "showCorrectAnswer": true, "type": "gapfill", "gaps": [{"scripts": {}, "variableReplacements": [], "precisionPartialCredit": 0, "minValue": "{p}", "type": "numberentry", "maxValue": "{p}", "precision": "3", "notationStyles": ["plain", "en", "si-en"], "showPrecisionHint": false, "correctAnswerStyle": "plain", "correctAnswerFraction": false, "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "mustBeReduced": false, "allowFractions": false, "showCorrectAnswer": true, "showFeedbackIcon": true, "marks": 1, "precisionType": "dp", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false}], "variableReplacementStrategy": "originalfirst", "marks": 0}, {"variableReplacements": [], "showFeedbackIcon": true, "prompt": "Determine the line widths within which \\(\\var{pr}\\)% of production will lie. Give your answers as integer values.
\n\nLower limit = [[0]]nm Upper limit = [[1]]nm
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\nA chip is rejected if its line width is found to be greater than \\(\\var{z_u}\\) nm or less than \\(\\var{z_l}\\) nm.
\n\n(a) \\(P(Reject)=P(\\var{z_l}>X\\,\\,or\\,\\,X>\\var{z_u})\\)
\n\\(=1 - P(Acceptable)\\)
\n\\(P(Acceptable)=P(\\var{z_l}<X<\\var{z_u})\\)
\n=\\(P\\left(\\frac{\\var{z_l}-\\var{m1}}{\\var{sd1}}<Z<\\frac{\\var{z_u}-\\var{m1}}{\\var{sd1}}\\right)\\)
\n=\\(P(Z<\\var{n})-P(Z<-\\var{n})\\)
\n=\\(P(Z<\\var{n})-(1-P(Z<\\var{n}))\\)
\n=\\(\\var{p1}-(1-\\var{p1})\\)
\n=\\(\\simplify{2*{p1}-1}\\)
\n\\(P(Reject)=1-\\simplify{2*{p1}-1}\\)
\n\\(P(Reject)=\\simplify{2-2*{p1}}\\)
\n\n\n(b) Find the values \\({k_1}\\) and \\({k_2}\\) such that \\(P({k_1}<X<{k_2})=\\simplify{{pr}*0.0100}\\)
\n\\(P\\left(\\frac{{k_1}-\\var{m1}}{\\var{sd1}}<Z<\\frac{{k_2}-\\var{m1}}{\\var{sd1}}\\right)=\\simplify{{pr}*0.0100}\\)
\nBut \\(P(-\\var{z1}<Z<\\var{z1})=\\simplify{{pr}*0.0100}\\)
\nSo \\(\\frac{{k_1}-\\var{m1}}{\\var{sd1}}=-\\var{z1}\\) and \\(\\frac{{k_2}-\\var{m1}}{\\var{sd1}}=\\var{z1}\\)
\n\\({k_1}-\\var{m1}=-\\simplify{{sd1}*{z1}}\\) and \\({k_2}-\\var{m1}=\\simplify{{sd1}*{z1}}\\)
\n\\({k_1}=\\simplify{{m1}-{sd1}*{z1}}\\) and \\({k_2}=\\simplify{{m1}+{sd1}*{z1}}\\)
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