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\$$X\$$ is normally distributed with mean \$$\\var{m1}\$$ nm and standard deviation \$$\\var{sd1}\$$ nm.

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A 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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(a)   \$$P(Reject)=P(\\var{z_l}>X\\,\\,or\\,\\,X>\\var{z_u})\$$

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\$$=1 - P(Acceptable)\$$

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\$$P(Acceptable)=P(\\var{z_l}<X<\\var{z_u})\$$

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=\$$P\\left(\\frac{\\var{z_l}-\\var{m1}}{\\var{sd1}}<Z<\\frac{\\var{z_u}-\\var{m1}}{\\var{sd1}}\\right)\$$

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=\$$P(Z<\\var{n})-P(Z<-\\var{n})\$$

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=\$$P(Z<\\var{n})-(1-P(Z<\\var{n}))\$$

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=\$$\\var{p1}-(1-\\var{p1})\$$

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=\$$\\simplify{2*{p1}-1}\$$

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\$$P(Reject)=1-\\simplify{2*{p1}-1}\$$

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\$$P(Reject)=\\simplify{2-2*{p1}}\$$

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(b)     Find the values \$${k_1}\$$ and \$${k_2}\$$ such that \$$P({k_1}<X<{k_2})=\\simplify{{pr}*0.0100}\$$

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\$$P\\left(\\frac{{k_1}-\\var{m1}}{\\var{sd1}}<Z<\\frac{{k_2}-\\var{m1}}{\\var{sd1}}\\right)=\\simplify{{pr}*0.0100}\$$

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But   \$$P(-\\var{z1}<Z<\\var{z1})=\\simplify{{pr}*0.0100}\$$

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So   \$$\\frac{{k_1}-\\var{m1}}{\\var{sd1}}=-\\var{z1}\$$   and   \$$\\frac{{k_2}-\\var{m1}}{\\var{sd1}}=\\var{z1}\$$

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\$${k_1}-\\var{m1}=-\\simplify{{sd1}*{z1}}\$$   and   \$${k_2}-\\var{m1}=\\simplify{{sd1}*{z1}}\$$

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\$${k_1}=\\simplify{{m1}-{sd1}*{z1}}\$$   and   \$${k_2}=\\simplify{{m1}+{sd1}*{z1}}\$$

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Calculate, correct to three decimal places, the probability that a resistor chosen at random will be rejected. [[0]]

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Determine the line widths within which \$$\\var{pr}\$$% of production will lie. Give your answers as integer values.

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Lower limit =  [[0]]nm         Upper limit = [[1]]nm

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The width of a line etched on an integrated circuit chip is normally distributed with mean \$$\\var{m1}\$$nm and standard deviation \$$\\var{sd1}\$$nm.

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A 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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