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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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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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\\(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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