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Once the standard deviation of the measurement apparatus has been estimated, we can set the measurement values below and above which we reject baskets. In this exercise, you can use the following values:

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$\\sigma_{West}$ = {sigmaWest}mm, the estimated value of the standard deviation when measuring width.

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$\\sigma_{Lest}$ = {sigmaLest}mm, the estimated value of the standard deviation when measuring length.

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Suppose the figure below represents a set of measurement values of the width $W$ of one particular basket which is just too small (45.9999999mm). If the measurement values are normally distributed, which percentage of the values will be in segments a t/m h? If you do not remember this, look it up in some reliable source.

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Which percentage of the values will be in segment a?

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Which percentage of the values will be in segment b?

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Which percentage of the values will be in segment c?

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Which percentage of the values will be in segment d?

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Which percentage of the values will be in segment e?

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Which percentage of the values will be in segment f?

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Which percentage of the values will be in segment g?

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Which percentage of the values will be in segment h?

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Look up what the requirements of the CEO of Cartonax w.r.t. the number of baskets that get approved when they are just too small or too large are. You can find this in the project 2 manual. 

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To meet the requirements of the CEO, below which measured value for $W$ should the baskets be rejected?

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Give your asnwer in mm.

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To meet the requirements of the CEO, above which measured value for $W$ should the baskets be rejected?

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Give your asnwer in mm.

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To meet the requirements of the CEO, below which measured value for $L$ should the baskets be rejected?

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Give your asnwer in mm.

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To meet the requirements of the CEO, above which measured value for $L$ should the baskets be rejected?

\n

Give your asnwer in mm.

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