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Once our measurement apparatus has been built, we can use it to remove wrongly produced baskets from the production line. However, as you have seen in the lecture, we need to consider the precision of our measurement apparatus. To meet the requirements of the CEO of Cartonax, we need to specify measurement values $L_{min}$, $W_{min}$ below which baskets get rejected and $L_{max}$, $W_{max}$ above which baskets get rejected. 

\n

Before these values can be set, the precision of the measurement apparatus has to be determined. In this exercise, you will practice with specifying this precision in terms of the standard deviation $\\sigma$. You will estimate the standard deviation from a sample of measurement values, just as you will have to do during the third lab.

\n

Remember, you can estimate the standard deviation $\\sigma$ from a sample using:

\n

$\\sigma_{est}=\\sqrt{\\frac{\\sum(x_i-\\mu_{est})^2}{n}}$

\n

In which $x_i$ are the individual measurement values, $\\mu_{est}$ is the mean measurement value and $n$ is the sample size. 

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Suppose you measure the width $W$ of some basked of which you know the exact value $W$ = 47,00mm ten times. Due to the inaccuracy of your measurement apparatus, you obtain the following values:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
measurementvalue (mm)
1{x1}
2{x2}
3{x3}
4{x4}
5{x5}
6{x6}
7{x7}
8{x8}
9{x9}
10{x10}
\n

What is the sample mean? 

\n

Give your answer in mm.

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Estimate the standard deviation of your measurement apparatus if it is used to measure the width $W$ by calculating $\\sigma_{West}$.

\n

Give your answer in mm.

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Suppose you measure the length $L$ of some basked of which you know the exact value $L$ = 76,00mm ten times. Due to the inaccuracy of your measurement apparatus, you obtain the following values:

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measurementvalue (mm)
1{l1}
2{l2}
3{l3}
4{l4}
5{l5}
6{l6}
7{l7}
8{l8}
9{l9}
10{l10}
\n

What is the sample mean? 

\n

Give your answer in mm.

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Estimate the standard deviation of your measurement apparatus when it is used to measure the length $L$ by calculating $\\sigma_{Lest}$.

\n

Give your answer in mm.

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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:

\n

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

\n

$\\sigma_{Lest}$ = {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.

\n

\n

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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\n

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. 

\n

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

\n

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?

\n

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?

\n

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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