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Proper blending of fine and coarse powders prior to copper
Chapter 4, Problem 132E(choose chapter or problem)
Proper blending of fine and coarse powders prior to copper sintering is essential for uniformity in the finished product. One way to check the homogeneity of the blend is to select many small samples of the blended powders and measure the proportion of the total weight contributed by the fine powders in each. These measurements should be relatively stable if a homogeneous blend has been obtained.
a Suppose that the proportion of total weight contributed by the fine powders has a beta distribution with \(\alpha=\beta=3 \text {. }\) Find the mean and variance of the proportion of weight contributed by the fine powders.
b Repeat part (a) if \(\alpha=\beta=2\).
c Repeat part (a) if \(\alpha=\beta=1\).
d Which of the cases—parts (a), (b), or (c)—yields the most homogeneous blending?
Questions & Answers
(1 Reviews)
QUESTION:
Proper blending of fine and coarse powders prior to copper sintering is essential for uniformity in the finished product. One way to check the homogeneity of the blend is to select many small samples of the blended powders and measure the proportion of the total weight contributed by the fine powders in each. These measurements should be relatively stable if a homogeneous blend has been obtained.
a Suppose that the proportion of total weight contributed by the fine powders has a beta distribution with \(\alpha=\beta=3 \text {. }\) Find the mean and variance of the proportion of weight contributed by the fine powders.
b Repeat part (a) if \(\alpha=\beta=2\).
c Repeat part (a) if \(\alpha=\beta=1\).
d Which of the cases—parts (a), (b), or (c)—yields the most homogeneous blending?
ANSWER:Step 1 of 4
a) We have to find mean and variance of weight contributed by the fine powders \(\alpha=\beta=3\)
Let Y be the weight contributed by the fine powders
Here Y follows the beta distribution
Given \(\alpha=3 \text { and } \beta=3\)
\(\begin{aligned}
\operatorname{Mean}(\mathrm{Y}) & =\alpha /(\alpha+\beta) \\
& =3 / 6 \\
& =0.5
\end{aligned}\)
\(\begin{aligned}
\operatorname{Variance}(\mathrm{Y}) & =\frac{\alpha \beta}{(\alpha+\beta)^{2}(\alpha+\beta+1)} \\
& =3(3) / 62(7) \\
& =9 / 252 \\
& =1 / 28 \\
& =0.0357
\end{aligned}\)
Hence if \(\alpha=\beta=3\) the mean(Y) is 0.5 and variance(Y) is 0.0357
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