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Softa [21]
4 years ago
7

An office supply company manufactures paper clips, and even tolerates a small proportion of those paper clips being ‘defective’

(or incorrectly shaped and/or twisted) in its outgoing product. (The company reasons that paper clips are so cheap, users will simply discard the occasional defective paper clip they might find in a box.) The average proportion of ‘defective’ paper clips is known to be 2% when the paper clip manufacturing process is ‘in control’. To monitor this issue, what should be the value of the upper control limit of a p-chart if the company plans to include 25 paper clips in each of its samples and use z-value of 3.0 to construct the chart? g
Mathematics
1 answer:
vovikov84 [41]4 years ago
4 0

Answer:

0.104 (10.4%)

Step-by-step explanation:

UCL = \bar{p}+z(\sigma)

\sigma = p(1-)) Vn = 1.02(1-202)) 5 = 0.028

\thereforeUCL = .02+(3x0.028) = 0.104

\thereforeUCL= 10.4%

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Step-by-step explanation:

Given

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First, we need to simplify f(x)

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f(x) = \frac{x(x - 2)}{x^2 - 2^2}

f(x) = \frac{x(x - 2)}{(x- 2)(x + 2)}

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f(x) = \frac{x}{x + 2}

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f(2.001) = \frac{2.001}{2.001 +2}

f(2.001) = \frac{2.001}{4.001}

f(2.001) = 0.500125

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f(x) = \frac{x}{x + 2}

f(2.0001) = \frac{2.0001}{2.0001 + 2}

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