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Rudik [331]
2 years ago
5

Consider a manufacturing process with a quality inspection station. In the past, 10% of parts are defective. As soon as one defe

ctive part is found, the process is stopped. If 6 parts have been inspected without finding a defective part, what is the probability that at least 9 total parts will be inspected before the process is stopped
Mathematics
1 answer:
Kruka [31]2 years ago
7 0

Answer:

0.9

Step-by-step explanation:

10% is equal to 0.1

The probability of having defective parts in a pile of parts is 0.1

Before the process is stopped, 1 part has to be defective.

In a pile of 9 parts, the probability that a part is defective 0.1 of 9, which is = 0.9 hence, approximately one (1) part will be defective in a pile of 9 parts and the process will be stopped.

Since there was no defective part among the first 6 parts, P(d) was 0

That is, probability of a defective part was zero.

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So the first thing we have to do is to calculate for the dilation factor. Taking point G as the reference point, we can see that the distance of point G from rectangle W’X’Y’Z’ is 1.5 while the distance from rectangle WXYZ is (1.5 + 7.5), therefore the dilation factor to use is:

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Since WX has an initial measure of 3 units, therefore the measure of W’X’ is:

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Answer:

<span>0.5 units</span>

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