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When circuit boards used in the manufacture of compact disk players are tested, the long-run percentage of defectives is 5$\%$ . Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board.(a) What is the approximate probability that at least 10$\%$ of the boards in the batch are defective?(b) What is the approximate probability that there are exactly 10 defectives in the batch?

(a) .0003 (exact: .00086) (b) .0888 (exact: .0963)

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

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When circuit boards used i…

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A certain batch contains $…

05:32

Suppose that only 10$\%$ o…

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The manufacturing of semic…

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Printed circuit cards are …

Okay, so let's define it X as the number of defective of defectives, and that's in the batch of 2 50 So X is approximately equal to the binomial 2 50 comma 0.5 and we can approximate X by a normal distribution. So since NP is equal to 12.5, which is greater than or equal to 10 this case it's greater than, um, the mean and standard deviation of X is the following. So the mean is 12.5 and the standard deviation is three point 446 If we consider the fact that 10% of 2 50 is 25 we could now fear out this probability probability of X being greater than or equal to 25 is equal to one, minus the probability of ex being less than 25 which is the same thing as one minus the probability of X being less than or equal to 24.5. And this is approximately equivalent to the following. And now we could simplify this even further to get the following and our final answer would be one minus nine 997 which is equal 2.3 and, um, the exact, uh, by no mo binomial probability from the software given in the textbook is 0.86 So either one of these works again. These are very small numbers, so this might seem like it's a big difference, but they're they're very, very close to one another. So this is a and now for B, um, we're using the same normal approximation with the continuity correction. So we have the probability of X equal to 10 the same thing as a probability being between 9.5 reclusive and 10.5 inclusive. Okay. And this is approximately equal to the following, just plugging it into our standard formula. And then we subtract the following. This is how we find the distribution it's actually move this to another line. Can be could add 9.5 minus 12.5 over 3.44 six. And this is all equivalent to 0.2 810 minus 2.192 to, and that is equivalent to 0.88 Just wanna put it out there of a software that the textbook uses gives you an answer of 0.963 This is how we do this problem, and an answer in this general ballpark would be correct.

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