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Samples of $20$ parts from a metal punching process are selected every hour. Typically, $1\%$ of the parts require rework. Let $X$ denote the number of parts in the sample of $20$ that require rework. A process problem is suspected if $X$ exceeds its mean by more than $3$ standard deviations.

(a) If the percentage of parts that require rework remains at $1\%$, what is the probability that $X$ exceeds its mean by more than $3$ standard deviations?

What I did was $(np(1-p))^{0.5}$, and I got $0.445$ for the standard deviation. However, the way the question was worded confused me. The solution proceeds to calculate the upper bound and finds the answer as $0.0169$. I do not understand how I can use $0.445$ to get $0.0169$. If someone could help, I would be really glad.

Solution is located here: https://www.slader.com/textbook/9781118539712-applied-statistics-and-probability-for-engineers-6th-edition/86/exercises/108/

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  • $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 28 at 8:47
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You got $\sigma = 0.445$. You also need $\mu = 0.2$. Then the probability that $X$ exceeds its mean by more than 3 standard deviations is $P(X> 0.2 + 3(0.445))$.

Since $X$ can only be integer, that is the same as $P(X \geq 2)$, which is the same as $1-P(X=0)-P(X=1)$.

$P(X=0) = (0.99)^{20}$ and $P(X=1) =20 (0.99)^{19} (0.01)$

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  • $\begingroup$ I have a quick question, first of all thanks for your time and explanation it is very clear and I see what I missed. I understand most of your answer but there is just one thing that I can't get, for P(X=0) I got 0.8179 as well but I didn't use binomial distribution, for the second one I couldn't get 0,1652, may I ask why I have to use binomial distribution? or why basic (0.99^19)*(0.01) doesn't work? $\endgroup$ – Aldo Sep 28 at 16:30
  • $\begingroup$ If you sample the 20 parts in an order, the probability that the first will require rework and the rest won’t is $(0.01)(0.99)^{19}$, the probability that the second will require rework and the rest won’t is $(0.99)(0.01)(0.99)^{18}$, etc. So the probability that any specific one of the 20 will require rework and the other 19 won’t is $(0.01)(0.99)^{19}$. There are 20 of these possible outcomes, and they are all mutually exclusive, so you add them all (i.e. multiply by 20) to get the desired probability. $\endgroup$ – Joe Sep 28 at 16:39
  • $\begingroup$ oh okay it makes sense thanks a lot Joe! $\endgroup$ – Aldo Sep 28 at 17:12
  • $\begingroup$ You're welcome! $\endgroup$ – Joe Sep 28 at 17:19
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The random variable $X$ denotes the number of parts that need rework. It is binomially distributed with parameters $n=20$ and $p$. You are told that the percentage of parts that require rework remains at $1\%$, so that $p=0.01$. Thus, $$\mathbb{E}[X] = np(1-p) = 20(0.01)(0.99) = 0.2$$

and $$\sigma_X = \sqrt{np(1-p)} =0.445.$$

We are asked to compute the probability that $X$ exceeds its mean by more than $3$ standard deviations, i.e., we are looking for $$\mathbb{P}(X > \mathbb{E}[X] +3\sigma_X )$$ and since $\mathbb{E}[X] +3\sigma_X = 0.2 + 3(0.445) =1.56$ the last expression is the equivalent to \begin{align*} \mathbb{P}(X > 1.56)&= \mathbb{P}(X \geq 2) \\ &= 1- \mathbb{P}(X<2)\\ &=1- \mathbb{P}(X=0)-\mathbb{P}(X=1) \end{align*} The first equality follows from the fact that $X$ is a discrete distribution taking on non-negative integer values. The terms in the last equality are obtained by evaluating the probability mass function for $X=0$ and $X=1$. Since $\mathbb{P}(X=0)= 0.8179$ and $\mathbb{P}(X=1)= 0.1652$ then $$\mathbb{P}(X > 1.56) = 1-0.8179-0.1652= 0.0169.$$

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