# Statistics question, mean vs standard deviation

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.

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Sep 28 at 8:47

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

• 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? – Aldo Sep 28 at 16:30
• 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. – Joe Sep 28 at 16:39
• oh okay it makes sense thanks a lot Joe! – Aldo Sep 28 at 17:12
• You're welcome! – Joe Sep 28 at 17:19

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 $$\mathbb{P}(X > 1.56) = 1-0.8179-0.1652= 0.0169.$$