Statistics question, mean vs standard deviation

Question

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/

Answer 1

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

Answer 2

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