Resistor with $500\Omega$ is made with tolerance $\pm 1\Omega$.
$a)$ Determine the spoilage percentage (that falls outside of the tolerance area) if the processing gives the standard deviation $\sigma =2\Omega$.
$b)$ Increase the field of tolerance such that the spoilage percentage is $10\%$ with standard deviation $\sigma = 2\Omega$.
Normal distribution should be used.
My attempt:
$a)$ $$P(X\le 499)+P(X\ge 501)=2P(X\ge 501)$$ $$=2(1-P(X\le 501))$$ $$=2(1-P(X\le 0.5)),\Phi(0.5)=0.6915,P(X\le 0.5)=0.6915$$ $$=2(1-0.6915)$$ $$=0.6915=61.7\%$$
How to solve for $b)$?
In my book’s solution, the field of tolerance is $[496,72,503,28]$. Could someone show the procedure for finding it?
You have a contratiction in your notation.
If $P(X\geq 501)$ then it is equal to $P(Z\geq 0.5)$, where $Z=\frac{X-\mu}{\sigma}=\frac{X-500}{2}$. $Z$ is normally standard distributed.
For $b)$ we use the standardized random variable as well.
$P(X\leq x)=\Phi\left(\frac{x-500}{2} \right)$. Thus
$$P(-z\leq Z \leq z)=2-2\Phi(z)$$ $$=2-2\Phi\left(\frac{x-500}{2}\right)=0.1\qquad |:2$$
$$1-1\Phi\left(\frac{x-500}{2}\right)=0.05\qquad |-0.05$$
$$0.95-1\Phi\left(\frac{x-500}{2}\right)=0\qquad $$
$$0.95=\Phi\left(\frac{x-500}{2}\right)\qquad $$
$$\Phi^{-1}(0.95)=\frac{x-500}{2}$$
$\Phi^{-1}(p)$ is inverse cumulative distribution function or quantile function of the standard normal distribution. The equation becomes
$$1.64=\frac{x-500}{2}$$
Calculate $x$. Finally the interval is $[2\mu-x,x]=[2\cdot 500-x, x]$.