Here is how I solved the problem:
$$ X\sim N(\mu=.13, \sigma^2=.005^2)\\ .12\le x\le .14 \\ \mu-2\sigma\le x \le \mu+2\sigma\\ $$ Using Tchebychev's inequality, I get $$ P(|x-\mu|\le 2\sigma)=1-\frac 1 {4} =.75\\ $$
However, the proposed solution to this problem goes as follows:
Why do these 2 approaches give significantly different answers? is it because of the $\phi_z$ estimate? Thanks!
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The key aspect is that Tchebyshev's Inequality gives an upper-bound for the probability while the solution uses the standard normal distribution to provide an approximate calculation of the probability that our random variable is within that given interval. In particular,
$$P(|x - \mu| \geq 2\sigma) \leq \frac{1}{4}$$ $$\implies 1-P(|x-\mu| \leq 2 \sigma) \leq \frac{1}{4}$$ $$\implies -P(|x-\mu| \leq 2 \sigma) \leq \frac{1}{4}-1 = \frac{-3}{4}$$ $$\implies P(|x-\mu| \leq 2 \sigma) \geq \frac{3}{4}.$$
If what you are looking for is a set of bounds, this is a (relatively poor) bound you can use. That being said, you may simply want to use the properties of the normal distribution to get a better approximation of the probability (as the solution suggests).
The Chebyshev Inequality is a very weak inequality. In your case, after the correction, it says that the probability a resistor is "within spec" is $\ge \frac{3}{4}$. (Note that it does not say that the probability is $\frac{3}{4}$, the Chebyshev Inequality is an inequality.)
If we know that the distribution is normal, we can get an essentially exact expression for the required probability. Note that the Chebyshev Inequality did not lie, the probability really is (a lot) greater than $\frac{3}{4}$. When we know quite a bit about the distribution of a random variable, one should avoid using the Chebyshev Inequality.