A component is defective if oversized. A sample of 460 components produced by a machine have a mean size of 7.2 cm and a standard deviation of 0.12 cm. The maximum size of an acceptable component is 7.46 cm. Assume a normal distribution.
(i) Determine how many components are defective.
(ii) Find how many components are acceptable.
Acceptable components:
$$P(X ≤ 7.46)$$ $$P(Z ≤ \frac{7.46-7.2}{0.12})$$ $$P(Z ≤ 2.17)$$ $$0.9850$$ $$0.9850*460$$ $$453$$
Defective components:
$$460 - 453 = 7$$
This seemed a bit to simple to me as I have been having a lot of trouble with these, is this correct?
As discussed in the comments, I do not believe this problem can be solved as stated. A sample is just a sample and might well fail to perfectly respect the properties of the underlying distribution.
To be more precise: Let $p$ be the probability that a given unit is defective. using the given normal distribution we see that $$p\sim 0.01513014$$
(Note: the OP's calculation gives $p=.015$ which is, of course, the same value with fewer significant places).
We can now use the binomial distribution to compute the probability, $p_k$, of getting exactly $k$ defective units. In general, of course, we have $$p_k=\binom {460}k\times p^k\times (1-p)^{460-k}$$
This is easy to compute, at least with mechanical assistance. The first few values are $$p_0=0.000900056,\,p_1=0.006360504,\,p_2=0.022425302,\,p_3=0.052595263$$ $$p_4=0.092313866,\,p_5=0.129337921,\,p_6=0.150678095,\,p_7=0.150131573,$$$$p_8=0.130600353,\,p_9=0.100763762,\,p_{10}=0.069814401,\,p_{11}=0.043876194$$ $$p_{12}=0.025220758,\,p_{13}=0.013352325,\,p_{14}=0.006549385,p_{15}=0.002991632$$ $$p_{16}=0.00127824$$
Collectively, that covers more than $99.9\%$ of the cases. We note that $7$ is not even the most probable (though it is very close).
Also worth noting: even if you "broaden the definition of $7$" to include $6,8$ it is still more likely than not that your actual sample will fall outside this strip.