So I decided to have a look back through Statistics Problems and Solutions (Second Edition) by Bassett et al., 1986 when I found this question:
Note that for the duration of this question, this will only be about part (a) of the question. Any other questions that I have about this problem will be in follow-up questions.
$4\text{B.}1\quad$ Length of manufactured bolts
Bolts are manufactured with a manufactured length of $5$cm, and it is known from past experience that the variance of the lengths of such bolts is $0.05\text{cm}^2$. A random sample of $10$ bolts is taken from a box containing a large amount of bolts, and their length (in cm) are found to be:
$$5.68, 5.13, 5.82, 5.71, 5.36, 5.52, 5.29, 5.57, 5.45, 5.39$$
(a) Find $95$% confidence limits for the mean length, $\mu$, of bolts in the box, stating clearly any assumptions made in deriving the limits.
(b) Without doing any formal test of a hypothesis, discuss whether $\mu=5$cm is a plausible hypothesis, given the result from part (a)
(c) By looking only at the number of bolts, out of $10$, whose lengths are greater than $5$, construct a formal test of the null hypothesis $\mathit{H}_0:\mu=5$cm, against the alternative $\mathit{H}_1:\mu\neq5$cm.
My attempt at solving part (a)
The variance from a sample of items from a group can be calculated as$$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N (x_i-\overline x)^2}$$$$\text{Where}$$$$\mathit{N}\text{ is the number of terms}$$$$x_i\text{ is one sample}$$$$\text{And }\overline x\text{is the mean of }\mathit{N}\text{ number of samples}$$To streamline the process a little (yes, I know, I'm sort of lazy), I decided to see if I could write an equation to see if I could determine the variance through Desmos. Long story short, it didn't work, and I ended up going back to the calculator.net standard deviation calculator to determine that the variance of the sample was $\approx0.044306666666667$, however, here is where we are going to make the assumption that the calculator is flawed and the variance of the sample is $0.05$. Using our current assumptions, it would nonetheless be correct to assume that $5$% confidence intervals are in fact determined by the formula $$s_{\overline x}=\frac{s}{\sqrt{N}}\approx0.06656325312563$$$$\text{Then, adding }s_{\overline x}\text{ to }s\text{ gets us }5.5586325312563$$$$\text{Then, we plug this info into the formula to get }(5.5586325312563+s_{\overline x})\pm\left(\frac{1.96(0.06656325312563)}{\sqrt{10}}\right)$$$$\text{Which gets us }\mu\approx5.4281685913, 5.68909650778$$My question
Is the solution that I reached correct, or what could I do to attain the correct solution more easily?
To clarify
Edit: As pointed out by Andrew Zhang in chat, I should have pointed out which type of distribution that I used to attain the answer. I used normal distribution for anyone wondering.