Find confidence levels from given intervals 
A set of $40$ data items, produces a confidence interval for the population mean of $94.93<\mu<105.07$. If $\sum x^2 = 424 375$, find the confidence level.

So the idea of confidence intervals is still rather new to me and one that isn't fully clear in my mind, would someone be able to give me some hints about solving this and explain what they are doing to solve it. 
I get that the confidence interval is given by $$\bar{x}-z\frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+z\frac{\sigma}{\sqrt{n}}$$ and that $z=\Phi^{-1}(c)$ where $c$ is the confidence level, however this is as far as my knowledge goes.
 A: Outline:
If $z=\Phi^{-1}(c)$, then let the $100(2c-1)\%$ confidence interval for $\mu$ be $\mu_1<\mu<\mu_2$. 
This is because the normal distribution (which is what we are assuming here) is two-tailed, so we have that $$P(X<z)=c,\quad P(X>z)=1-c$$ and we seek the region $P(-z<X<z)=c-(1-c)=2c-1$ due to symmetry. Here, $X$ is just a random variable.
Therefore, using the standard inequalities for confidence intervals, we obtain the following equalities $$\bar x-z\frac\sigma{\sqrt n}=\mu_1\tag1$$ $$\bar x+z\frac\sigma{\sqrt n}=\mu_2.\tag2$$ Adding the two yields $\bar x=(\mu_1+\mu_2)/2$, or that $$\sum x_i=\frac n2(\mu_1+\mu_2).\tag3$$ Similarly, performing $(2)-(1)$ results in $$z\frac\sigma{\sqrt n}=\frac{\mu_2-\mu_1}2\implies z=\frac{(\mu_2-\mu_1)\sqrt n}{2\sigma}\tag4$$ but recall that the unbiased estimator for the variance is $$\sigma^2=\frac{\sum x_i^2-\frac1n\left(\sum x_i\right)^2}{n-1}.\tag5$$ You now have enough information to calculate $z$, and thus $c=\Phi(z)$.

Solution:

 The confidence level is $$c=\Phi\left(\frac{\mu_2-\mu_1}2\sqrt{\frac{n(n-1)}{\sum\limits_{i=1}^nx_i^2-n\left(\frac{\mu_1+\mu_2}2\right)^2}}\right)\tag6$$ and substituting $n=40$, $\mu_1=94.93$, $\mu_2=105.07$ and $\sum\limits_{i=1}^nx_i^2=424375$ into $(6)$, we obtain $c=0.9002$. Thus the "strength" of confidence is $100(2\times0.9002-1)\%\approx80\%$.

A: To match the confidence interval with your numerical data, we get
$$\bar{x}-z\frac{\sigma}{\sqrt{n}} = 94.93$$
and
$$\bar{x}+z\frac{\sigma}{\sqrt{n}} = 105.07$$
You are interested in the confidence level, which you could know if you get $z$, i.e. via the normal tail. Subtracting both equations you get
$$2z\frac{\sigma}{\sqrt{n}} =105.07-94.93=10.14 $$
so
$$z = \frac{10.14}{2\frac{\sigma}{\sqrt{n}}}$$. Since your sample size consists of 40 elements, then $n = 40$, hence
$$z = \frac{10.14}{2\frac{\sigma}{\sqrt{40}}}$$
and your $\sigma$ could be estimated using an unbiased estimator
$$\sigma^2=\frac{\sum_{i=1}^{40} x_i^2-\frac{1}{n}\left(\sum_{i=1}^{40} x_i\right)^2}{n-1} = \frac{424375-\frac{1}{40}\left(\sum_{i=1}^{40} x_i\right)^2}{40-1}$$
The sum $\sum x_i$ is nothing other than $\sum_{i=1}^{40} x_i = n \bar{x}$ which could be obtained by adding the first two equations:
$$2\bar{x} = 105.07+94.93 =200 $$
hence $$\bar{x} =100$$
so
$$\sum_{i=1}^{40} x_i = n \bar{x} = 40(100) = 4000$$
Replacing we get
$$\sigma= \sqrt{\frac{424375-\frac{1}{40}\left( 4000 \right)^2}{40-1}}$$
So 
$$z = \frac{10.14}{2\frac{\sqrt{\frac{424375-\frac{1}{40}\left( 4000 \right)^2}{40-1}}}{\sqrt{40}}}$$
