I am attempting to comprehend finding the confidence interval given the probability. My classmate said the answer is $1.4024$ but I am not sure how he got to that answer. Am I right to assume that a $Z$-table is necessary?
$\mu = 24.1 \\ \sigma = 6.4 \\ \text{sample size} = 80 \\ \sigma_x= 0.7155$
The question: An interval $[μ_x − a, μ_x + a]$ symmetric about $μ_x$, satisfying $P(X ∈ [μX − a, μX + a]) = 0.95$
Find $a$
My classmate said $a$ is $1.4024$ but I am not sure how he got that result based on the given information. May anyone please provide assistance?
How to get approximate confidence intervals with the CLT
Let $X_{1},\ldots,X_{n}$ be IID with mean $\mu_{X}$ and variance $\sigma_{X}^{2} < \infty$. Let $\overline{X}_{n}\equiv(X_{1}+\cdots+X_{n})/n$ be the sample mean. The central limit theorem (CLT) says that the distribution of $$ \frac{\sqrt{n}}{\sigma_{X}}\left(\mu_{X}-\overline{X}_{n}\right) $$ converges to a standard normal as $n \rightarrow \infty$. Letting $Z$ be a standard normal random variable, the above suggests that it is reasonable to make the approximation \begin{align} \mathbb{P}\left(\left|Z\right|\leq z\right)&\approx\mathbb{P}\left(\frac{\sqrt{n}}{\sigma_{X}}\left|\mu_{X}-\overline{X}_{n}\right|\leq z\right)\\&=\mathbb{P}\left(\overline{X}_{n}-\frac{\sigma_{X}}{\sqrt{n}}z\leq\mu_{X}\leq\overline{X}_{n}+\frac{\sigma_{X}}{\sqrt{n}}z\right). \end{align} Given a level $1-\alpha$ (e.g., $\alpha=0.05$ gives $1-\alpha=0.95$), note that $$ 1-\alpha=\mathbb{P}\left(\left|Z\right|\leq z\right)\implies z=z_{1-\alpha/2} $$ where $z_{\beta}$ is the $\beta$ quantile of a normal distribution (you can find values of $z_\beta$ in a z-table, as you suggest). In conclusion, we have shown that an approximate confidence interval for $\mu_{X}$ is $$ \overline{X}_{n}\pm\frac{\sigma_{X}}{\sqrt{n}}z_{1-\alpha/2}. $$
Your problem
In your case, $\overline{X}_{n}=24.1$, $\sigma_X=0.7155$, $n=80$, $\alpha=0.05$, and $z_{0.975}=1.96$. Therefore, the confidence interval for $\mu_X$ is $$ 24.1\pm\underbrace{\frac{0.7155}{\sqrt{80}}1.96}_a. $$
I don't think your friend's answer is correct: he never used the sample size. If their answer were correct, adding more samples would not narrow the confidence interval: it would stay the same size.
Note: It's unclear what $\sigma$ and $\sigma_X$ are in your question. I assumed $\sigma$ was a sample variance and $\sigma_X$ was the true variance. If you know the true variance, you should use that over the sample variance, which would be, in this context, an estimate of the true variance.