a)
The mean of the sample of patients visiting on Mondays is
\[ \bar{x} = \frac{1}{n}(\sum x_i) = \var{precround(monday_mean,2)} \text{ (to 2 d.p.)} \]
The standard deviation is
\[ s = \sqrt{\frac{1}{n}\sum \left(x_i-\bar{x}\right)^2} = \var{precround(monday_sd,2)} \text{ (to 2 d.p.)} \]
b)
The sample mean $\mu_m$ follows a $t$-distribution with $n-1 = \var{num_monday-1}$ degrees of freedom centred on the population mean, with a standard error of $\frac{s}{\sqrt{n}} = \simplify{{precround(monday_sd,2)}/sqrt({num_monday})} = \var{precround(monday_se,2)}$.
That means that the 95% confidence interval for the population mean is as follows:
\begin{array}{lrcl}
&\left[ \bar{x} - s \times t_{\var{num_monday-1},0.975} \right. \kern{-0.5em}&,&\kern{-0.5em} \left. \bar{x} + s \times t_{\var{num_monday-1},0.975} \right] \\
= & \left[ \simplify[]{{precround(monday_mean,2)} - {precround(monday_se,2)}*{precround(monday_t,2)}} \right. \kern{-0.5em}&,&\kern{-0.5em} \left. \simplify[]{{precround(monday_mean,2)} + {precround(monday_se,2)}*{precround(monday_t,2)}} \right] \\
= & \left[ \var{precround(monday_low,2)} \right. \kern{-0.5em}&,&\kern{-0.5em} \left. \var{precround(monday_high,2)} \right]
\end{array}
c)
The 95% confidence intervals for the two population means do not overlap, so they must be significantly different.