$\backslash$table$[[,$Days,$40,,],[-3\mathrm{.}050505,$standard deviation,$46,=(40-46\mathrm{.}04)/1\mathrm{.}98],[-1\mathrm{.}98,$z$-$score,$-2,=(6\mathrm{.}04/$K$13),],[1.$ you get substract $40-46\mathrm{.}04=6\mathrm{.}04],[2.$ then you divide $6\mathrm{.}04/3\mathrm{.}05=-1\mathrm{.}98,,,],[$variance$,0\mathrm{.}0239,=$NOR,M$.$S$.$DIST$($M$,$TRUE$)]]$

where did the $-1\mathrm{.}98$ came from can someone help me please, im having trouble explaining this to the teacher. preferable answer before $12$ pm today.