Current multifactor productivity for $640$ work hours per month $=0\mathrm{.}244$ loaves$/$dollar

$640$x$8$hours $=51201600$x$0\mathrm{.}50=8006505120+800+650=65701600/6570=0\mathrm{.}243531202$

After increasing the number of work hours to $992$ per month, the multifactor productivity $=0\mathrm{.}252$ loaves$/$dollar$.992$x$8$hours$=7936--1600+(55\%$x$1600)=1600+880=24806502480/(7936+650+(2480$x$0\mathrm{.}50))=0\mathrm{.}252391614$

The percentage increase in productivity $=3\mathrm{.}28\%$

$(0\mathrm{.}252-0\mathrm{.}244)/(0\mathrm{.}244)$x$100=3\mathrm{.}28\%$

The specifications for a plastic liner for a concrete highway project calls for thickness of $3\mathrm{.}0$ mm plus or minus $\backslash$pm $0\mathrm{.}10$ mm$.$ The standard deviation of the process is estimated to be $0\mathrm{.}020$ mm$.$

The upper specification limit for this product $=3\mathrm{.}1$ mm $($round your response to one decimal place$).$

The lower specification limit for this product $=2\mathrm{.}9$mm $($round your response to one decimal place$).$

The process is known to operate at a mean thickness of $3\mathrm{.}0$ mm$.$ The process capability index

$($Upper C Subscript pkCpk$)=1\mathrm{.}667$

The upper specification limit lies about $5$ standard deviations from the centerline $($mean thickness$).$

How is the answer $5$ for standard deviation?