If we were given mean as 40 and standard deviation as 2 and asked the proportion of shirts manufactured in each category, the following solution will apply
Small
"0<X<37"
Since "\\mu = 40" and "\\sigma = 2"
"= P(0<X<37)\\\\\n= P(0-40<X-\\mu<37-40)\\\\\n= P(\\frac{0-40}{2}<\\frac{X-\\mu}{\\sigma}<\\frac{37-40}{2})\\\\\n= P(-20<Z<-1.5)\\\\\n=0.0668\\\\\n\\implies P_{small}= 0.0668*100=7 \\%"
Medium
"37<X<40.5"
Since "\\mu = 40" and "\\sigma = 2"
"= P(37<X<40.5)\\\\\n= P(37-40<X-\\mu<40.5-40)\\\\\n= P(\\frac{37-40}{2}<\\frac{X-\\mu}{\\sigma}<\\frac{40.5-40}{2})\\\\\n= P(-1.5<Z<0.25)\\\\\n=0.5319\\\\\n\\implies P_{medium}= 0.5319*100=53 \\%"
Large
"40.5<X<44"
Since "\\mu = 40" and "\\sigma = 2"
"= P(40.5<X<44)\\\\\n= P(40.5-40<X-\\mu<44-40)\\\\\n= P(\\frac{40.5-40}{2}<\\frac{X-\\mu}{\\sigma}<\\frac{44-40}{2})\\\\\n= P(0.25<Z<2)\\\\\n=0.3785\\\\\n\\implies P_{large}= 0.3785*100=38 \\%"
Extra-Large
"X>44"
Since "\\mu = 40" and "\\sigma = 2"
"= P(X>44)\\\\\n= P(X-\\mu>44-40)\\\\\n= P(\\frac{X-\\mu}{\\sigma}>\\frac{44-40}{2})\\\\\n= P(Z>2)\\\\\n=0.0228\\\\\n\\implies P_{extra-large}= 0.0228*100=2 \\%"
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