a)

Converting to $\operatorname{N}(0,1)$

$$\begin{eqnarray*}P(X \lt \var{lower})&=&P\left(Z \lt \frac{\var{lower}-\var{m}}{\var{s}}\right)\\&=&P(Z\lt \var{z})=1-P(Z \lt \var{-z})=1-\var{normalcdf(-z,0,1)}\\&\approx&\var{prop/100}\end{eqnarray*}$$

Hence the proportion is $\var{prop}$% to 1 decimal place.

b)

The proportion of men more than $\var{upper1}$ inches tall is given by finding the probability:

$$\begin{align} \operatorname{P}(X>\var{upper1} &= 1 - \operatorname{P}(X<\var{upper1})\\ &= 1 - \operatorname{P}(Z<\dfrac{\var{upper1}-\var{m}}{\var{s}})\\ &= 1 - \var{w2}\\ &\approx \var{prop1/100} \end{align}$$

Hence the proportion is $\var{prop1}$% to 1 decimal place.

1. Converting to $\operatorname{N}(0,1)$

$\simplify[all,!collectNumbers]{P(X < {lower}) = P(Z < ({lower} -{m}) / {s}) = P(Z < {lower-m}/{s}) = 1 -P(Z < {m-lower}/{s})} = 1 -\var{p} = \var{precround(1 -p,4)}$ to 4 decimal places.

2. Converting to $\operatorname{N}(0,1)$

$\simplify[all,!collectNumbers]{P(X > {upper}) = P(Z > ({upper} -{m}) / {s}) = P(Z > {upper-m}/{s}) = 1 -P(Z < {upper-m}/{s})} = 1-\var{p1} = \var{precround(1 -p1,4)}$ to 4 decimal places.

Converting to $\operatorname{N}(0,1)$:

a)

\[\begin{eqnarray*}P(\var{lower1} \lt X \lt \var{upper1})&=&P(X \lt \var{upper1})- P(X \lt \var{lower1})\\&=&P\left(Z \lt \frac{\var{upper1}-\var{m}}{\var{s}}\right)-P\left(Z \lt \frac{\var{lower1}-\var{m}}{\var{s}}\right)\\&=&P(Z \lt \var{zu})-P(Z \lt \var{zl})=\var{precround(u,4)}-\var{precround(l,4)},\\&=&\var{p}\end{eqnarray*}\]  (to 3 decimal places.)

b)

We need to find $x$ such that:

\[\begin{eqnarray*}P(X \lt x) &\le& \var{percentile/100}\\ \Rightarrow P\left(Z \le \frac{x-\var{m}}{\var{s}}\right) &\le& \var{percentile/100}\\ \Rightarrow \frac{x-\var{m}}{\var{s}} &\le& \Phi^{-1}(\var{percentile/100})=\var{precround(normalinv(percentile/100,0,1),4)}\\ \Rightarrow x &\le&\var{precround(normalinv(percentile/100,0,1),4)}\times\var{s}+\var{m}=\var{perX}\end{eqnarray*}\] to the nearest whole number.

Hence $X=\var{perX}$ gives the $\var{percentile}$% percentile.