Let $X$ represent the height.
$E(x) =$ $5$ feet $10$ inches
$SD(X) = 2$
$P(X > 6) = P(X^* > 1)$
$X^*$ is $X$ in standard units.
Why isn't it $P(X^* \gt \frac{6 - (5+\frac{10}{12})}{2}) = P(X^* \gt \frac{1}{12})$
References:
1 Answer
Because you have $SD(X)=2$ inches! (Read the text in the second image).
Your calculations would be OK if you had $SD(X)=2$ ft. But in this case you have $$\frac{6\text{ ft}-(5\text{ ft}+10\text{ in} \cdot \tfrac{1\text{ ft}}{12\text{ in}})}{2\text{ in}\cdot \tfrac{1\text{ ft}}{12\text{ in}}}=\frac{\tfrac2{12}\text{ ft}}{\tfrac2{12}\text{ ft}}=1,$$ (without units). Or in inches $$\frac{6 \text{ ft}\cdot \tfrac{12\text{ in}}{1\text{ ft}}-(5\text{ ft}\cdot \tfrac{12\text{ in}}{1\text{ ft}}+10\text{ in})}{2\text{ in}}=\frac{2\text{ in}}{2\text{ in}}=1,$$ (again a dimensionless quantity).
