For a normally distributed variable B variance is (0.714)^2 and mean is x.
If P( x < b < 3.2) = 0.475 find x.
My working:
P ( x < b < 3.2) = P ( b < 3.2 )- P ( b < * x* )
P ( b < 3.2 ) - P ( b < * x* ) = 0.475
P ( z < [ 3.2 - x / 0.714 ] ) - P ( z < 0 ) = 0.475
P ( z < [ 3.2 - x / 0.714 ] ) - 0.5000 = 0.475
P ( z < [ 3.2 - x / 0.714 ] ) = 0.975
phi ( [ 3.2 - x / 0.714 ] ) = phi (1.960)
3.2 - x / 0.714 = 1.960
x = 1.8005
However this is incorrect, it should be 4.60. In the second to last line if 1.960 is negative this is what you get.
What did I do wrong so that 1.960 is positiv instead of negative?
Your result is right. You can check it by using your result. $B$ is disributed as
$B\sim\mathcal N(1.8005,0.714^2)$. Thus
$$P ( 1.8005 < B < 3.2)=\Phi\left( \frac{3.2-1.8005}{0.714} \right)-\Phi\left( \frac{1.8005-1.8005}{0.714} \right)$$
$$\Phi\left( 1.96 \right)-\Phi\left( 0 \right)=0.975-0.5=0.475$$
You get the result of $\mu=4.6 \ $ if $$\color{blue}{P( 3.2 < B < x^*) = 0.475}$$
$$\Phi\left( \frac{x^*-x^*}{0.714} \right)-\Phi\left( \frac{3.2-x^*}{0.714} \right)=0.475$$
$$0.5-\Phi\left( \frac{3.2-x^*}{0.714} \right)=0.475$$
$$-\Phi\left( \frac{3.2-x^*}{0.714} \right)=-0.025$$
$$ \frac{3.2-x^*}{0.714} =\Phi^{-1}(0.025)$$
$$ \frac{3.2-x^*}{0.714} =-1.96$$
$$\Rightarrow x^*=\mu=4.59974\approx 4.6$$