If the diameters of ball bearings are normally distributed, determine the percentage with diameters between $0.610$ and $0.618$ inches. If the diameters of ball bearings are normally distributed with mean $0.6140$ inches and standard deviation
$0.0025$ inches, determine the percentage of ball bearings with diameters 


*

*Between $0.610$ and $0.618$ inches inclusive, 

*Greater than $0.617$
    inches, 

*Less than $0.608$ inches, 

*Equal to $0.615$ inches.



My solution:
$$\mathbb P(-1.6\leq Z\leq1.6) = 89.04\%$$ 
The book's answer was $(1)~93%$ 
Can someone offer some help?
 A: *

*If $X$ is such that $E(X)=0.6140$ and $\sigma(X)=0.0025$, then the
right answer is $89.04\%$. One may write $$ \begin{align}
P(0.610<X\leq0.618)&=P\left(\frac{0.610-0.6140}{0.0025}<\frac{X-0.6140}{0.0025}\leq\frac{0.618-0.6140}{0.0025}\right)
\\&=P\left(-1.6<Z\leq 1.6\right)
\\&=\Pi\left(1.6\right)-\Pi\left(-1.6\right)
\\&=2\times\Pi\left(1.6\right)-1 \\&=0.8904\cdots \end{align} $$
giving $89.04\%$ as the right answer.

*If $X$ is such that $E(X)=0.6140$ and $\sigma^2(X)=0.0025$, that is
$\color{red}{\sigma(X)=0.05}$, then let's try to understand where $93\%$ could come from. One may write $$ \begin{align}
    P(0.610<X\leq0.618)&=P\left(\frac{0.610-0.6140}{0.05}<\frac{X-0.6140}{0.05}\leq\frac{0.618-0.6140}{0.05}\right)
    \\&=P\left(-0.08<Z\leq 0.08\right)
    \\&=\Pi\left(0.08\right)-\Pi\left(-0.08\right)
    \\&=2\times\Pi\left(0.08\right)-1 \\&=0.0638\cdots \end{align} $$
and noticing that $1-0.0638=0.9362\approx \color{red}{93.6\%}$. But I don't see why this should be the answer.
