If the diameters of ball bearings are normally distributed, determine the percentage with diameters between $0.610$ and $0.618$ inches.
- 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.
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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.