Standard deviation in normal distribution

Question

A manufacturer uses a machine to make metal rods.The diameter of the rods follow a normal distribution with a mean of 1cm and a standard deviation of 0.02cm

If the standard deviation of the diameters of the rods produced in the process can be adjusted,what should the new standard deviation be so that 90% of the rods produced will have diameters between 0.98cm and 1.02cm

My working is

$ X~(1,σ^2)$

$p(0.98-1/σ < Z < 1.02-1/σ)=0.90$

$P(-0.02/σ < Z < 0.02/σ)= (0.90-0.5)$ $P(-0.02/σ < Z < 0.02/σ)= (0.4)$

FROM THE TABLE

$P(-0.02/σ < Z < 0.02/σ)= (1.28)$

I'm stuck after that...help pls

Answer 1

You made a minor simplification mistake. With $X\sim N(1,\sigma^2)$ $$ 0.9=Pr(0.98<X\leq 1.02)=Pr\left(\frac{0.98-1}{\sigma}<\frac{X-1}{\sigma}\leq\frac{1.02-1}{\sigma}\right)=Pr\left(\frac{-0.02}\sigma<Z\leq\frac{0.02}{\sigma}\right) $$ where $Z=(X-1)/\sigma$ has a standard normal distribution, the CDF of which we denote by $\Phi$. Then, $$ 0.9=\Phi(0.02/\sigma)-\Phi(-0.02/\sigma)=\Phi(0.02/\sigma)-(1-\Phi(0.02/\sigma))=2\Phi(0.02/\sigma)-1 $$ so that $$ \Phi(0.02/\sigma)=0.95. $$ Now look up $0.02/\sigma$ from a table.