# Normal distribution problem of adjusting S.D

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.

What I tried: −0.02/σ < Z ≤ 0.02/σ

My confusion is what should I be equating this to?

• Seems someone is trying to make this problem harder (or more confusing) than it needs to be by asking for the new 'margin of error' to be equal to the old $\sigma.$ – BruceET Sep 15 '18 at 22:44

If $X \sim \mathsf{Norm}(\mu = 1, \sigma),$ then $P\left(-1.645 < \frac{X-1}{\sigma} < 1.645\right) = 0.90.$

qnorm(c(.05,.95))
 -1.644854  1.644854


Currently, you have $P(|X-1| < 1.645\sigma = 1.645(.02) = 0.0329) = 0.90.$

When the process is improved, you want $P(|X-1| < 1.645\sigma = 0.02) = 0.90.$

• So we have two unknown variables, right? – user218970 Sep 16 '18 at 16:46
• Not my intent. In my last sentence, there is $1.645\sigma = 0.02;$ seems to me that solving for $\sigma$ does it, for the new process. Weren't you are given $\sigma = 0.02$ for the original process? – BruceET Sep 16 '18 at 20:10
• Thank you, so this means that σ = 0.012? Am I correct? – user218970 Sep 17 '18 at 9:42