How to reduce the standard deviation to ensure 99% of rods are within tolerance?

beanryu
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Homework Statement


Plastic rods are cut into nominal length of 6 inches. Actual lengths are normally distributed about a mean of 6 inches and their standard deviation is 0.06 inch.

Question: To what value does the standard deviation need to be reduced if 99% of the rods must be within tolerance?

Homework Equations


sd=standard deviation
u=mean
P(a<X<=b)=F((b-u)/(sd))-F((a-u)/(sd))

The Attempt at a Solution


since they want the possibility of rods to be between u+sd and u-sd to be 0.99, b=u+sd and a=u-sd
and the equation will become
P(a<X<=b)=F((u+sd-u)/(sd))-F((u-sd-u)/(sd))
F(1)-F(-1) doesn't equal to 0.99.

Am I misinterpreting the word tolerance?
I don't know what else to try... please help thank you!
 
Last edited:
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nevermind i misread the problem...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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