Question

The diameter of shafts turned on a lathe are considered to be normally distributed, with a mean of 1.00 in. and a standard deviation of 0.01 in. There is a 99% chance that the diameter will be below what value?

Answer 1

Given that,

mean = $\mu$ = 1.00

standard deviation = $\sigma$ = 0.01

Using standard normal table

P(Z < z ) = 99%

P(Z < z ) = 0.99

P(Z < 2.326 ) = 0.99

z = 2.326

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 2.326 * 0.01 + 1.00

x = 1.02 in.