Question

An iron ball diameter measurement is installed automatically in a factory. The gauge will only pass the ball diameter of 1.50 ± d cm. It is known that the ball of the factory production has a diameter that is normally distributed with a mean diameter of 1.50 and a standard deviation of 0.2 cm. If it is desired that 95% of the production passes the selection, should the value be determined?

Answer 1

let x=diameter of a ball

$X \backsim N(1.5,0.2)$

this can be convert to standard normal by,

$z=\frac{x-\bar{x}}{\sigma}=\frac{x-1.5}{0.2}\\$

here task is to find a find a z value such that,

$P(-z<Z<z)\le0.95\\ 2*P(0<Z<z)\le0.95\\ P(0<Z<z)\le0.475$

using table,

$z=1.96$

$z=\frac{x-1.5}{0.2}\\ when\ z=1.96\\ 1.96=\frac{x-1.5}{0.2}\\ x_{max}=1.5+0.2*1.96=1.5+0.392\\ when\ z=-1.96\\ -1.96=\frac{x-1.5}{0.2}\\ x_{min}=1.5-0.2*1.96=1.5-0.392\\$

Therefore to pass 95 of ball,diameter should be $1.5\pm 0.392$ .

value=0.392