Automated manufacturing operations are quite precise but still vary, often with distributions that are close to normal. the width in inches of slots cut by a milling machine follows approximately the n(0.8750,0.0011) n(0.8750,0.0011) distribution. the specifications allow slot widths between 0.8725 and 0.8775 inch. what proportion (±0.001) of slots meet these specifications?

0.977

   So we have a population with the mean being 0.8750 and the standard deviations being 0.0011. So let's see how many standard deviations we need to be off by to exceed the specifications.

 Low end
 (0.8725 - 0.8750)/ 0.0011 = -0.0025/0.0011 = -2.272727273
High end
 (0.8775 - 0.8750)/ 0.0011 = 0.0025/0.0011 = 2.272727273

   So we need to be within 2.272727273 deviations of the mean. Let's use a standard normal table to look up that value, which is 0.48848, which is half the percentage. So 0.48848 * 2 = 0.97696, rounding to 3 digits gives 0.977

Could you make it more clear?