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notsponge
2 months ago
10

Automated manufacturing operations are quite precise but still vary, often with distributions that are close to Normal. The widt

h 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?
Mathematics
1 answer:
PIT_PIT [12.4K]2 months ago
8 0
<span>0.977
The mean for our population is 0.8750, and the standard deviation is 0.0011. Let's determine how many standard deviations we need 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

We require a range within 2.272727273 deviations from the mean. Referring to a standard normal table, the corresponding value is 0.48848, representing half the percentage. Therefore, 0.48848 * 2 = 0.97696, rounding it to three digits yields 0.977 </span>
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